Electromagnetism

Class 11 Physics, unit 12, book pages 219–245

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This unit studies how electric currents and moving charges produce magnetic fields, and how magnetic fields act on currents and moving charges. It covers the force on a current-carrying conductor and on a moving charge, the circular motion of charged particles and the velocity selector, the field patterns of a straight wire, a coil and a solenoid, and the forces between parallel conductors. It then introduces magnetic flux and flux linkage, Faraday's and Lenz's laws of electromagnetic induction and their use in the seismometer.

Learning outcomes (15)

Define and explain magnetic fields.
State that a force might act on a current-carrying conductor placed in a magnetic field.
Use the equation $F = B I L sin θ$ (with directions as interpreted by Fleming's left-hand rule) to solve problems.
Define magnetic flux density (as the force acting per unit current per unit length on a wire placed at right angles to the magnetic field) and use $F = B Q v sin θ$ to solve problems.
Describe the motion of a charged particle moving in a uniform magnetic field perpendicular to the direction of motion of the particle.
Explain how electric and magnetic fields can be used in velocity selection.
Sketch magnetic field patterns due to the currents in a long straight wire, a flat circular coil and a long solenoid.
State that the magnetic field due to the current in a solenoid is increased by a ferrous core.
Explain the origin of the forces between current-carrying conductors and determine the direction of the forces.
Define magnetic flux (as the product of the magnetic flux density and the cross-sectional area perpendicular to the direction of the magnetic flux density).
Use $Φ = B A$ to solve problems.
Use the concept of magnetic flux linkage.
Explain experiments that demonstrate Faraday's and Lenz's laws: (a) that a changing magnetic flux can induce an e.m.f. in a circuit; (b) that the induced e.m.f. is in such a direction as to oppose the change producing it; (c) the factors affecting the magnitude of the induced e.m.f.
Use Faraday's and Lenz's laws of electromagnetic induction to solve problems.
Explain how seismometers make use of electromagnetic induction to detect earthquakes, specifically in terms of: (i) any movement or vibration of the rock on which the seismometer rests (buried in a protective case) results in relative motion between the magnet and the coil (suspended by a spring from the frame); (ii) the emf induced in the coil is directly proportional to the displacement associated.

12.1Magnetic fields

Magnetism is the study of how a magnetic field, generated by a moving charged particle, affects other charged particles or permanent magnets. Electromagnetism is the study of the magnetic effects of current. This unit deals with the laws and phenomena related to electromagnetism.

Magnetic field: The region or space surrounding a magnet in which a compass needle, a small magnet, a piece of ferromagnetic material or a moving charged particle can experience a magnetic force.

A permanent bar magnet, a current-carrying conductor, a fluctuating electric field and a moving charge can all create a magnetic field. The magnetic force between magnets acts over a distance: the magnets do not touch, they act on each other through the interaction of their magnetic fields (a non-contact force).

Michael Faraday gave the alternative explanation. Each magnet sets up a magnetic field in the space around it, as in Fig. 12.1. When you place another magnet in that field, it responds to the field at its own location. This interaction of fields is the cause of the force between the magnets, and the field lines are the picture we draw of it.

The idea is the same as for the other fields met in earlier classes. A gravitational field exerts the force of gravity on other masses, and an electric field around charges exerts an electric force on other charges. In the same way a magnetic field exerts a force on other magnets such as compass needles, on magnetic materials and on moving electric charges.

Figure 12.1: Magnetic field of a bar magnet. Outside the magnet the lines run from the north pole to the south pole, and inside the magnet from south to north, so every line is a closed loop.

You can visualise the magnetic field of a magnet by sprinkling iron filings near it (Fig. 12.1 (b)). The iron particles become magnetised in the field and stick together along the field lines, so the filings trace out the pattern of the lines.

One key difference between magnetic and electric field lines is that magnetic field lines form complete loops. Outside a magnet they exit from the north pole and enter the south pole; inside the magnet they go from the south pole to the north pole. This is why there are no magnetic monopoles: the north and south poles do not exist independently. The magnetic field is strongest at the poles of a bar magnet.

Properties of magnetic field lines

Magnetic field lines never cross each other.
The density of the magnetic field lines indicates the strength of the magnetic field.
Magnetic field lines always form closed loops.
Magnetic field lines always start from the north pole and enter at the south pole (outside the magnet).
Like magnetic poles repel, and unlike magnetic poles attract each other.

12.1.1Magnetic field due to current in a long straight wire

In 1820 Hans Oersted first described the magnetic field due to current in a wire.

Experiment

Take a piece of copper wire that passes vertically through a horizontal piece of cardboard, as in Fig. 12.2. Place small magnetic compass needles on the cardboard along a circle with the wire at its centre.

Figure 12.2: Compass needles placed round a vertical current-carrying wire point along the tangent to the circle, in the direction of the magnetic field.

What the experiment shows

With no current, all the compass needles point in the north-south direction.
When a heavy current passes through the wire, the needles set themselves along the tangent to the circle. A circular magnetic field in the anticlockwise direction is produced round the current-carrying conductor (Fig. 12.3 (a)).
When the direction of the current is reversed, the direction of $B$ is also reversed and becomes clockwise (Fig. 12.3 (b)).
If the current through the wire is stopped, all the needles again point in the north-south direction.

Figure 12.3: The field round a current-carrying conductor is circular. (a) Current out of the page gives an anticlockwise field. (b) Reversing the current gives a clockwise field.

We conclude that:

A magnetic field is set up only around a current-carrying conductor.
The lines of force are circular, and their direction depends upon the direction of the current.

Direction of the magnetic field

The direction of the magnetic field is found with the right-hand rule (Fig. 12.3 (c)).

Right-hand rule for a straight wire: If the wire is grasped in the fist of the right hand with the thumb pointing in the direction of the (conventional) current, then the curled fingers indicate the direction of the magnetic field.

12.2Magnetic force on a current-carrying conductor in a uniform magnetic fieldOften examined

Consider a straight current-carrying conductor with concentric circular field lines in the clockwise direction, due to a current flowing into the page (Fig. 12.4 (a)). Fig. 12.4 (b) shows the magnetic field of a permanent magnet, from its north pole to its south pole.

When this conductor is placed in the external field of the permanent magnet, the two fields interact (Fig. 12.4 (c)). On one side of the conductor the two fields point the same way and support each other, so the net magnetic field is stronger there. On the other side the two fields are opposite in direction, so the net field is weaker.

Figure 12.4: A current into the page in the field of a magnet (north pole below, south pole above). On the left the two fields add and the net field is stronger, on the right they oppose and it is weaker, so the force

F

acts to the right.

A magnetic force acts on the current-carrying conductor from the stronger towards the weaker region of the magnetic field. In this case the conductor tends to move towards the right, where the net field is weak. The force is perpendicular to both the length of the conductor and the external magnetic field.

Derivation Force on a current-carrying conductor in a magnetic field

Aim. Find the force $F$ on a conductor of length $L$ , carrying a current $I$ , placed in a uniform magnetic field $B$ at an angle $θ$ to the field.

Assumptions.

The conductor of length $L$ carries a current $I$ and lies on two conducting rails in a uniform magnetic field $B$ (Fig. 12.5).
When the magnetic force acts on the conductor, it begins to move on the rails in the direction of the force.
$θ$ is the angle between $B$ and the length $L$ of the conductor.

Figure 12.5: A conductor of length

L

carrying a current

I

, resting on two conducting rails in a uniform magnetic field

B

(shown as crosses, into the page). The force

F

pushes the rod along the rails.

Experiments with different currents show that the force is directly proportional to the current flowing in the conductor.
$F \propto I$
A longer rod in the field (a larger length $L$ inside the field) feels a larger force: the force is directly proportional to the length of the conductor.
$F \propto L$
A stronger external field gives a larger force: the force is directly proportional to the strength of the field.
$F \propto B$
The force depends on the angle between the conductor and the field: it is directly proportional to the sine of that angle.
$F \propto sin θ$
Combine the four relations.
$F \propto I L B sin θ or F = k I L B sin θ$
Here $k$ is the constant of proportionality. In the SI system its value is $1$ (the unit of $B$ is chosen so that this is so).
$F = B I L sin θ (12.1 a)$
In vector form, with $L$ a vector of length $L$ along the conductor in the direction of the current, the force is perpendicular to both $L$ and $B$ .
$F = I L B sin θ \hat{n} = I (L \times B) (12.1 b)$

Result

F = B I L sin θ, F = I (L \times B)

The magnetic force is perpendicular to both the length of the conductor and the magnetic field. It is large for a strong field, a large current and a long conductor, and it depends on how the conductor is turned relative to the field.

Maximum and minimum force

Maximum force. The force is maximum when $θ = 9 0^{\circ}$ , that is, when the conductor is placed perpendicular to the magnetic field.

F_{m a x} = I L B sin 9 0^{\circ} = I L B (1) = I L B

Minimum force. The force is minimum (zero) when the conductor is placed parallel ( $θ = 0^{\circ}$ ) or antiparallel ( $θ = 18 0^{\circ}$ ) to the magnetic field.

θ = 0^{\circ} : F_{m i n} θ = 18 0^{\circ} : F_{m i n} = I L B sin 0^{\circ} = I L B (0) = 0 = I L B sin 18 0^{\circ} = 0

Thus a current-carrying conductor does not experience a force in a magnetic field if the angle between $B$ and $L$ is $0^{\circ}$ or $18 0^{\circ}$ .

Force on a conductor against the angle

120

150

180

0.25

0.5

0.75

1.25

θ (^{\circ})

F / (B I L)

F_{m a x} = B I L

0

0

F = B I L sin θ

Variation of the magnetic force on a conductor with the angle

θ

between the conductor and the field. The force is zero at

0^{\circ}

and

18 0^{\circ}

and has its maximum value

B I L

9 0^{\circ}

12.2Fleming's left-hand ruleOften examined

The direction of the force on a current-carrying conductor is found with Fleming's left-hand rule.

Fleming's left-hand rule: If we hold the thumb, forefinger and middle finger of the left hand mutually perpendicular, the forefinger points in the direction of the magnetic field, the middle finger points in the direction of the electric current, and then the thumb indicates the direction of the magnetic force experienced by the conductor.

Figure 12.6: Fleming's left-hand rule. Thumb: thrust or motion (force). Forefinger: magnetic field. Middle finger: current.

Example 12.1: Lifting a wire with a magnetic field

A wire having a mass per unit length of

0.400 g c m^{- 1}

carries a

5.0 A

current horizontally to the south. What is (a) the direction of force and (b) the magnitude of the minimum magnetic field perpendicular to the length of the conductor needed to lift this wire vertically upward?

Given

$\frac{m}{L} = 0.400 g c m^{- 1} = \frac{0.400 \times 1 0 ^{- 3} kg}{1 \times 1 0 ^{- 2} m} = 0.04 kg m^{- 1}$ mass per unit length
$I = 5.0 A$ current, towards the south
$g = 9.8 m s^{- 2}$

To find

direction of the force
minimum $B$ perpendicular to the wire

(a) To lift the wire the magnetic force must point upward. Apply Fleming's left-hand rule: the middle finger points south (current) and the thumb points up (force), so the forefinger, which gives the field, must point east. The field must therefore be directed towards the east (Fig. below).
(b) To just lift the wire, the magnetic force must be equal to the weight of the wire.
$F_{B} = F_{g} \Rightarrow I L B sin θ = m g$
Solve for $B$ . The field is perpendicular to the wire, so $θ = 9 0^{\circ}$ and $sin θ = 1$ (this gives the minimum field).
$B = \frac{( m / L ) g}{I sin θ}$
Substitute the values.
$B = \frac{( 0.04 ) ( 9.8 )}{( 5.0 ) sin 9 0 ^{\circ}} = 0.0784 T$

Directions for Example 12.1: the current

I

flows south, the field

B

points east, and the force

F

on the wire points upward.

(a) field towards the east; (b) B = 0.0784 T

Assignment 12.1: Aluminium rod between the poles of a magnet

A light aluminium rod rests across two brass strips between the poles of a horseshoe magnet (north pole above the rod, south pole below). A current is passed through the rod from the two brass strips, which are connected through a switch (open) to a power supply. (a) Draw the direction of the current when the switch is closed. (b) State which way the rod moves when the switch is closed. Give a reason for your answer. (c) State the effect on the movement of the rod when (i) the current is increased, (ii) the current is reversed.

(a) The current leaves the positive terminal of the cell (the longer line of the cell symbol) and returns to the negative terminal. In the figure the longer line is on the left, so the current flows up the left brass strip, along the rod from left to right, and back down the right brass strip to the switch and the cell.
(b) The field of the horseshoe magnet runs from the north pole (above the rod) to the south pole (below it), that is, downward. The rod carries current from left to right across this field, so a force acts on it. By Fleming's left-hand rule (forefinger down, middle finger to the right) the thumb points along the brass strips, away from you, towards the curved end of the magnet, perpendicular to both the rod and the field. The rod therefore rolls along the strips in that direction.
Reason: a current-carrying conductor in a magnetic field feels a force $F = B I L sin θ$ , directed at right angles to both the field and the current.
(c) (i) A larger current gives a larger force ( $F \propto I$ ), so the rod accelerates faster and moves more quickly along the strips. (ii) Reversing the current reverses the direction of the force, so the rod rolls the opposite way.

12.3Magnetic induction or magnetic flux densityOften examined

A current-carrying conductor positioned at right angles to a magnetic field experiences a magnetic force. This force can be used to quantify the magnetic flux density. From $F = B I L sin θ$ with $θ = 9 0^{\circ}$ :

B = \frac{F}{I L}

12.3a

B is the flux density of a uniform field, $F$ the force on the conductor, $I$ the current and $L$ the length of the conductor in the field.

Magnetic flux density (magnetic induction): The force acting per unit current per unit length on a conductor placed at right angles to the magnetic field. It is a vector quantity.
Tesla (T): The SI unit of magnetic flux density. If a $1 m$ long conductor carrying a current of $1 A$ , placed perpendicular to a magnetic field, experiences a force of $1 N$ , then the magnetic induction is one tesla.

1 T = \frac{1 N}{( 1 A ) ( 1 m ) sin 9 0 ^{\circ}} = 1 N A^{- 1} m^{- 1}

Another unit of $B$ used in the CGS system is the gauss (G). The relation between tesla and gauss is

1 T = 10000 G = 1 0^{4} G or 1 G = 1 0^{- 4} T

12.4Magnetic force acting on a moving charge in a uniform magnetic fieldOften examined

It is verified by experiment that moving charges produce magnetic fields around them. When a moving charged particle enters an external magnetic field, such as that of a permanent magnet, its own magnetic field interacts with the field of the magnet. Because of this interaction the net magnetic field becomes stronger on one side of the charge than on the other (just as for a current-carrying conductor in a field). The magnetic force therefore acts on the moving charge from the stronger towards the weaker field region, and the particle is deflected from its path (Fig. 12.7).

The charged particle is deflected perpendicular to both the magnetic field and its velocity, because the net force acting on it is perpendicular to $B$ and $v$ . When a charge enters a magnetic field it experiences a force only if the following conditions are fulfilled:

The charge must be moving, because no magnetic force acts on a stationary charge.
The velocity of the moving charge must have a component that is perpendicular to the direction of the magnetic field.

Figure 12.7: Trajectories of charged particles in a magnetic field directed into the page. The positive charge is deflected one way and the negative charge the other way.

Force on a moving charge

Consider a positive point charge $+ q$ moving with velocity $v$ at an angle $θ$ to the magnetic field $B$ (Fig. 12.8). The force experienced by the charge is expressed as a vector (cross) product:

F_{B} = q (v \times B)

12.2

For a negative charge, with $q$ the size of the charge, the magnetic force is

F_{B} = - q (v \times B)

12.3

The negative sign shows that the direction of the magnetic force on a negative charge is opposite to that on a positive charge. The magnitude of the force on a charge $q$ moving with velocity $v$ in a field $B$ is

F_{B} = q v B sin θ

$θ$ is the angle between $v$ and $B$ .

Figure 12.8: A charge moving with velocity

v

at an angle

θ

to the magnetic field

B

. Only the component

v sin θ

, perpendicular to

B

, produces the force.

Maximum force

The force is maximum when the charged particle moves perpendicular to $B$ ( $θ = 9 0^{\circ}$ ):

F_{B} = q v B sin 9 0^{\circ} \Rightarrow F_{B (m a x)} = q v B

In this case the magnetic force is maximum and the charged particle moves in a circular path.

Minimum force

When $v$ and $B$ are parallel ( $θ = 0^{\circ}$ ): $F_{B} = q v B sin 0^{\circ} = q v B (0) = 0$ .
When $v$ and $B$ are antiparallel ( $θ = 18 0^{\circ}$ ): $F_{B} = q v B sin 18 0^{\circ} = q v B (0) = 0$ .
When the charged particle is at rest ( $v = 0$ ): $F_{B} = q (0) B sin θ = 0$ .
If $q = 0$ (a neutral particle): $F_{B} = (0) v B sin θ = 0$ .

This means that when the charged particle moves at an angle of $0^{\circ}$ or $18 0^{\circ}$ to the magnetic field, $F = 0$ and the particle moves in a straight path.

Figure 12.9: When the angle between

v

and

B

is other than

0^{\circ}

9 0^{\circ}

18 0^{\circ}

, the charged particle moves in a helical (spiral) path along the field.

If the angle between $B$ and $v$ is other than $0^{\circ}$ , $9 0^{\circ}$ or $18 0^{\circ}$ , the charged particle moves in a helical (spiral) path, as in Fig. 12.9. Its motion has both a component parallel to $B$ (which is not changed, so the particle drifts along the field) and a component perpendicular to $B$ (which makes it turn in a circle).

Direction of the force

The direction of the force on the moving charge is perpendicular to both the magnetic field $B$ and the velocity $v$ . For a positive charge, right-hand rule II and Fleming's left-hand rule for the force on a current-carrying conductor can be used, with the direction of the current replaced by the velocity of the charged particle.

For instance, if a positive particle moves from the bottom of the setup shown in Fig. 12.10, an outward force acts on it. The direction of the magnetic force on a negatively charged particle is the opposite, that is inward.

Figure 12.10: The magnetic force on a moving charge has a direction perpendicular to both the velocity

v

and the magnetic field

B

Example 12.2: Electron in a magnetic field

An electron is accelerated through

3600 V

from rest and then enters a uniform

2.70 T

magnetic field. What are (a) the maximum and (b) the minimum values of the magnetic force this particle experiences?

Given

$Δ V = 3600 V$ accelerating potential difference
$B = 2.70 T$
$e = 1.6 \times 1 0^{- 19} C$ charge of the electron
$m = 9.1 \times 1 0^{- 31} kg$ mass of the electron

To find

$F_{B (m a x)}$
$F_{B (m i n)}$

The electron gains kinetic energy equal to the work done on it by the potential difference, so $\frac{1}{2} m v^{2} = e Δ V$ . Solve for the speed.
$v = \frac{2 e Δ V}{m}$
Put in the values.
$v = \frac{2 ( 1.6 \times 1 0 ^{- 19} ) ( 3600 )}{9.1 \times 1 0 ^{- 31}} = 3.556 \times 1 0^{7} m s^{- 1}$
(a) The force is maximum when the electron moves perpendicular to the field ( $θ = 9 0^{\circ}$ ).
$F_{B (m a x)} = e v B = (1.60 \times 1 0^{- 19}) (3.556 \times 1 0^{7}) (2.70) = 1.54 \times 1 0^{- 11} N$
(b) The force is minimum (zero) when the electron moves along the field, parallel ( $0^{\circ}$ ) or antiparallel ( $18 0^{\circ}$ ) to it.
$F_{B (m i n)} = e v B sin 0^{\circ} = 0 or F_{B (m i n)} = e v B sin 18 0^{\circ} = 0$

F_{B (m a x)} = 1.54 \times 1 0^{- 11} N, F_{B (m i n)} = 0

Assignment 12.2: Magnetic force does no work

Prove that the magnetic force is not responsible for doing work on a charged particle moving in a circular path in a magnetic field.

In a short time $Δ t$ the particle is displaced by $Δ s = v Δ t$ , along its velocity.
The work done by the magnetic force in this displacement is the scalar product of force and displacement.
$Δ W = F_{B} \cdot Δ s = q (v \times B) \cdot v Δ t$
The vector $v \times B$ is perpendicular to $v$ , so the angle between $F_{B}$ and $Δ s$ is $9 0^{\circ}$ and $cos 9 0^{\circ} = 0$ .
$Δ W = F_{B} Δ s cos 9 0^{\circ} = 0$
The work done in every small step is zero, so the total work done by the magnetic force is zero. The kinetic energy and the speed of the particle do not change; the force only changes the direction of the velocity.

Worked example: force on a proton and on an electron in the same field

A proton and an electron each move to the right at

3.0 \times 1 0^{6} m s^{- 1}

in a uniform magnetic field of

0.50 T

that points out of the page. (a) Find the size and direction of the magnetic force on each when they move perpendicular to the field. (b) Find the force on the proton if it moves at

3 0^{\circ}

to the field.

A charge moving to the right in a field coming out of the page (dots). The force on the proton points down and the force on the electron, moving the same way, points up; the sizes are equal.

Given

$v = 3.0 \times 1 0^{6} m s^{- 1}$
$B = 0.50 T$
$q = 1.6 \times 1 0^{- 19} C$ size of the charge of each particle

To find

(a) $F$ and its direction for the proton and the electron
(b) $F$ for the proton at $3 0^{\circ}$

(a) The motion is perpendicular to the field, $θ = 9 0^{\circ}$ , so $sin θ = 1$ and the force is the maximum value.
$F = q v B sin 9 0^{\circ} = (1.6 \times 1 0^{- 19}) (3.0 \times 1 0^{6}) (0.50) = 2.4 \times 1 0^{- 13} N$
Direction for the proton: $F = q (v \times B)$ . Sweep the right hand from $v$ (right) towards $B$ (out of the page): the thumb points down the page.
The electron has the same size of charge but is negative, so its force has the same size and the opposite direction: up the page.
(b) At $3 0^{\circ}$ only the part of the velocity at right angles to the field counts, through $sin 3 0^{\circ} = 0.5$ .
$F = q v B sin 3 0^{\circ} = (2.4 \times 1 0^{- 13}) (0.5) = 1.2 \times 1 0^{- 13} N$

F = 2.4 \times 1 0^{- 13} N (proton down, electron up), F_{3 0^{\circ}} = 1.2 \times 1 0^{- 13} N

12.5Motion of a charged particle in a uniform magnetic fieldOften examined

Consider a charge q that enters a uniform magnetic field B with velocity v, perpendicular to the field (Fig. 12.11). The magnetic field produced by the charged particle interacts with the field of the permanent magnet. A magnetic force therefore acts on the charged particle and it begins to move in a circular path of radius r. We now find the radius, the angular speed, the period and the frequency of this circular motion.

Derivation Radius, period and frequency of the circular path

Aim. Find the radius r, the angular speed \omega, the period T and the frequency f of a particle of mass m and charge q moving with speed v in a plane perpendicular to a uniform field B.

Assumptions.

The velocity \mathbf{v} is perpendicular to \mathbf{B}, so the magnetic force has its maximum magnitude.
The magnetic force is the only force that acts, so it supplies the centripetal force for the circular motion.

Figure 12.11: A magnetic force acts on a charged particle and moves it into a circular path. The field B is into the page; the force F is always directed towards the centre, at right angles to the velocity v.

The magnitude of the maximum magnetic force is
F_{B(\max)}=qvB\qquad(12.4)
A particle of mass m moving with speed v in a circle of radius r needs a centripetal force
F_c=\dfrac{mv^{2}}{r}\qquad(12.5)
The magnetic force provides the required centripetal force, so the two are equal.
F_c=F_{B(\max)}\qquad(12.6)
Put the values from Eq. (12.4) and Eq. (12.5) in Eq. (12.6).
\dfrac{mv^{2}}{r}=qvB
Cancel one factor of v.
\dfrac{mv}{r}=qB\qquad(12.7)
Solve for the radius of the circular path.
r=\dfrac{mv}{qB}\qquad(12.8)
Put v=r\omega in Eq. (12.7).
\dfrac{m\,r\omega}{r}=qB\quad\text{or}\quad m\omega=qB
Solve for the angular speed. This is the cyclotron frequency of a charged particle moving in a circle in a magnetic field, in terms of angular motion.
\omega=\dfrac{qB}{m}\qquad(12.9)
The time period is T=\dfrac{2\pi}{\omega} (Eq. 12.10). Put the value of \omega from Eq. (12.9).
T=\dfrac{2\pi}{qB/m}=\dfrac{2\pi m}{qB}\qquad(12.11)
The frequency is the reciprocal of the period.
f=\dfrac{1}{T}=\dfrac{qB}{2\pi m}\qquad(12.12)

Resultr=\dfrac{mv}{qB},\qquad \omega=\dfrac{qB}{m},\qquad T=\dfrac{2\pi m}{qB},\qquad f=\dfrac{qB}{2\pi m}

For a given particle and field the period and frequency do not involve the speed v or the radius r (they depend only on q/m and B). A faster particle goes round a larger circle.

Example 12.3: Proton in a circular orbit

A proton is moving in a circular orbit of radius 12\,\mathrm{cm} in a uniform 0.40\,\mathrm{T} magnetic field perpendicular to the velocity of the proton. Find the speed of the proton.

Given

r = 12\,\mathrm{cm}=0.12\,\mathrm{m}
B = 0.40\,\mathrm{T}
e = 1.60\times10^{-19}\,\mathrm{C} charge of the proton
m_p = 1.67\times10^{-27}\,\mathrm{kg} mass of the proton

To find

speed v

The magnetic force supplies the centripetal force, so r=\dfrac{m_pv}{eB}. Solve for v.
v=\dfrac{eBr}{m_p}
Substitute the numerical values.
v=\dfrac{(1.6\times10^{-19})(0.40)(0.12)}{1.67\times10^{-27}}\approx4.6\times10^{6}\,\mathrm{m\,s^{-1}}

v\approx4.6\times10^{6}\,\mathrm{m\,s^{-1}}

Assignment 12.3: Electron in a weak field

In a uniform magnetic field with a strength of 1.20\times10^{-3}\,\mathrm{T}, calculate the radius of the orbit of the circular path of an electron moving at a speed of 2.0\times10^{7}\,\mathrm{m\,s^{-1}}.

Given

B = 1.20\times10^{-3}\,\mathrm{T}
v = 2.0\times10^{7}\,\mathrm{m\,s^{-1}}
e = 1.6\times10^{-19}\,\mathrm{C}
m = 9.1\times10^{-31}\,\mathrm{kg} mass of the electron

To find

radius r

Use Eq. (12.8).
r=\dfrac{mv}{eB}
Substitute the values.
r=\dfrac{(9.1\times10^{-31})(2.0\times10^{7})}{(1.6\times10^{-19})(1.20\times10^{-3})}=\dfrac{1.82\times10^{-23}}{1.92\times10^{-22}}\approx9.5\times10^{-2}\,\mathrm{m}

r\approx9.5\times10^{-2}\,\mathrm{m}\approx9.5\,\mathrm{cm}

Worked example: an electron moving in a circle in a weak field

An electron moves at 4.0\times10^{6}\,\mathrm{m\,s^{-1}} at right angles to a uniform field of 2.0\,\mathrm{mT} directed into the page. Find the radius of its circular path, the time for one revolution and the frequency. What happens to the radius and the time period if the speed is doubled?

An electron entering a field directed into the page (crosses) moves on a circle. At the lowest point it moves to the left and the magnetic force F points up, towards the centre, so the electron circulates clockwise.

Given

v = 4.0\times10^{6}\,\mathrm{m\,s^{-1}}
B = 2.0\,\mathrm{mT}=2.0\times10^{-3}\,\mathrm{T}
e = 1.6\times10^{-19}\,\mathrm{C}
m = 9.1\times10^{-31}\,\mathrm{kg}

To find

r
T and f
effect of doubling v

The magnetic force supplies the centripetal force, which gives the radius.
r=\dfrac{mv}{eB}=\dfrac{(9.1\times10^{-31})(4.0\times10^{6})}{(1.6\times10^{-19})(2.0\times10^{-3})}=\dfrac{3.64\times10^{-24}}{3.2\times10^{-22}}=1.14\times10^{-2}\,\mathrm{m}
Time period of one revolution.
T=\dfrac{2\pi m}{eB}=\dfrac{2\pi(9.1\times10^{-31})}{3.2\times10^{-22}}=1.79\times10^{-8}\,\mathrm{s}
Frequency is the reciprocal of the time period.
f=\dfrac{1}{T}=\dfrac{eB}{2\pi m}=5.6\times10^{7}\,\mathrm{Hz}
Doubling v: the radius is proportional to v, so it doubles to 2.28\,\mathrm{cm}. The expression for T contains only m, e and B, so T does not change.
r\propto v,\qquad T=\dfrac{2\pi m}{eB}\ \text{(no }v\text{)}
Check: the faster electron travels a path twice as long in the same time, because the circumference 2\pi r has also doubled.

\begin{aligned}r&=1.14\,\mathrm{cm}\\ T&=1.79\times10^{-8}\,\mathrm{s}\\ f&=5.6\times10^{7}\,\mathrm{Hz}\\ v\to2v:&\ r\to2r,\ T\ \text{unchanged}\end{aligned}

12.6Velocity selector methodOften examined

A velocity selector is a device used as a velocity filter for charged particles. It has mutually perpendicular electric and magnetic fields. A charged particle enters with velocity \mathbf{v} perpendicular to both the electric and the magnetic field, so both an electric force and a magnetic force act on it. At one particular velocity the electric and the magnetic force on the particle are equal in magnitude but opposite in direction, so their net effect cancels out and the charged particle moves in a straight path, undeflected (Fig. 12.12).

Derivation Speed of the undeflected particles

Aim. Find the speed v of the particles that pass straight through crossed electric and magnetic fields.

Assumptions.

The electric field E and the magnetic field B are perpendicular to each other and to the velocity of the particle.
The particle moves in a straight line, so the net force on it is zero.

Figure 12.12: Velocity selector. A particle with v=E/B feels no net force and goes straight through the slit. A slower particle is deflected in the direction of the electric force, and a faster one the other way.

For no deflection the electric force and the magnetic force must be equal (and opposite).
F_E=F_B
Put F_E=qE and F_B=qvB.
qE=qvB
Cancel q and solve for v.
v=\dfrac{E}{B}\qquad(12.13)

Resultv=\dfrac{E}{B}

If a beam of charged particles moving at different velocities passes through the velocity selector, only the particles whose velocity equals the ratio of the electric field to the magnetic field pass straight through the device. The result does not depend on the charge or the mass of the particles.

Particles moving slower than this speed are deflected in the direction of the electric force, because the magnetic force on them is less than the electric force. On the contrary, the particles that have greater speeds are deflected in the direction of the magnetic force, because the magnetic force on them is greater than the electric force.

Worked example: choosing the speed of a velocity selector

The plates of a velocity selector are 1.0\,\mathrm{cm} apart and a potential difference of 400\,\mathrm{V} is applied across them. A uniform magnetic field of 0.20\,\mathrm{T} is applied at right angles to the electric field. (a) What speed passes through undeflected? (b) Does the answer change for an electron or for a heavier ion? (c) What happens to a particle of the same charge moving at 1.0\times10^{5}\,\mathrm{m\,s^{-1}}?

A positive particle entering a velocity selector. The electric force qE (down) and the magnetic force qvB (up) are opposite, and they cancel only when v=E/B, so the particle follows the straight path (accent).

Given

V = 400\,\mathrm{V}
d = 1.0\,\mathrm{cm}=1.0\times10^{-2}\,\mathrm{m}
B = 0.20\,\mathrm{T}

To find

(a) undeflected speed v
(b) dependence on charge and mass
(c) behaviour at 1.0\times10^{5}\,\mathrm{m\,s^{-1}}

First find the electric field between the plates, from E=V/d.
E=\dfrac{400}{1.0\times10^{-2}}=4.0\times10^{4}\,\mathrm{V\,m^{-1}}
(a) A particle goes straight when the electric and magnetic forces cancel: qE=qvB.
v=\dfrac{E}{B}=\dfrac{4.0\times10^{4}}{0.20}=2.0\times10^{5}\,\mathrm{m\,s^{-1}}
(b) The charge q cancels from qE=qvB and the mass never appears. So an electron, a proton or a heavy ion with the same speed all pass through; the selector sorts by speed only. (A reversed charge reverses both forces, so they still cancel.)
(c) At 1.0\times10^{5}\,\mathrm{m\,s^{-1}} the particle is slower than E/B, so the magnetic force qvB is smaller than the electric force qE. The net force is in the direction of the electric force, and the particle is deflected that way (down in the figure).

v=2.0\times10^{5}\,\mathrm{m\,s^{-1}}\ (\text{independent of }q\text{ and }m)

Lorentz force

Lorentz force: The vector sum of the electric and the magnetic forces acting on a charged particle in electric and magnetic fields applied in the same region.

\mathbf{F}=\mathbf{F}_E+\mathbf{F}_B=q\,\mathbf{E}+q\,(\mathbf{v}\times\mathbf{B})

12.7.1Magnetic field due to current carrying straight conductorOften examined

The electric current flowing through a wire produces a magnetic field around it in the form of concentric circles (Fig. 12.13). The strength of the magnetic field depends upon the magnitude of the current and the distance of the point from the current-carrying conductor. The strength of the magnetic field produced by the current in a long, straight wire is given by

B=\dfrac{\mu_0 I}{2\pi r}

I is the current, r is the distance of the point from the wire, and the constant is \mu_0=4\pi\times10^{-7}\,\mathrm{Wb\,A^{-1}\,m^{-1}}.

Figure 12.13: Magnetic field lines for a current-carrying straight wire. They are concentric circles round the wire; the direction follows the right-hand rule.

Example 12.4: Field of a power line

A horizontal power line carries a current of 100\,\mathrm{A} in an east to west direction. What is the magnitude and direction of the magnetic field due to the current 2.0\,\mathrm{m} below the line?

Given

I = 100\,\mathrm{A}
r = 2.0\,\mathrm{m}
\mu_0 = 4\pi\times10^{-7}\,\mathrm{Wb\,A^{-1}\,m^{-1}}

To find

magnetic field B (magnitude and direction)

Use the formula for the field of a long straight wire.
B=\dfrac{\mu_0I}{2\pi r}
Substitute the values.
B=\dfrac{4\pi\times10^{-7}\times100}{2\pi\times2.0}=1.0\times10^{-5}\,\mathrm{T}
Direction: grasp the wire with the right hand, thumb along the current (towards the west). At a point below the wire the curled fingers point towards the south. (Looking along the current, towards the west, the field circulates clockwise, and below the wire a clockwise circulation goes to your left, which is south.)

B=1.0\times10^{-5}\,\mathrm{T},\ \text{directed towards the south}

12.7.2Magnetic field due to current carrying flat circular coil

The pattern of magnetic field lines for a current-carrying coil is shown in Fig. 12.14. Consider a circular coil of radius r carrying a current. When the current passes through the coil, a magnetic field is produced. The field at the centre of a single current-carrying circular loop is given by

B=\dfrac{\mu_0 I}{2r}

I is the current and r the radius of the loop.

Figure 12.14: Magnetic field lines for a current-carrying coil. The loop is seen edge-on: the current comes out of the page at the top (a dot) and goes into the page at the bottom (a cross). Near each wire the lines are circles, and through the middle of the loop they run straight, in one direction.

12.7.3Magnetic field due to current carrying solenoidOften examined

A solenoid is a current-carrying coil that produces a magnetic field, as shown in Fig. 12.15. The magnetic field of a solenoid is similar to the magnetic field of a bar magnet, with the north pole at one end of the coil and the south pole at the other end, depending on the direction of the current I. The field inside the solenoid is strong and uniform compared with the field outside. The mathematical expression for the magnetic field of a solenoid is

B=\mu_0\,n\,I

n=N/L is the number of turns N per unit length L of the solenoid.

Figure 12.15: Solenoid. Inside the coil the field lines are straight, parallel and uniform. Outside they spread out and return to the other end, as for a bar magnet (north pole at the right-hand end, south pole at the left).

Right-hand rule for a solenoid: Curl the fingers of the right hand in the direction of the current, and the erect thumb points in the direction of the north pole.

Assignment 12.4: Field inside a solenoid

A 15.0\,\mathrm{cm} long solenoid has 300 turns of wire and a current of 5.0\,\mathrm{A} flows through it. How strong is the magnetic field inside the solenoid?

Given

L = 15.0\,\mathrm{cm}=0.150\,\mathrm{m}
N = 300
I = 5.0\,\mathrm{A}
\mu_0 = 4\pi\times10^{-7}\,\mathrm{Wb\,A^{-1}\,m^{-1}}

To find

B inside the solenoid

Find the number of turns per unit length.
n=\dfrac{N}{L}=\dfrac{300}{0.150}=2000\,\mathrm{m^{-1}}
Use B=\mu_0nI.
B=(4\pi\times10^{-7})(2000)(5.0)=1.26\times10^{-2}\,\mathrm{T}

B\approx1.26\times10^{-2}\,\mathrm{T}

12.8Magnetic field due to the current in a solenoid is increased by a ferrous core

Ferrous materials are those that contain iron; the term ‘ferrous’ comes from the Latin word ‘ferrum’, which means iron. Iron can be easily magnetised when placed in an external magnetic field.

In the absence of a core, the magnetic lines of force start to diverge by curving sharply and immediately outside the coil. If a ferrous material (such as an iron rod) is placed inside the solenoid (Fig. 12.17), the strength of the magnetic field increases significantly. This happens because the ferrous core becomes magnetised by the field generated by the solenoid; the process is called induction. If the wire of the solenoid is tightly wound on the iron core, the magnetic field lines in the cross-section remain parallel and close to each other, which is why the new magnetic field is uniform and strong there.

The relative permeability \mu_r of a ferrous (iron) core is very high. When an iron core is inserted in a solenoid it increases the strength of the magnetic field many times, given by the formula

B_{\text{(iron cored solenoid)}}=\mu_r\,B_{\text{(air cored)}}=\mu_r\,\mu_0\,n\,I

Figure 12.18: The magnetic field of a solenoid with an iron core is stronger. (a) Air core: B=\mu_0nI. (b) Iron core (shaded): B=\mu_r\mu_0nI, so more lines pass through the coil.

Common ferrous materials used in cores include iron, silicon steel and various iron alloys. Each of these has its own characteristic relative permeability.

12.9Magnetic forces between current-carrying conductorsOften examined

The attractive force between two parallel wires carrying current in the same direction is magnetic in nature. When two parallel current-carrying conductors are placed close to each other, each conductor produces its own magnetic field. Each wire is in the magnetic field of the other, which causes the wires to experience a force that is at right angles to both the current and the magnetic field.

Currents in the same direction: attraction

In the region between two parallel wires carrying currents in the same direction, the magnetic fields point in opposite directions, which weakens the net magnetic field. The magnetic field on the outer sides of the wires is stronger than between the wires. Magnetic force acts on a current-carrying conductor from a stronger towards a weaker field region, so the two wires attract each other (Fig. 12.19 (a)).

Note that when the force is attractive (currents in the same direction) there is a neutral point between the wires where the magnetic fields are equal in magnitude but opposite in direction. If the currents in the two wires are equal, the neutral point is midway between the wires (assuming the medium has uniform permeability); otherwise the neutral point is closer to the wire with the smaller current.

Currents in opposite directions: repulsion

The magnetic force acting per unit length on each current-carrying wire is repulsive when the currents in the two wires are in opposite directions. In the region between two parallel wires carrying currents in opposite directions (Fig. 12.19 (b)), the magnetic fields of the two wires are in the same direction and support each other, so the net field is stronger there. The field on the outer sides of the wires is weaker than the field between the wires. The force on each wire acts from the stronger towards the weaker region, so the wires repel each other.

Figure 12.19: Interaction of two current-carrying conductors (seen in cross-section). (a) Currents in the same direction: the fields cancel between the wires at the neutral point and the forces are attractive. (b) Currents in opposite directions: the fields add between the wires and the forces are repulsive.

Derivation Force per unit length between two parallel wires

Aim. Find the magnitude of the force between two long parallel wires of length L, carrying currents I_1 and I_2, placed a distance r apart.

Assumptions.

The two wires are parallel and long, a distance r apart, and each lies in the field of the other (Fig. 12.19 (c)).
The field of each wire at the position of the other is perpendicular to that wire, so \theta=90^\circ and F=BIL.

Figure 12.19 (c): Magnitude of the force between two current-carrying conductors. B_1 (into the page) acts at wire 2 and B_2 (out of the page) at wire 1, so F_{12} and F_{21} are equal and opposite.

The magnetic field of the first wire at the second wire (distance r) is
B_1=\dfrac{\mu_0I_1}{2\pi r}\qquad(12.14)
The magnetic field of the second wire at the first wire is
B_2=\dfrac{\mu_0I_2}{2\pi r}\qquad(12.15)
The force exerted by the first wire on the second is the force on a current I_2 in the field B_1.
F_{12}=B_1I_2L\qquad(12.16)
Put the value of B_1 from Eq. (12.14) in Eq. (12.16).
F_{12}=\left(\dfrac{\mu_0I_1}{2\pi r}\right)I_2L
Divide by L to get the force acting per unit length of the conductor.
\dfrac{F_{12}}{L}=\dfrac{\mu_0I_1I_2}{2\pi r}\qquad(12.17)
Similarly, the force exerted by the second wire on the first is
F_{21}=B_2I_1L\qquad(12.18)
Put the value of B_2 from Eq. (12.15).
F_{21}=\left(\dfrac{\mu_0I_2}{2\pi r}\right)I_1L\quad\text{or}\quad\dfrac{F_{21}}{L}=\dfrac{\mu_0I_1I_2}{2\pi r}

Result\dfrac{F}{L}=\dfrac{\mu_0I_1I_2}{2\pi r}

F_{12} and F_{21} are equal in magnitude and opposite in direction, so these forces make an action and reaction pair. The force per unit length is proportional to the product of the currents and inversely proportional to the distance between the wires.

12.10Magnetic fluxOften examined

Magnetic flux: The number of magnetic lines of force passing through a certain element of area.

Magnetic flux \Phi_B is the scalar product of the uniform magnetic field \mathbf{B} and the vector area \mathbf{A}. (The vector area \mathbf{A} has the magnitude of the area and is directed along the normal to the surface.)

\Phi_B=\mathbf{B}\cdot\mathbf{A}

If \theta is the angle between the magnetic field \mathbf{B} and the vector area \mathbf{A} (Fig. 12.20 (a)), then the magnitude of the magnetic flux is

\Phi_B=BA\cos\theta

Figure 12.20 (a): Magnetic flux through a surface. The loop is seen edge-on; \theta is the angle between the field B and the area vector A (the normal to the surface).

Unit. The unit of magnetic flux is the weber (Wb). One weber is given by 1\,\mathrm{Wb}=1\,\mathrm{N\,m\,A^{-1}}.

Special cases

Maximum flux. If the surface (plane) is held perpendicular to the magnetic field lines, the vectors \mathbf{B} and \mathbf{A} are parallel to each other (\theta=0^\circ), Fig. 12.20 (b). Then \Phi_B=BA\cos0^\circ=BA(1)=BA.
Minimum (zero) flux. If the surface is held parallel to the magnetic field lines, the vectors \mathbf{B} and \mathbf{A} are perpendicular to each other (\theta=90^\circ), Fig. 12.20 (b). Then \Phi_B=BA\cos90^\circ=BA(0)=0.

Figure 12.20 (b): Maximum and minimum magnetic flux. (a) A surface perpendicular to the field lines has maximum flux. (b) A surface parallel to the field lines has zero flux.

Magnetic flux against the angle

Flux through a loop of area A in a field B against the angle \theta between \mathbf{B} and the area vector \mathbf{A}. The flux is maximum (BA) at 0^\circ, zero at 90^\circ and negative beyond it.

Magnetic flux through a curved surface or in a non-uniform magnetic field

When a curved surface is placed in a non-uniform magnetic field (Fig. 12.20 (c)), we divide the curved surface into a number of small elements. The net magnetic flux can be found by adding the magnetic flux through each element:

\Phi_B=\Sigma\,\mathbf{B}\cdot\Delta\mathbf{A}

Figure 12.20 (c): Magnetic flux through a curved surface. The surface is divided into small elements; for each element \Delta A the field B makes an angle \theta with the area vector.

Magnetic flux density

Magnetic flux density B is also defined as the maximum magnetic flux per unit area:

B=\dfrac{\Phi}{A}

The unit of magnetic flux density is \mathrm{Wb\,m^{-2}}, which is equal to the tesla (T).

Example 12.5: Flux through a loop in the xy-plane

In a certain region, the magnetic field is given by \mathbf{B}=(4\,\hat{\mathbf{i}}+6\,\hat{\mathbf{k}})\,\mathrm{Wb\,m^{-2}}. How much magnetic flux passes through a 2.0\,\mathrm{m^2} area loop in this region if the loop lies flat in the xy-plane?

Given

\mathbf{B} = (4\,\hat{\mathbf{i}}+6\,\hat{\mathbf{k}})\,\mathrm{Wb\,m^{-2}}
\Delta\mathbf{A} = (2.0\,\hat{\mathbf{k}})\,\mathrm{m^{2}} the area vector of a loop lying flat in the xy-plane is along the z-axis

To find

magnetic flux \Phi_B

Use \Phi_B=\mathbf{B}\cdot\Delta\mathbf{A}.
\Phi_B=(4\,\hat{\mathbf{i}}+6\,\hat{\mathbf{k}})\cdot(2\,\hat{\mathbf{k}})
In the dot product \hat{\mathbf{i}}\cdot\hat{\mathbf{k}}=0 and \hat{\mathbf{k}}\cdot\hat{\mathbf{k}}=1, so only the z-component of \mathbf{B} contributes.
\Phi_B=4(2)(0)+6(2)(1)=12\,\mathrm{Wb}

\Phi_B=12\,\mathrm{Wb}

Assignment 12.5: Flux through a disc

A magnetic field of 0.8\,\mathrm{T} passes perpendicular to a disc with a radius of 2\,\mathrm{cm}. Find the magnetic flux through the disc.

Given

B = 0.8\,\mathrm{T}
r = 2\,\mathrm{cm}=0.02\,\mathrm{m}
\theta = 0^\circ the field is perpendicular to the disc, so parallel to its area vector

To find

flux \Phi_B

Area of the disc.
A=\pi r^{2}=\pi(0.02)^{2}=1.26\times10^{-3}\,\mathrm{m^{2}}
Flux, with \cos0^\circ=1.
\Phi_B=BA\cos0^\circ=(0.8)(1.26\times10^{-3})\approx1.0\times10^{-3}\,\mathrm{Wb}

\Phi_B\approx1.0\times10^{-3}\,\mathrm{Wb}

12.11Magnetic flux linkage

Magnetic flux linkage: The product of the magnetic flux and the number of turns of the coil.

Flux linkage is a quantity commonly used for solenoids and coils, which are made of N turns of wire. The magnetic flux through a single turn is usually very small. The magnetic flux of a single turn can be increased by increasing the magnetic flux density and by increasing the area of the loop.

The magnetic flux linkage refers to the number of turns N on a coil multiplied by the magnetic flux \Phi of one turn. This gives the equation

N\Phi=NBA

If the cross-sectional area A of the loop and the magnetic flux density B are not perpendicular, the equation becomes

\text{Magnetic flux linkage}=BAN\cos\theta

\theta is the angle between the vector area \mathbf{A} and the magnetic flux density \mathbf{B} (Fig. 12.21).

Figure 12.21: Magnetic flux linkage. A coil of N turns, each of area A, in a field B. \theta is the angle between the area vector (the normal) and B.

Units. The flux linkage \Phi N has the units of weber turns (Wb turns).

Example 12.6: Flux linkage of a solenoid

A solenoid with a circular cross-sectional area of 0.80\,\mathrm{m^2} and 300 turns is facing perpendicular to a magnetic field with magnetic flux density of 4\,\mathrm{mT}. Determine the magnetic flux linkage for this solenoid.

Given

A = 0.80\,\mathrm{m^{2}}
N = 300\ \text{turns}
B = 4\,\mathrm{mT}=4\times10^{-3}\,\mathrm{T}
\theta = 0^\circ the solenoid faces the field, so \cos\theta=1

To find

flux linkage \Phi N

Use the formula for the flux linkage.
\Phi N=BAN
Substitute the values.
\Phi N=(4\times10^{-3}\,\mathrm{T})\times(0.80\,\mathrm{m^{2}})\times300=0.96\ \mathrm{Wb\ turns}

\Phi N=0.96\ \mathrm{Wb\ turns}

Assignment 12.6: Flux linkage of a coil

A solenoid contains 200 turns of wire. Each piece of wire has a cross-sectional area of 0.004\,\mathrm{m^2}. If the magnetic flux density is 12.0\,\mathrm{mT}, calculate the magnetic flux linkage.

Given

N = 200
A = 0.004\,\mathrm{m^{2}}
B = 12.0\,\mathrm{mT}=12.0\times10^{-3}\,\mathrm{T}

To find

flux linkage \Phi N

Take the field perpendicular to the area (\cos\theta=1) and use \Phi N=BAN.
\Phi N=(12.0\times10^{-3})(0.004)(200)=9.6\times10^{-3}\ \mathrm{Wb\ turns}

\Phi N=9.6\times10^{-3}\ \mathrm{Wb\ turns}

12.12Faraday's lawOften examined

Electromagnetic induction: The phenomenon in which a change in magnetic flux causes an induced emf in a conductor.

The basic requirement for electromagnetic induction is a change in the magnetic flux linking the conductor (or coil).

Faraday's law of electromagnetic induction

Faraday's law: The average e.m.f. induced in a conducting coil of N turns is directly proportional to the rate of change of magnetic flux through the conductor.

If N is the number of turns of the coil and \varepsilon is the induced e.m.f., then

\text{induced e.m.f.}\propto\text{rate of change of magnetic flux}=\dfrac{\text{total change in magnetic flux}}{\text{total time}}

For a coil of N turns:

\text{induced e.m.f.}\propto N\dfrac{\Delta\Phi}{\Delta t}\qquad\text{so}\qquad\varepsilon=k\,N\dfrac{\Delta\Phi}{\Delta t}

Here k is a constant whose value is 1 in SI units. So

\varepsilon=N\,\dfrac{\Delta\Phi}{\Delta t}

This equation is called Faraday's law of electromagnetic induction and it is used to determine the magnitude of the induced emf. The induced emf always opposes the change in magnetic flux. The direction of the induced emf is given by Lenz's law (section 12.13), which is included by a negative sign:

\varepsilon=-N\,\dfrac{\Delta\Phi}{\Delta t}

The negative sign indicates that the direction of the induced current is such that it opposes the change in flux. Faraday's law is the fundamental law that describes the production of emf due to change of magnetic flux in various circuits and devices. For example, AC devices such as induction motors, induction generators or transformers, as well as DC devices such as DC motors and DC generators, depend on it.

Experiments on electromagnetic induction

Experiment 1: a moving magnet and a coil. In this experiment we use a bar magnet and a coil connected to a sensitive galvanometer, as in Fig. 12.22.

Figure 12.22: Magnet moving and coil at rest. The galvanometer shows no current when there is no relative motion, and currents in opposite directions when the magnet moves towards and away from the coil.

Case I: when there is no relative motion between the magnet and the coil, the galvanometer shows no current.
Case II: when the bar magnet is moved towards the coil, the magnetic flux changes and induces an emf. As a result an induced current flows in the coil.
Case III: when the bar magnet is moved away from the coil, the magnetic flux again changes, inducing an emf and an induced current in the coil, but in the opposite direction.

Experiment 2: emf induced by relative motion of a coil in a magnetic field. In this experiment the coil is rotated in a magnetic field (Fig. 12.23). The change in magnetic flux is given by \Delta\Phi=\Delta\left(BA\cos\theta\right). During the rotation the angle \theta between the magnetic field and the vector area \mathbf{A} of the coil changes, so the magnetic flux of the coil changes and induces an emf in the coil. The induced current flows through the circuit and produces a deflection in the galvanometer. This is the basic principle of an electric generator.

Figure 12.23: A coil is rotated in a magnetic field (into the page). The angle between the field and the area vector of the coil keeps changing, so the flux changes and an emf is induced.

Experiment 3: emf induced by a changing magnetic field (the transformer effect). In this experiment two coils are placed close to each other (Fig. 12.24). The primary coil P is connected in series with a battery through a switch and a rheostat, while the secondary coil S is connected only to a galvanometer.

Figure 12.24: By changing the magnetic field of the primary coil, the magnetic flux through the secondary coil changes and an emf is induced in it.

When the switch of coil P is closed, a momentary current is induced in coil S.
When the switch of P is opened, a current is produced in coil S in the opposite direction.
Opening or closing the switch changes the current, and so the magnetic field of coil P; the magnetic flux through coil S changes and the changing flux induces an emf in coil S.
If the current in the primary coil is varied with the help of the rheostat, the magnetic field of coil P and the flux through coil S change. The changing flux induces an emf in coil S. The induced emf and induced current remain in the coil as long as the magnetic flux through it changes. This is the basic principle of the working of a transformer.

Points to remember about induced emf

An electromotive force (emf) is induced when the magnetic flux linking a coil changes (the flux either increases or decreases).
A constant magnetic flux cannot produce an emf in a conductor. The cause of the induced emf is a change in magnetic flux.
The magnitude of the induced emf depends on the rate of change of magnetic flux.
The induced emf is independent of the resistance and resistivity of the circuit and depends only on the rate of change of magnetic flux through the conductor.
The induced current (I=\varepsilon/R) depends on the resistance and resistivity of the circuit.
When the magnetic flux through a conductor changes, an induced emf is always produced, but an induced current flows only when the circuit is closed.

Factors affecting the magnitude of the induced emf

The induced emf depends on the number of turns in a coil.
The induced emf depends on the speed of the movement of the conductor through the magnetic field.
The induced emf depends on the length of the conductor inside the magnetic field.
The induced emf depends on the rate of change of magnetic flux through the conductor.
The induced emf can be increased by increasing the strength of the external magnetic field.
The induced emf can be increased by increasing the area of the coil.

Example 12.7: Average emf and current in a coil

A coil of resistance 50\,\Omega is placed in a magnetic field and the initial magnetic flux is 2\,\mathrm{mWb}. The coil has 50 turns and a galvanometer of 100\,\Omega resistance is connected in series with it. Find the average emf and the current if the coil is moved in 1/10th s from the given magnetic field to another field where the final magnetic flux is 4\,\mathrm{mWb}.

Given

N = 50
\Phi_1 = 2\,\mathrm{mWb}=2\times10^{-3}\,\mathrm{Wb} initial flux
\Phi_2 = 4\,\mathrm{mWb}=4\times10^{-3}\,\mathrm{Wb} final flux
R_1 = 50\,\Omega resistance of the coil
R_2 = 100\,\Omega resistance of the galvanometer
\Delta t = \dfrac{1}{10}\,\mathrm{s}=0.1\,\mathrm{s}

To find

average emf \varepsilon
average current I

By Faraday's law the magnitude of the induced emf is \varepsilon=N\dfrac{\Delta\Phi}{\Delta t}, with \Delta\Phi=\Phi_2-\Phi_1.
\varepsilon=N\left(\dfrac{\Phi_2-\Phi_1}{\Delta t}\right)
Put in the values.
\varepsilon=50\times\left(\dfrac{4\times10^{-3}-2\times10^{-3}}{0.1}\right)=1\,\mathrm{V}
The galvanometer is connected in series with the coil, so the total resistance is the sum.
R_{eq}=R_1+R_2=50+100=150\,\Omega
The current is the emf divided by the total resistance.
I=\dfrac{\varepsilon}{R_{eq}}=\dfrac{1}{150}=6.67\times10^{-3}\,\mathrm{A}=6.67\,\mathrm{mA}

\varepsilon=1\,\mathrm{V},\qquad I=6.67\,\mathrm{mA}

Assignment 12.7: Emf in a shrinking loop (printed as Assignment 12.6)

A loop of wire is placed in a uniform magnetic field that is perpendicular to the plane of the loop. The strength of the magnetic field is 0.6\,\mathrm{T}. The area of the loop begins to shrink at a constant rate of 0.8\,\mathrm{m^2\,s^{-1}}. What is the magnitude of the emf induced in the loop while it is shrinking?

Given

B = 0.6\,\mathrm{T} constant
\dfrac{\Delta A}{\Delta t} = 0.8\,\mathrm{m^{2}\,s^{-1}}
N = 1 a single loop

To find

induced emf \varepsilon

The field is perpendicular to the loop (\cos\theta=1), so the flux is \Phi=BA. Only the area changes, so \Delta\Phi=B\,\Delta A.
\Phi=BA\quad\Rightarrow\quad\Delta\Phi=B\,\Delta A
Apply Faraday's law for one turn.
\varepsilon=\dfrac{\Delta\Phi}{\Delta t}=B\,\dfrac{\Delta A}{\Delta t}=(0.6)(0.8)=0.48\,\mathrm{V}

\varepsilon=0.48\,\mathrm{V}

Worked example: induced emf from a flux-time graph

The magnetic flux through a coil of 100 turns varies with time as in the graph below: it rises uniformly from zero to 60\,\mathrm{mWb} in 0.30\,\mathrm{s}, stays constant for the next 0.30\,\mathrm{s} and then falls uniformly to zero in 0.20\,\mathrm{s}. Find the induced emf in each interval. If the total resistance of the circuit is 10\,\Omega, find the current in the first interval.

Flux through the coil against time

The induced emf is proportional to the slope of the flux-time graph: 20\,\mathrm{V} while the flux rises (slope 0.2\,\mathrm{Wb\,s^{-1}}), zero while it is constant and 30\,\mathrm{V} in the opposite sense while it falls.

Given

N = 100
\Phi_{\max} = 60\,\mathrm{mWb}=6.0\times10^{-2}\,\mathrm{Wb}
R = 10\,\Omega

To find

emf in the three intervals
current in the first interval

Faraday's law gives the size of the emf from the rate of change of flux, \varepsilon=N\Delta\Phi/\Delta t, which is N times the slope of the graph.
Interval 1 (0 to 0.30 s): the flux increases by 0.060\,\mathrm{Wb}.
\varepsilon_1=100\times\dfrac{0.060}{0.30}=20\,\mathrm{V}
Interval 2 (0.30 s to 0.60 s): the flux is constant, so \Delta\Phi=0 and no emf is induced.
\varepsilon_2=0
Interval 3 (0.60 s to 0.80 s): the flux decreases by 0.060\,\mathrm{Wb} in 0.20\,\mathrm{s}.
\varepsilon_3=100\times\dfrac{0.060}{0.20}=30\,\mathrm{V}
By Lenz's law the induced current opposes the change. The flux rises in interval 1 and falls in interval 3, so the emf in interval 3 acts in the opposite direction to that in interval 1.
Current in interval 1.
I=\dfrac{\varepsilon_1}{R}=\dfrac{20}{10}=2.0\,\mathrm{A}

\varepsilon_1=20\,\mathrm{V},\quad\varepsilon_2=0,\quad\varepsilon_3=30\,\mathrm{V}\ (\text{reversed});\quad I_1=2.0\,\mathrm{A}

12.13Lenz's lawOften examined

In 1834 the Russian physicist Heinrich Lenz discovered that the polarity of the induced emf always leads to an induced current that opposes the change which induces the emf.

Lenz's law: The direction of the induced current is always such as to oppose the change that causes the current.

Condition for its application. We can apply Lenz's law directly to closed loops because it refers to induced currents and not induced emf. However, if the loop is not closed, we imagine it being closed and then, from the direction of the induced current, we can find the polarity of the induced emf.

Explanation with a bar magnet and a coil

When a bar magnet is pushed towards a coil connected to a galvanometer, an emf is induced because the magnetic flux through the coil changes, and an induced current flows through it (Fig. 12.25 (a)). The direction of the current in the coil is determined by Lenz's law.

Figure 12.25: (a) The bar magnet is pushed towards the coil, so the near face of the coil becomes a north pole and repels it. (b) The bar magnet is pulled away from the coil, so the near face becomes a south pole and attracts it.

Direction of the induced current

Pushing the magnet in. The cause of the induced current is the push of the north pole of the magnet towards the coil. This increases the magnetic flux through the coil.
The induced current must oppose this push (and the increase of flux), so it makes a magnetic field of its own that pushes the magnet back. This is possible only if the left side of the coil (the side facing the magnet) acts as a north pole and the right side as a south pole.
Applying the right-hand rule to these poles, the current in the coil as seen from the magnet is anticlockwise (Fig. 12.25 (a)).
Pulling the magnet away. The induced current opposes the pull by creating a south pole towards the bar magnet. The flux through the coil now decreases, and to oppose this decrease the coil attracts the magnet: the left side of the coil acts as a south pole and the right side as a north pole.
The south pole of the coil attracts the north pole of the bar magnet and opposes its motion. The direction of the induced current in the coil reverses, that is, the current flows clockwise by the right-hand rule (Fig. 12.25 (b)).

Worked example: polarity of the coil in four cases of Lenz's law

A coil is joined to a galvanometer. A bar magnet is moved towards or away from it, with either its north pole or its south pole facing the coil. For each of the four cases state whether the near face of the coil becomes a north or a south pole, and sketch the case in which the south pole is pulled away.

Lenz's law: the induced current flows so that its magnetic field opposes the change of flux. So first decide whether the flux is increasing or decreasing, then ask which pole the coil must show to resist that.
Case: south pole pulled away. The field of the magnet at the coil points towards the south pole, and it is getting weaker as the magnet leaves, so the flux decreases.
To oppose a decrease, the coil makes a field in the same direction (towards the magnet). A coil's field leaves from its north face, so the near face is a north pole.
Check by forces: a north pole on the coil attracts the south pole of the magnet and tries to hold it back, which opposes the motion, as Lenz's law requires.
Repeat the reasoning for the other three cases to complete the table: approaching poles are met by a like pole (repulsion), receding poles by an unlike pole (attraction).

The south pole of a magnet is pulled away from a coil. The flux (towards the magnet) falls, so the coil makes a field in the same direction: its near face becomes a north pole (accent) that attracts the receding south pole.

The four cases. The coil always opposes the motion of the magnet.
Pole near coil	Motion	Flux through coil	Near face of coil	Force on magnet
N	towards	increases	N	repulsion
N	away	decreases	S	attraction
S	towards	increases	S	repulsion
S	away	decreases	N	attraction

\text{S pole pulled away: near face of the coil is a north pole (attracts the magnet)}

12.13Lenz's law and law of conservation of energyOften examined

The law of conservation of energy states that energy can neither be created nor destroyed, but it can be changed from one form to another.

Lenz's law states that the direction of the current is such that it opposes the change in the magnetic flux. So an effort is required to do work against the opposing forces. This work leads to changes in magnetic flux and hence an electric current is induced. Thus mechanical energy is converted into an equivalent amount of electrical energy, which is in accordance with the law of conservation of energy.

Consider an example: the N-pole of a magnet is approaching the coil, as in Fig. 12.26. A repulsive force acts on the bar magnet due to the current induced in the coil. The result is that the motion of the magnet is opposed. The mechanical energy spent in overcoming this opposition is converted into electrical energy, which appears in the coil. We have to spend mechanical energy to induce electrical energy. Thus Lenz's law is in accordance with the law of conservation of energy.

12.13Fleming's right-hand rule

To find the direction of the induced current, Fleming's right-hand rule may be used, as shown in Fig. 12.27.

Fleming's right-hand rule: Stretch out the forefinger, middle finger and thumb of your right hand so that they are at right angles to one another. If the forefinger points in the direction of the magnetic field and the thumb in the direction of motion of the conductor, then the middle finger will point in the direction of the induced current.

Figure 12.27: Fleming's right-hand rule to find the direction of the induced current.

When the conductor moves upwards at right angles to a stationary magnetic field, the direction of the induced current is from right to left. If the motion of the conductor is downward, the direction of the induced current is from left to right.

12.14Seismometer

Around 1906 a Russian Empire nobleman and inventor named Gallitzin was the first to create a seismometer based on Faraday's law of electromagnetic induction. A seismometer is a device that records seismic waves emitted during earthquakes, explosions or other earth-shaking events. Electromagnetic sensors in seismographs convert ground movements into electrical signals.

Construction

A frame is made on which wire is wound, and the ends of this frame are connected with springs tied to the side walls. A permanent magnet is placed between this frame and its coil in such a way that the coil can oscillate freely (Fig. 12.28).

Figure 12.28: Seismometer. The magnet is fixed to the casing, which rests on the ground; the coil on its frame hangs from springs and can move relative to the magnet.

Working

When the Earth vibrates because of an earthquake, there is relative motion between the magnet and the coil.
This changes the magnetic flux through the coil. According to Faraday's law of electromagnetic induction an emf and a current are induced in the coil.
The induced current is given to an external circuit where the values are recorded and a graph is plotted.
During a high-intensity earthquake the relative velocity between the coil and the magnet increases, and the amplitude of the vibration of the coil increases.

The induced emf in the coil is directly proportional to the displacement of the coil within the magnetic field. When the coil moves faster, the rate of change of displacement of the coil in the magnetic field increases. As a result the rate of change of magnetic flux through the coil increases, and the induced emf increases.

Formula sheet

Force on a current-carrying conductor

F=BIL\sin\theta

B: magnetic flux density, I: current, L: length of conductor in the field, \theta: angle between \mathbf{L} and \mathbf{B}

Unit: \mathrm{N}

Force on a conductor (vector form)

\mathbf{F}=I\,(\mathbf{L}\times\mathbf{B})

\mathbf{L}: length vector along the direction of the current

Unit: \mathrm{N}

Magnetic flux density

B=\dfrac{F}{IL}

F: force on a conductor placed at right angles to the field, I: current, L: length

Unit: \mathrm{T}=\mathrm{N\,A^{-1}\,m^{-1}}

Force on a moving charge

\mathbf{F}_B=q\,(\mathbf{v}\times\mathbf{B}),\quad F_B=qvB\sin\theta

q: charge, v: speed, B: field, \theta: angle between \mathbf{v} and \mathbf{B}

Unit: \mathrm{N}

Radius of the circular path

r=\dfrac{mv}{qB}

m: mass, v: speed, q: charge, B: field perpendicular to v

Unit: \mathrm{m}

Cyclotron (angular) frequency

\omega=\dfrac{qB}{m}

q: charge, B: field, m: mass

Unit: \mathrm{rad\,s^{-1}}

Time period in a magnetic field

T=\dfrac{2\pi m}{qB}

m: mass, q: charge, B: field

Unit: \mathrm{s}

Frequency in a magnetic field

f=\dfrac{qB}{2\pi m}

q: charge, B: field, m: mass

Unit: \mathrm{Hz}

Velocity selector

v=\dfrac{E}{B}

E: electric field, B: magnetic field (perpendicular to each other and to v)

Unit: \mathrm{m\,s^{-1}}

Lorentz force

\mathbf{F}=q\mathbf{E}+q\,(\mathbf{v}\times\mathbf{B})

q: charge, \mathbf{E}: electric field, \mathbf{B}: magnetic field, \mathbf{v}: velocity

Unit: \mathrm{N}

Field of a long straight wire

B=\dfrac{\mu_0I}{2\pi r}

I: current, r: distance from the wire, \mu_0=4\pi\times10^{-7}\,\mathrm{Wb\,A^{-1}\,m^{-1}}

Unit: \mathrm{T}

Field at the centre of a circular loop

B=\dfrac{\mu_0I}{2r}

I: current, r: radius of the loop

Unit: \mathrm{T}

Field inside a solenoid

B=\mu_0nI,\quad n=\dfrac{N}{L}

n: turns per unit length, N: number of turns, L: length, I: current

Unit: \mathrm{T}

Solenoid with a ferrous core

B=\mu_r\mu_0nI

\mu_r: relative permeability of the core

Unit: \mathrm{T}

Force per unit length between parallel wires

\dfrac{F}{L}=\dfrac{\mu_0I_1I_2}{2\pi r}

I_1,I_2: currents in the wires, r: distance between them

Unit: \mathrm{N\,m^{-1}}

Magnetic flux

\Phi_B=\mathbf{B}\cdot\mathbf{A}=BA\cos\theta

B: flux density, A: area, \theta: angle between \mathbf{B} and the area vector \mathbf{A}

Unit: \mathrm{Wb}

Flux through a curved surface

\Phi_B=\Sigma\,\mathbf{B}\cdot\Delta\mathbf{A}

sum over small surface elements \Delta\mathbf{A}

Unit: \mathrm{Wb}

Flux density from flux

B=\dfrac{\Phi}{A}

\Phi: flux through area A held perpendicular to the field

Unit: \mathrm{Wb\,m^{-2}}=\mathrm{T}

Magnetic flux linkage

N\Phi=BAN\cos\theta

N: number of turns, \Phi: flux through one turn

Unit: \mathrm{Wb\ turns}

Faraday's law

\varepsilon=N\,\dfrac{\Delta\Phi}{\Delta t}

\varepsilon: magnitude of the induced emf, N: turns, \Delta\Phi/\Delta t: rate of change of flux

Unit: \mathrm{V}

Faraday's law with Lenz's law

\varepsilon=-N\,\dfrac{\Delta\Phi}{\Delta t}

the negative sign shows that the induced current opposes the change of flux

Unit: \mathrm{V}

Induced current

I=\dfrac{\varepsilon}{R}

R: total resistance of the closed circuit

Unit: \mathrm{A}

Key terms

Magnetic field: The region of space around a magnet in which a compass needle, a small magnet, ferromagnetic material or a moving charged particle experiences a magnetic force.
Magnetic field lines: Lines drawn to picture a magnetic field. They never cross, form closed loops, run from the north pole to the south pole outside a magnet, and are closest together where the field is strongest.
Magnetic monopole: An isolated north or south pole. None exists: the north and south poles of a magnet do not exist independently.
Right-hand rule for a straight wire: Grasp the wire in the right fist with the thumb along the conventional current; the curled fingers give the direction of the magnetic field.
Catapult field: The combined magnetic field of a current-carrying wire and an external magnet, which makes the wire move from the stronger to the weaker side of the field.
Fleming's left-hand rule: With thumb, forefinger and middle finger of the left hand at right angles: forefinger = field, middle finger = current, thumb = force on the conductor.
Magnetic flux density (magnetic induction): The force acting per unit current per unit length on a conductor placed at right angles to the magnetic field, B=F/IL. A vector; unit tesla.
Tesla: The SI unit of magnetic flux density: the flux density in which a 1\,\mathrm{m} conductor carrying 1\,\mathrm{A} at right angles to the field feels a force of 1\,\mathrm{N}. 1\,\mathrm{T}=10^{4}\,\mathrm{G}.
Right-hand rule for the force on a moving charge: Sweep from the velocity towards the magnetic field with the right hand through the smaller angle; the thumb gives the direction of the force on a positive charge.
Helical path: The spiral path of a charged particle whose velocity makes an angle other than 0^\circ, 90^\circ or 180^\circ with the magnetic field.
Cyclotron frequency: The angular frequency \omega=qB/m of a charged particle moving in a circle in a magnetic field.
Velocity selector: A device with mutually perpendicular electric and magnetic fields that lets only particles of speed v=E/B pass through undeflected.
Lorentz force: The vector sum of the electric and magnetic forces on a charged particle, \mathbf{F}=q\mathbf{E}+q(\mathbf{v}\times\mathbf{B}).
Solenoid: A current-carrying coil that produces a magnetic field like that of a bar magnet, strong and uniform inside.
Right-hand rule for a solenoid: Curl the fingers of the right hand in the direction of the current; the erect thumb points to the north pole.
Ferrous core: A core containing iron placed inside a solenoid. It becomes magnetised by induction and multiplies the field by the relative permeability \mu_r.
Relative permeability: The factor \mu_r by which a core increases the magnetic field of a solenoid, B=\mu_r\mu_0nI.
Neutral point: A point between two parallel wires carrying currents in the same direction where the two fields are equal and opposite.
Magnetic flux: The number of magnetic field lines passing through an area: \Phi=\mathbf{B}\cdot\mathbf{A}=BA\cos\theta. Scalar; unit weber.
Weber: The SI unit of magnetic flux, 1\,\mathrm{Wb}=1\,\mathrm{N\,m\,A^{-1}}.
Magnetic flux linkage: The product of the magnetic flux and the number of turns of a coil, N\Phi; unit Wb turns.
Electromagnetic induction: The production of an induced emf in a conductor by a change of the magnetic flux linking it.
Faraday's law: The magnitude of the induced emf in a coil of N turns is directly proportional to the rate of change of magnetic flux linking it: \varepsilon=N\Delta\Phi/\Delta t.
Lenz's law: The direction of the induced current is always such as to oppose the change that causes the current.
Fleming's right-hand rule: With thumb, forefinger and middle finger of the right hand at right angles: forefinger = field, thumb = motion of the conductor, middle finger = induced current.
Seismometer: A device that records seismic waves; a coil on springs moves relative to a magnet during an earthquake and the emf induced in the coil is proportional to its displacement.

Multiple choice questionsp. 241–242

Encircle the correct option. The book prints no answer key for these questions; each key below was worked out from the physics.

12025 Annual I, Set D, MCQ 14

Which of the following is the SI unit of magnetic induction?

Ahenry
BteslaAnswer
Cweber
Dfarad

Why

The tesla is the SI unit of magnetic flux density (magnetic induction). It is defined from F=BIL\sin\theta: 1\,\mathrm{T}=1\,\mathrm{N\,A^{-1}\,m^{-1}}. The henry is the unit of inductance, the weber of magnetic flux and the farad of capacitance.

A 0.5\,\mathrm{m} long straight wire carrying a current of 3.2\,\mathrm{A} kept at right angle to a uniform magnetic field of 2.0\,\mathrm{T}. The force acting on the wire will be:

A2\,\mathrm{N}
B2.4\,\mathrm{N}
C1.2\,\mathrm{N}
D3.2\,\mathrm{N}Answer

Why

The wire is at right angles to the field, so \theta=90^\circ and \sin\theta=1. Then F=BIL\sin\theta=2.0\,\mathrm{T}\times3.2\,\mathrm{A}\times0.5\,\mathrm{m}\times1=3.2\,\mathrm{N}. The length is 0.5\,\mathrm{m}, not 2\,\mathrm{m}, so the other options come from wrong multiplications of B, I and L.

An electron and proton enter a magnetic field with same velocity, which of the following is the ratio of their acceleration?

A\dfrac{m_e}{m_p}
B\dfrac{m_p}{m_e}Answer
C\left(\dfrac{m_e}{m_p}\right)^{1/2}
D\left(\dfrac{m_p}{m_e}\right)^{1/2}

Why

The magnetic force is F=qvB\sin\theta, and with \theta=90^\circ the acceleration is a=\dfrac{qvB}{m}. The electron and proton carry charges of the same magnitude e and enter with the same v and B, so only the mass differs.

\frac{a_e}{a_p}=\frac{evB/m_e}{evB/m_p}=\frac{m_p}{m_e}\approx1836

Option A is the reciprocal of the answer. The square roots in C and D do not appear in the expression.

42026 Annual I, Set B, MCQ 10

If charge particle enters a magnetic field an angle 60^\circ, then its path will be:

Aelliptical
Bstraight line
ChelicalAnswer
Dcircular

Why

The velocity has two components: v\cos60^\circ parallel to B (unchanged, since the force is zero along B) and v\sin60^\circ perpendicular to B (which makes the particle circle). The combined motion is a spiral, a helix, as in Fig. 12.9. The path is a straight line only at 0^\circ or 180^\circ and a circle only at 90^\circ.

An electron is revolving in circular path of radius r in a magnetic field. If the magnetic field is halved, then new radius will be:

Ar/2
Br/4
C2rAnswer
D4r

Why

The radius of the circle is r=\dfrac{mv}{qB}. The mass, speed and charge do not change, so r\propto 1/B. Halving B doubles the radius: r'=\dfrac{r}{1/2}=2r. Options A and B would follow if B were doubled or quadrupled, not halved.

Which of the following particles in motion cannot be deflected by magnetic field?

Aelectron
Bproton
Calpha-particle
DneutronAnswer

Why

A magnetic force F=qv\times B needs a charge q. The electron, proton and alpha-particle are charged, so they are deflected when they move across a field. The neutron has no charge, so F=0 and it is not deflected.

A proton and an \alpha-particle moving with same velocity enter a magnetic field perpendicularly. Which one of the following is the ratio of radius of circular path of proton to \alpha-particle?

A1/2Answer
B1
C2
D4

Why

The radius is r=\dfrac{mv}{qB}. The proton has mass m_p and charge e; the \alpha-particle has mass \approx4m_p and charge 2e. The velocity and field are the same for both.

\frac{r_p}{r_\alpha}=\frac{m_p/e}{4m_p/2e}=\frac{1}{2}

The \alpha-particle has the larger radius because its mass-to-charge ratio is twice that of the proton.

A charged particle is moving in uniform magnetic field in a circular path. The radius of circular path is R. If the energy of the particle is doubled, then the new radius will be:

A\dfrac{R}{\sqrt2}
B\dfrac{R}{2}
C\sqrt2\,RAnswer
D2R

Why

The kinetic energy is K=\tfrac12mv^2, so mv=\sqrt{2mK} and the radius is R=\dfrac{mv}{qB}=\dfrac{\sqrt{2mK}}{qB}. For fixed m, q and B, R\propto\sqrt{K}.

\frac{R'}{R}=\sqrt{\frac{2K}{K}}=\sqrt2\quad\Rightarrow\quad R'=\sqrt2\,R

A positively charged particle moving with velocity of 3.1\times10^5\,\mathrm{m\,s^{-1}} normal to uniform magnetic field of magnitude 5.4\times10^{-5}\,\mathrm{T}. Particle experiences force of 8.1\times10^{-16}\,\mathrm{N}, the charge on particle is:

A4.8\times10^{-10}\,\mathrm{C}
B1.6\times10^{-19}\,\mathrm{C}
C6.4\times10^{-17}\,\mathrm{C}
D4.8\times10^{-17}\,\mathrm{C}Answer

Why

The velocity is normal to B, so \theta=90^\circ and F=qvB.

q=\frac{F}{vB}=\frac{8.1\times10^{-16}}{(3.1\times10^{5})(5.4\times10^{-5})}=\frac{8.1\times10^{-16}}{16.7}\approx4.8\times10^{-17}\,\mathrm{C}

A proton is moving along the axis of a solenoid carrying a current, as shown in figure. The magnetic force on proton will be:

Aradially inward.
Bradially outward.
Cno force acts.Answer
Dradially outward.

Why

Inside a solenoid the field is uniform and runs along the axis. The proton moves along the same axis, so its velocity is parallel to B (\theta=0^\circ), and F=qvB\sin0^\circ=0. The proton keeps moving in a straight line and no force acts. A radial force would need a velocity across the field, so B and D are wrong.

Note. Misprint in the book: options B and D are both printed as "radially outward".

Magnetic flux will be maximum when:

Amagnetic field is perpendicular to the plane area.Answer
Bmagnetic field lies parallel to the plane area.
Carea is held at an angle of 45^\circ.
Darea is held at an angle of 60^\circ.

Why

The flux is \Phi_B=BA\cos\theta, where \theta is the angle between B and the area vector \mathbf A (the normal to the plane). The flux is largest when \theta=0^\circ, that is, when B is perpendicular to the plane of the area, so \Phi_B=BA.

If B lies parallel to the plane, \theta=90^\circ and \Phi_B=0. A plane at 45^\circ or 60^\circ to the field gives a flux less than BA in each case.

The cause of induced emf is:

Aconstant magnetic flux.
Bincrease in magnetic flux only.
Cdecrease in magnetic flux only.
Dchange in magnetic flux.Answer

Why

By Faraday's law, an emf is induced whenever the magnetic flux linking a conductor changes. A constant flux induces no emf (A). A change may be an increase (B) or a decrease (C), so each of those is only part of the answer. The general cause is a change in flux, whichever direction it takes.

Lenz's law is in accordance with the law of conservation of:

Acharge
BenergyAnswer
Cmomentum
Dangular momentum

Why

Lenz's law says the induced current opposes the change that causes it. Work must be done against this opposing force to keep the magnet or coil moving, and that mechanical energy appears as electrical energy in the circuit. This is the law of conservation of energy, as the book explains in section 12.13.

A metallic circular ring is suspended by a string and is kept in a vertical plane. When a magnet is pushed towards the ring, then it will:

Aget displaced towards the magnet.
Bremains stationary.
Cget displaced away from the magnet.Answer
Dnothing can be said.

Why

Pushing the north pole towards the ring increases the flux through it, so an induced current flows in the ring. By Lenz's law this current opposes the approach of the magnet. The ring acts as a magnet whose face towards the magnet is also a north pole, so the ring is repelled and swings away from the magnet.

Option A would mean the ring helps the approach, which Lenz's law forbids. Option B is wrong because a current is induced, so the ring is pushed. Option D is wrong because the direction follows from Lenz's law.

Short questionsp. 243

13 marks

What are the various ways to create a magnetic field.

Solution

A magnetic field is created in four ways:

a permanent bar magnet, which sets up a field in the space around it (Fig. 12.1);
a current-carrying conductor. A current in a long straight wire produces concentric circular field lines, with the direction given by the right-hand rule;
a moving charge, because a moving charge is equivalent to a current;
a fluctuating electric field.

So a magnet and a current are the two common sources, while a moving charge and a changing electric field are the general cases.

23 marks

In what way is a magnetic field different from an electric field?

Solution

An electric field exists around charges and exerts a force on any charge in it, whether the charge is at rest or moving. A magnetic field is set up by moving charges, currents or permanent magnets, and it exerts a force only on moving charges, currents and magnetic materials. A charge at rest feels no magnetic force.

Direction of the force. The magnetic force on a moving charge is perpendicular to both the velocity and the field, \vec F=q(\vec v\times\vec B), so it changes the direction of motion but not the speed. The electric force acts along the field.
Field lines. Magnetic field lines always form complete closed loops, with no isolated north or south poles (magnetic monopoles). Electric field lines start on positive charges and end on negative charges.
Source. A magnetic field is always produced by a current or a moving charge, while an electric field is produced by any charge.

33 marks

Why may a current-carrying conductor experience a force in a magnetic field?

Solution

A current-carrying conductor produces its own magnetic field, made of circular lines around the wire. This field combines with the external field of the magnet (Fig. 12.4).

On one side of the wire the two fields point in the same direction, so they strengthen each other and the net field is stronger. On the other side they point in opposite directions, so the net field is weaker. The conductor is pushed from the stronger region towards the weaker region, so the force acts across the field and the wire moves.

The size of the force is F=BIL\sin\theta, where \theta is the angle between the conductor and the field. It is greatest when the conductor is at right angles to the field.

43 marks

Can a charged particle move in a straight line through some region of space, in which magnetic field is non-zero? Explain.

Solution

Yes. The magnetic force is F=qvB\sin\theta, where \theta is the angle between the velocity \vec v and the field \vec B. If the particle moves parallel (\theta=0^\circ) or antiparallel (\theta=180^\circ) to the field, then \sin\theta=0 and F=0 even though B\neq0. The particle keeps moving in a straight line.

A charge can also pass straight through a non-zero field when an electric field balances the magnetic force. In a velocity selector qE=qvB, so the net force is zero and the particle is undeflected (Fig. 12.12). Its speed is then v=E/B.

53 marks

When is the magnetic force on a moving charge in a magnetic field maximum, and when is it minimum?

Solution

The force on a charge is F=qvB\sin\theta, where \theta is the angle between \vec v and \vec B.

Maximum: when the charge moves perpendicular to the field, \theta=90^\circ, so \sin\theta=1 and F_{\max}=qvB. The particle then moves in a circle.
Minimum (zero): when the charge moves parallel or antiparallel to the field, \theta=0^\circ or 180^\circ, so \sin\theta=0 and F=0. The charge then moves in a straight line. A charge at rest (v=0) also has zero force.

63 marks

In a region with a homogeneous magnetic field B directed normal to the plane into the paper, an alpha particle and a proton are travelling in its plane. What will be the ratio of the radii of the two particles' field trajectories if they have equal linear momenta?

Solution

Both particles move perpendicular to \vec B, so each follows a circle of radius r=\dfrac{mv}{qB}=\dfrac{p}{qB}. The momentum p=mv and the field B are the same for both, so r\propto\dfrac{1}{q}.

\frac{r_\alpha}{r_p}=\frac{p/(2eB)}{p/(eB)}=\frac{q_p}{q_\alpha}=\frac{e}{2e}=\frac{1}{2}

The ratio of the radii of the alpha particle to the proton is 1:2. The proton, which has half the charge of the alpha particle, moves on the larger circle.

72025 Annual I, Set B, Q.2 (ix)3 marks

The kinetic energy of a charged particle moving in uniform magnetic field does not change. Why? Explain.

Solution

The magnetic force on a moving charge is always perpendicular to its velocity, \vec F=q(\vec v\times\vec B). A force perpendicular to the displacement does no work, because the work done is W=\vec F\cdot\vec d and the displacement is along the velocity, so \vec F\cdot\vec d=0.

Since no work is done, the change in kinetic energy is zero, \Delta K=W=0, so \tfrac12mv^2 stays constant. The speed does not change. Only the direction of the velocity, and hence the momentum, changes, which is why the particle moves in a circle.

83 marks

Describe the changes in the magnetic field inside a solenoid carrying a constant current I under the conditions of (a) doubling the length of the solenoid while maintaining the same number of turns, and (b) doubling the number of turns while maintaining the same length.

Solution

Inside a long solenoid the field is B=\mu_0 nI, where n=N/L is the number of turns per unit length. The field depends on N/L only, so changing N or L changes n.

**(a) Length doubled, N the same:** n=N/L is halved, so B is halved: B'=\dfrac{B}{2}.
(b) Turns doubled, length the same: n=N/L is doubled, so B is doubled: B'=2B.

In both cases the field stays uniform and its direction is unchanged, since the current is the same.

93 marks

Explain how 1 ampere is defined using concept of force between two long parallel current-carrying wires.

Solution

Two long parallel wires carrying currents I_1 and I_2 a distance r apart attract (same direction) or repel (opposite directions). The force per unit length on each wire is, from the book's equation for the force between parallel conductors,

\frac{F}{L}=\frac{\mu_0I_1I_2}{2\pi r}

Take I_1=I_2=I and r=1\,\mathrm{m}, with \mu_0=4\pi\times10^{-7}\,\mathrm{Wb\,A^{-1}\,m^{-1}}. One ampere is the steady current which, flowing in each of two long straight parallel wires placed 1\,\mathrm{m} apart in vacuum, produces a force of exactly 2\times10^{-7}\,\mathrm{N} per metre of each wire. Substituting I=1\,\mathrm{A} gives \dfrac{F}{L}=\dfrac{4\pi\times10^{-7}\times1\times1}{2\pi\times1}=2\times10^{-7}\,\mathrm{N\,m^{-1}}.

103 marks

A circular loop of radius R carrying current I lies in the XY-plane with its centre at origin (as shown in figure). What is the magnetic flux through the XY-plane?

Solution

The flux through the XY-plane is zero. The loop's field lines pass up through the inside of the loop, so they cross the XY-plane inside the loop, all in the same direction along z.

Magnetic field lines always form closed loops, so every line that goes up through the loop must return downwards through the plane outside the loop. The outward and inward contributions cancel exactly, and the net flux through the whole plane is \Phi_B=0.

113 marks

A suspended magnet is vibrating freely in a horizontal plane. When a metal plate is brought beneath the magnet, the oscillations of the magnet are significantly dampened. Describe the cause of damping by using Lenz's law.

Solution

As the magnet swings, its flux through the metal plate changes, so currents are induced in the plate. By Lenz's law, these induced currents flow in a direction that opposes the change in flux that produces them.

The induced currents therefore set up a field that opposes the motion of the magnet. The magnet is pulled back against its motion, so each swing loses energy, and the oscillations die out quickly. The energy lost by the magnet is converted into electrical energy in the plate and then into heat.

123 marks

A thin metallic ring is dropped into a vertical bar magnet. Does current in the ring flow clockwise or counterclockwise when seen from above?

Solution

The figure is not given, so take the north pole of the magnet to be its upper end. As the ring falls towards this north pole, the field lines of the magnet pass upwards through the ring and become denser, so the upward flux through the ring increases.

By Lenz's law the induced current must create a field that opposes this increase, so the induced field inside the ring points downwards. A current that gives a downward field inside the ring flows clockwise when seen from above (right-hand rule). If the south pole were at the top, the flux would be downward and the current would be counterclockwise.

Note. The book gives no figure for this question, so the answer assumes the north pole is at the top of the magnet. With the south pole at the top, the sense is reversed.

132026 Annual I, Set B, Q.2 (iii) OR3 marks

Show that \varepsilon and \dfrac{\Delta\phi}{\Delta t} have the same units.

Solution

By Faraday's law, the magnitude of the induced emf is \varepsilon=N\dfrac{\Delta\phi}{\Delta t}, with N a pure number, so the units of \varepsilon are those of \dfrac{\Delta\phi}{\Delta t}.

\left[\frac{\Delta\phi}{\Delta t}\right]=\frac{\text{weber}}{\text{second}}=\mathrm{Wb\,s^{-1}}

The weber is 1\,\mathrm{Wb}=1\,\mathrm{N\,m\,A^{-1}}. Since 1\,\mathrm{A}=1\,\mathrm{C\,s^{-1}} and 1\,\mathrm{N\,m\,C^{-1}}=1\,\mathrm{V}, we have 1\,\mathrm{Wb\,s^{-1}}=\dfrac{\mathrm{N\,m}}{\mathrm{A\,s}}=\dfrac{\mathrm{N\,m}}{\mathrm{C}}=1\,\mathrm{V}. So \varepsilon and \Delta\phi/\Delta t both have the unit of volt, and they have the same units.

Comprehensive questionsp. 243–244

Explain that magnetic field is an example of a field of force produced either by current-carrying conductors or by permanent magnets.

Solution

Field of force

A magnetic field is the region or space around a magnet in which a compass needle, a small magnet, a piece of ferromagnetic material, or a moving charged particle experiences a magnetic force. The force acts without contact, just as gravitational and electric forces do. Michael Faraday explained such forces by saying that each magnet sets up a field in the space around it. When a second magnet is placed in that field, it responds to the field at its own position, and this interaction of fields produces the force between them.

Sources of the field

Permanent magnet. A bar magnet, or a horseshoe magnet, sets up a field outside it from the north pole to the south pole. Iron filings sprinkled near a magnet show the pattern (Fig. 12.1).
Current-carrying conductor. Oersted showed in 1820 that a magnetic needle turns when a current passes through a nearby wire. A long straight wire carrying a current I produces circular field lines around it. The field strength at distance r is B=\dfrac{\mu_0 I}{2\pi r}, with \mu_0=4\pi\times10^{-7}\,\mathrm{Wb\,A^{-1}\,m^{-1}}. The direction is given by the right-hand rule: with the thumb along the current, the fingers curl in the direction of \vec B.
Other sources. A moving charge, which is equivalent to a current, and a fluctuating electric field also create a magnetic field.

Figure 12.13: The field lines of a straight wire carrying current out of the page are circles around the wire. The field is strongest close to the wire.

Why the field is a field of force

The field of a current or a magnet exerts a force on other currents, magnets, compass needles, magnetic materials and moving charges placed in it. A compass needle aligns itself along the field, and a current-carrying conductor placed in the field of a magnet is pushed (Fig. 12.4). The force is always a result of the field at the position of the object, which is what makes the magnetic field a field of force.

Properties of the field lines

Field lines never cross each other.
The density of field lines indicates the strength of the field.
Field lines always form closed loops. Outside the magnet they go from north to south, and inside they go from south to north. This is why no isolated magnetic poles (monopoles) exist.
Like poles repel and unlike poles attract.

Explain that a magnetic force acts on a current-carrying conductor placed in a magnetic field. Derive an expression for it.

Solution

The magnetic force on a conductor

When a conductor carrying a current is placed in a magnetic field, the field of the current interacts with the external field. The net field is stronger on one side of the wire and weaker on the other, and the conductor is pushed from the stronger region towards the weaker region. Figure 12.4 of the book shows this for a wire in the field of a bar magnet. The conductor moves when it is free to do so, which shows that a force acts on it.

Derivation of F=BIL\sin\theta

Derivation Force on a straight conductor in a uniform field

Aim. Find the magnetic force F on a straight conductor of length L carrying a current I in a uniform magnetic field B, when the conductor makes an angle \theta with the field.

Assumptions.

The field B is uniform over the length of the conductor.
The conductor is straight and its length L lies entirely in the field.

Figure 12.5: A conductor carrying a current to the right, at right angles to a field into the page. The force on it is upward, perpendicular to both the wire and the field.

Experiments on conductors on rails show that the force is directly proportional to the current in the conductor.
F\propto I
The force is directly proportional to the length of the conductor in the field.
F\propto L
The force is directly proportional to the strength of the external field.
F\propto B
The force depends on the angle \theta between the conductor and the field. It is proportional to \sin\theta, which is zero for \theta=0^\circ and 180^\circ and largest for \theta=90^\circ.
F\propto\sin\theta
Combine the four proportionalities into one relation, with a constant k of proportionality.
F=k\,ILB\sin\theta
In SI units the constant has the value k=1. This sets the unit of B: for \theta=90^\circ and L=1\,\mathrm{m}, I=1\,\mathrm{A}, the field is 1\,\mathrm{T} when the force is 1\,\mathrm{N}.
k=1

ResultF=BIL\sin\theta

In vector form \vec F=I(\vec L\times\vec B). The force is perpendicular to both the conductor and the field. It is maximum at \theta=90^\circ (F=BIL) and zero at \theta=0^\circ or 180^\circ.

Direction of the force

Fleming's left-hand rule gives the direction. Stretch the thumb, forefinger and middle finger of the left hand mutually at right angles. The forefinger points along the field B and the middle finger along the current I. The thumb then points along the force on the conductor (Fig. 12.6). In diagrams a dot shows a vector pointing out of the plane, and a cross shows one pointing into the plane.

32026 Annual I, Set B, Q.4 OR

Explain that a magnetic force acts on a charged particle in a uniform magnetic field.

Solution

The force on a moving charge

A charge at rest feels no magnetic force. A charge moving with velocity \vec v in a magnetic field \vec B feels a force \vec F=q(\vec v\times\vec B), which is perpendicular to both \vec v and \vec B (equation 12.2). A negative charge feels the force in the opposite direction (equation 12.3).

Derivation of F=qvB\sin\theta

Derivation Force on a moving charge

Aim. Find the magnetic force on a charge q moving with speed v at an angle \theta to a uniform field B, by treating the moving charge as a current.

Assumptions.

A charge q passes a fixed point in a time t, so it forms a current I=q/t in the conductor it represents.
In time t the charge travels a distance L=vt in the field, at the angle \theta.

For a positive charge moving to the right in a field into the page, the force is upward, perpendicular to both \vec v and \vec B (right-hand rule II).

A moving charge is equivalent to a current I=\dfrac{q}{t} over the length L=vt it travels in the field.
I=\frac{q}{t},\qquad L=vt
Use the force on a current-carrying conductor, F=BIL\sin\theta, from the previous part.
F=BIL\sin\theta
Substitute the current and the length.
F=B\left(\frac{q}{t}\right)(vt)\sin\theta
The time t cancels, so the force does not depend on how long the charge takes to cross the field.
F=qvB\sin\theta

ResultF=qvB\sin\theta

The force is largest when the charge moves at 90^\circ to the field (F=qvB). It is zero when the charge moves parallel or antiparallel to the field, and then it goes in a straight line.

Direction of the force

Use the right-hand rule II, or Fleming's left-hand rule with the current replaced by the velocity of the positive charge. Curl the fingers of the right hand from \vec v towards \vec B through the smaller angle; the thumb then points along the force on a positive charge. Figure 12.10 of the book shows a positive charge deflected outwards, and a negative charge deflected in the opposite direction.

Effect on the motion

Because the force is always perpendicular to the velocity, it changes only the direction of motion, not the speed. The particle moves in a circle when \theta=90^\circ and in a helix when the angle is between 0^\circ and 90^\circ. The circular motion is treated in Question 5.

Describe the concept of magnetic flux (\Phi_B) as scalar product of magnetic field (B) and area (A) using the relation \Phi_B=B\cdot A.

Solution

Definition

The magnetic flux through a surface is the number of magnetic field lines passing through that area. It is the scalar product of the uniform magnetic field \vec B and the vector area \vec A, where \vec A is a vector of magnitude equal to the area and direction along the normal to the surface:

\Phi_B=\vec B\cdot\vec A

Magnitude

If \theta is the angle between \vec B and \vec A (Fig. 12.20 a), then

\Phi_B=BA\cos\theta

Figure 12.20 (a): The magnetic flux through a plane depends on the angle \theta between \vec B and the area vector \vec A, which is the normal to the plane.

Special cases

Maximum flux: when the plane is held perpendicular to the field lines, so \vec A is parallel to \vec B (\theta=0^\circ): \Phi_B=BA.
Zero flux: when the plane is parallel to the field lines, so \vec A is perpendicular to \vec B (\theta=90^\circ): \Phi_B=0.

Curved surface and non-uniform field

For a curved surface in a non-uniform field, divide the surface into small elements \Delta\vec A. The total flux is the sum of the fluxes through the elements: \Phi_B=\sum\vec B\cdot\Delta\vec A.

Units and flux density

The SI unit of magnetic flux is the weber, with 1\,\mathrm{Wb}=1\,\mathrm{N\,m\,A^{-1}}. The magnetic flux density is the flux per unit area, B=\dfrac{\Phi_B}{A}, so the unit of B is \mathrm{Wb\,m^{-2}}, which is the tesla. Flux is a scalar, while B is a vector.

Describe quantitatively the path followed by a charged particle shot into a magnetic field in a direction perpendicular to the field.

Solution

Qualitative path

When a charged particle enters a uniform magnetic field with its velocity perpendicular to the field, the magnetic force is always at right angles to the velocity. The force therefore acts as a centripetal force that changes the direction of the velocity without changing its speed. The particle moves in a circle in the plane perpendicular to \vec B.

Radius of the path

Derivation Radius of the circular path

Aim. Find the radius r of the circle described by a charge q of mass m moving with speed v perpendicular to a uniform field B.

Assumptions.

The velocity is perpendicular to \vec B, so \theta=90^\circ.
Only the magnetic force acts, and it provides the centripetal force.

Figure 12.11: The magnetic force on a positive charge acts towards the centre and moves the particle in a circle of radius r.

The magnetic force on the charge is maximum for \theta=90^\circ.
F_B=qvB
For a particle of mass m moving in a circle of radius r, the centripetal force is F_c=\dfrac{mv^2}{r}.
F_c=\frac{mv^2}{r}
The magnetic force provides the centripetal force.
\frac{mv^2}{r}=qvB
Cancel one factor v and rearrange for r.
r=\frac{mv}{qB}

Resultr=\dfrac{mv}{qB}

The radius is proportional to the momentum mv and inversely proportional to qB. A faster particle moves on a larger circle, and a stronger field gives a smaller circle.

Angular velocity, period and frequency

Substitute v=r\omega into \dfrac{mv}{r}=qB to get m\omega=qB, so the angular velocity is

\omega=\frac{qB}{m}

This is the cyclotron frequency in angular form. The period and the frequency are

T=\frac{2\pi}{\omega}=\frac{2\pi m}{qB},\qquad f=\frac{qB}{2\pi m}

62025 Annual I, Set D, Q.6

Describe the concept of magnetic force between current-carrying conductors. Derive an expression for the magnetic force acting per unit length on each current-carrying parallel conductor when current is flowing through them.

Solution

The concept

Currents in the same direction: the field between the wires is weakened and the field on the outer sides is stronger, so the wires are pulled towards each other. The force is attractive. A neutral point, where the two fields cancel, lies between the wires.
Currents in opposite directions: the field between the wires is strengthened and the outer field is weaker, so the wires are pushed apart. The force is repulsive.

Derivation of the force per unit length

Derivation Force per unit length between two parallel wires

Aim. Find the force per unit length on each of two long parallel wires, carrying currents I_1 and I_2 and separated by a distance r.

Assumptions.

Both wires are long and straight, and of length L in the region of interest.
The wires are in vacuum (or a medium with permeability \mu_0), and r\gg the wire diameter.

Figure 12.19 (c): Two wires carrying currents in the same direction attract each other. F_{12} is the force on wire 2 due to wire 1, and F_{21} the force on wire 1 due to wire 2.

Wire 1 produces a field at the position of wire 2: B_1=\dfrac{\mu_0I_1}{2\pi r}.
B_1=\frac{\mu_0 I_1}{2\pi r}
Wire 2 is perpendicular to B_1, so the force on the length L of wire 2 is F_{12}=B_1I_2L\sin90^\circ.
F_{12}=B_1I_2L
Substitute B_1 into the force.
F_{12}=\frac{\mu_0I_1}{2\pi r}I_2L
Divide by L to get the force per unit length on wire 2.
\frac{F_{12}}{L}=\frac{\mu_0I_1I_2}{2\pi r}
By the same method, wire 2 produces B_2=\dfrac{\mu_0I_2}{2\pi r} at wire 1, so F_{21}=B_2I_1L and \dfrac{F_{21}}{L}=\dfrac{\mu_0I_1I_2}{2\pi r}.
\frac{F_{21}}{L}=\frac{\mu_0I_1I_2}{2\pi r}

Result\frac{F}{L}=\frac{\mu_0I_1I_2}{2\pi r}

The two forces are equal in magnitude and opposite in direction, so they form an action and reaction pair. The force is attractive when the currents are in the same direction and repulsive when they are opposite. This result defines the ampere (Question 9 of the short questions).

Neutral point

For currents in the same direction, the fields of the two wires are equal and opposite at one point between them. This neutral point is midway if I_1=I_2, and nearer to the smaller current if the currents differ.

72025 Annual I, Set B, Q.2 (vii) OR

State and explain Faraday's law of electromagnetic induction.

Solution

Statement

Faraday's law: the average emf induced in a conducting coil of N turns is directly proportional to the rate of change of magnetic flux through the coil.

\varepsilon=N\frac{\Delta\Phi}{\Delta t}

Here \varepsilon is the magnitude of the induced emf, N the number of turns and \Delta\Phi/\Delta t the rate of change of the flux. The constant of proportionality is 1 in SI units. The negative sign in \varepsilon=-N\dfrac{\Delta\Phi}{\Delta t} shows the direction of the induced current, which is given by Lenz's law (Question 8).

Explanation

An emf is induced whenever the magnetic flux linking a conductor changes. A constant flux induces nothing, so the change is the cause of induction. The induced emf is independent of the resistance of the circuit, while the induced current I=\varepsilon/R depends on it.

Experiments that show the law

Experiment 1: a bar magnet is moved towards or away from a coil connected to a galvanometer. There is no current when the magnet is at rest. Moving it towards the coil gives a current in one direction, and moving it away gives a current in the opposite direction (Fig. 12.22).
Experiment 2: a coil is rotated in a magnetic field. The angle between \vec B and the area of the coil changes, so the flux changes by \Delta\Phi=\Delta(BA\cos\theta) and an emf is produced (Fig. 12.23). This is the principle of the electric generator.
Experiment 3: a changing current in a primary coil changes the flux through a nearby secondary coil, and an emf is induced in the secondary. This is the principle of the transformer (Fig. 12.24).

Factors affecting the induced emf

the number of turns N in the coil;
the rate of change of flux, which depends on the speed of the motion or of the change in field;
the strength of the field and the area of the coil.

Example 12.7 (from the book)

A coil of resistance 50\,\Omega is placed in a magnetic field, and its initial flux is 2\,\mathrm{mWb}. The coil has 50 turns and a galvanometer of 100\,\Omega resistance is connected in series with it. The coil is moved in \dfrac{1}{10}\,\mathrm{s} to a field where the final flux is 4\,\mathrm{mWb}. Find the average emf and the current.

Apply Faraday's law with N=50, \Phi_1=2\times10^{-3}\,\mathrm{Wb}, \Phi_2=4\times10^{-3}\,\mathrm{Wb} and \Delta t=0.1\,\mathrm{s}.
\varepsilon=N\frac{\Phi_2-\Phi_1}{\Delta t}=50\times\frac{(4-2)\times10^{-3}}{0.1}=1\,\mathrm{V}
The galvanometer and the coil are in series, so the total resistance is R=50\,\Omega+100\,\Omega=150\,\Omega.
R=150\,\Omega
The current is I=\dfrac{\varepsilon}{R}.
I=\frac{1}{150}=6.67\times10^{-3}\,\mathrm{A}

\varepsilon=1\,\mathrm{V},\qquad I=6.67\,\mathrm{mA}

Describe how the direction of an induced current can be predicted by Lenz's law and how Lenz's law obeys the principle of conservation of energy.

Solution

Lenz's law

Lenz's law: the direction of the induced current is always such that it opposes the change that produces it. The induced current sets up its own field, and this field opposes the change in the flux through the circuit.

Predicting the direction

Decide whether the flux through the circuit is increasing or decreasing.
Find the direction of the induced field that would oppose this change: opposite to the field if the flux increases, the same as the field if it decreases.
Use the right-hand rule for a coil: curl the fingers in the direction of the induced current, and the thumb points out of the coil's north pole face, in the direction of the induced field.

Consider a north pole moved towards a coil (Fig. 12.25 a). The flux through the coil increases. The induced field must oppose this, so the face of the coil towards the magnet becomes a north pole, which repels the magnet. Viewed from the magnet's side, the current in the coil is anticlockwise.

Figure 12.25: A north pole approaches a coil. The coil face towards the magnet becomes a north pole, so the induced field opposes the approach.

If the magnet is pulled away, the flux decreases, and the coil face towards the magnet becomes a south pole, which attracts the magnet and opposes the pull. Fleming's right-hand rule gives the same result for a conductor moving across a field: with the thumb in the direction of the motion and the forefinger in the direction of the field, the middle finger gives the direction of the induced current (Fig. 12.27).

Lenz's law and conservation of energy

To push the magnet towards the coil we must do work against the repulsive force that the induced current exerts. This mechanical energy is converted into electrical energy in the circuit, and then into heat in the resistance of the coil. Energy is conserved.

If Lenz's law were reversed, the induced current would help the approach. The magnet would then speed up, induce a larger current and speed up further, creating energy from nothing. Lenz's law prevents this and is therefore a statement of the conservation of energy.

Numerical problemsp. 244–245

A horizontal power line carries a current of 600\,\mathrm{A} from south to north. Earth's magnetic field 5\times10^{-5}\,\mathrm{T} is directed toward the north and inclined downward at 70^\circ to the horizontal. Find the magnitude of the magnetic force on 100\,\mathrm{m} of the line due to Earth's magnetic field.

Solution

Given

I = 600\,\mathrm{A}
L = 100\,\mathrm{m}
B = 5\times10^{-5}\,\mathrm{T}
\theta = 70^\circ angle between \vec B and the wire, which runs north

To find

The wire runs north, the same direction as the horizontal part of \vec B. That part is parallel to the wire and gives no force. The angle between \vec B and the wire is \theta=70^\circ, so use F=BIL\sin\theta.
F=BIL\sin\theta
Substitute the values.
F=(5\times10^{-5})(600)(100)\sin70^\circ=3.0\times0.940
Evaluate.
F=2.82\,\mathrm{N}

F=2.82\,\mathrm{N}

Book answer. 2.82\,\mathrm{N}

A proton is accelerated in a cyclotron in which the magnetic induction is 0.75\,\mathrm{Wb\,m^{-2}}. Find the cyclotron frequency.

Solution

Given

B = 0.75\,\mathrm{Wb\,m^{-2}}
q = 1.6\times10^{-19}\,\mathrm{C}
m_p = 1.67\times10^{-27}\,\mathrm{kg}

To find

The cyclotron frequency of a charged particle in a magnetic field is f=\dfrac{qB}{2\pi m}.
f=\frac{qB}{2\pi m}
Substitute for a proton.
f=\frac{(1.6\times10^{-19})(0.75)}{2\pi(1.67\times10^{-27})}
Evaluate.
f=1.14\times10^{7}\,\mathrm{Hz}

f\approx1.14\times10^{7}\,\mathrm{Hz}\;(11.4\,\mathrm{MHz})

Book answer. 11.4\times10^{6}\,\mathrm{Hz}

A velocity selector consists of electric and magnetic fields described by the expressions \vec E=E\,\hat k and \vec B=B\,\hat j. With B=18.0\,\mathrm{mT}, find the value of E such that electron of energy 850\,\mathrm{eV} moving in the negative x direction is undeflected.

Solution

Given

B = 18.0\,\mathrm{mT}=1.80\times10^{-2}\,\mathrm{T}
K = 850\,\mathrm{eV}
m_e = 9.11\times10^{-31}\,\mathrm{kg}
e = 1.6\times10^{-19}\,\mathrm{C}

To find

E
\text{direction of } \vec E

Convert the kinetic energy to joules: K=850\times1.6\times10^{-19}\,\mathrm{J}=1.36\times10^{-16}\,\mathrm{J}.
K=1.36\times10^{-16}\,\mathrm{J}
Find the speed from K=\tfrac12mv^2.
v=\sqrt{\frac{2K}{m_e}}=\sqrt{\frac{2(1.36\times10^{-16})}{9.11\times10^{-31}}}=1.73\times10^{7}\,\mathrm{m\,s^{-1}}
The electron is undeflected when the electric force balances the magnetic force, eE=evB, so E=vB.
E=vB
Substitute. The magnetic force on the electron points along +\hat k, so the electric field must also point along +\hat k (the electric force on the electron is opposite to \vec E).
E=(1.73\times10^{7})(1.80\times10^{-2})=3.11\times10^{5}\,\mathrm{N\,C^{-1}}

E=3.11\times10^{5}\,\mathrm{N\,C^{-1}}\ \text{along } +\hat k

Book answer. 311\times10^{3}\,\mathrm{N\,C^{-1}}

Two long straight parallel wires are 0.6\,\mathrm{m} apart and carry current 4\,\mathrm{A} and 8\,\mathrm{A}. Find the point lying between two wires at perpendicular distance where resultant magnetic field will be zero (neutral point).

Solution

Given

d = 0.6\,\mathrm{m}
I_1 = 4\,\mathrm{A}
I_2 = 8\,\mathrm{A} currents in the same direction, so the fields between the wires oppose each other

To find

x
\text{distance from the 4 A wire}

Let x be the distance from the 4 A wire, so 0.6-x is the distance from the 8 A wire. The fields are equal and opposite at the neutral point, so B_1=B_2.
\frac{\mu_0 I_1}{2\pi x}=\frac{\mu_0 I_2}{2\pi(0.6-x)}
Cancel \mu_0/2\pi and substitute the currents.
\frac{4}{x}=\frac{8}{0.6-x}
Cross-multiply.
4(0.6-x)=8x
Solve for x.
2.4=12x\quad\Rightarrow\quad x=0.2\,\mathrm{m}
The distance from the 8 A wire is 0.6-0.2=0.4\,\mathrm{m}.
0.6-x=0.4\,\mathrm{m}

0.2\,\mathrm{m}\ \text{from the 4 A wire, and}\ 0.4\,\mathrm{m}\ \text{from the 8 A wire}

Book answer. 0.2\,\mathrm{m} and 0.4\,\mathrm{m} respectively

Note. The neutral point lies between the wires only when the currents flow in the same direction. The point is nearer the smaller current.

A coil of wire has 100 loops. Each loop has an area of 2.0\times10^{-3}\,\mathrm{m^2}. A magnetic field is perpendicular to the surface of each loop at all times. If the magnetic field is changed from 0.2\,\mathrm{T} to 0.6\,\mathrm{T} in 0.1\,\mathrm{s}, find the average emf induced in the coil during this time.

Solution

Given

N = 100
A = 2.0\times10^{-3}\,\mathrm{m^2}
B_1 = 0.2\,\mathrm{T}
B_2 = 0.6\,\mathrm{T}
\Delta t = 0.1\,\mathrm{s}

To find

\varepsilon
\text{average emf}

The field is perpendicular to each loop, so the flux through one loop is \Phi=BA and its change is \Delta\Phi=A\,\Delta B.
\Delta\Phi=A(B_2-B_1)
Faraday's law for N loops gives the average emf.
\varepsilon=N\frac{\Delta\Phi}{\Delta t}=N\frac{A(B_2-B_1)}{\Delta t}
Substitute the values.
\varepsilon=100\times\frac{(2.0\times10^{-3})(0.6-0.2)}{0.1}
Evaluate.
\varepsilon=0.8\,\mathrm{V}

\varepsilon=0.8\,\mathrm{V}

Book answer. 0.8\,\mathrm{V}

A 2\,\mathrm{m} long wire carrying a current of 15\,\mathrm{A} is placed in a uniform magnetic field of 0.50\,\mathrm{T}. If the wire makes an angle of 60^\circ with the direction of magnetic field, find the magnitude of the magnetic force acting on the wire.

Solution

Given

L = 2\,\mathrm{m}
I = 15\,\mathrm{A}
B = 0.50\,\mathrm{T}
\theta = 60^\circ

To find

Use the force on a current-carrying wire at an angle to the field.
F=BIL\sin\theta
Substitute the values.
F=(0.50)(15)(2)\sin60^\circ=15\times0.866
Evaluate.
F=12.99\,\mathrm{N}

F=12.99\,\mathrm{N}

Book answer. 12.99\,\mathrm{N}

A particle carrying a charge of 1\,\mu\mathrm{C} and moving with velocity of 2\times10^{6}\,\mathrm{m\,s^{-1}} enters in a magnetic field of 0.5\,\mathrm{T} at angle of 60^\circ with the field. Calculate the magnitude of magnetic force acting on it.

Solution

Given

q = 1\,\mu\mathrm{C}=1\times10^{-6}\,\mathrm{C}
v = 2\times10^{6}\,\mathrm{m\,s^{-1}}
B = 0.5\,\mathrm{T}
\theta = 60^\circ

To find

Use the magnetic force on a moving charge.
F=qvB\sin\theta
Substitute the values.
F=(1\times10^{-6})(2\times10^{6})(0.5)\sin60^\circ
Evaluate.
F=1\times0.5\times0.866=0.866\,\mathrm{N}

F=0.866\,\mathrm{N}

Book answer. 0.866\,\mathrm{N}

A single circular loop has radius 2\,\mathrm{cm}. The plane of the loop lies at 40^\circ to a uniform magnetic field of 0.2\,\mathrm{T}. Find the magnitude of the magnetic flux through the loop.

Solution

Given

r = 2\,\mathrm{cm}=0.02\,\mathrm{m}
B = 0.2\,\mathrm{T}
\alpha = 40^\circ angle between the plane of the loop and \vec B

To find

\Phi_B

The area of the loop is A=\pi r^2.
A=\pi(0.02)^2=1.257\times10^{-3}\,\mathrm{m^2}
The flux depends on the angle between \vec B and the area vector, which is the normal to the plane. The normal makes 90^\circ-40^\circ=50^\circ with the field.
\theta=90^\circ-40^\circ=50^\circ
Use \Phi_B=BA\cos\theta.
\Phi_B=(0.2)(1.257\times10^{-3})\cos50^\circ
Evaluate.
\Phi_B=1.62\times10^{-4}\,\mathrm{Wb}

\Phi_B=1.62\times10^{-4}\,\mathrm{Wb}

Book answer. 1.62\times10^{-4}\,\mathrm{Wb}

Note. The angle used in the flux formula is between \vec B and the normal (50^\circ), not the 40^\circ given for the plane. Using 40^\circ directly would give 1.93\times10^{-4}\,\mathrm{Wb}.

Find the magnetic force acting on a charged particle of charge 2\,\mu\mathrm{C} in a magnetic field of 2\,\mathrm{T} acting in y-direction when particle velocity is (2\hat i+3\hat j)\times10^{6}\,\mathrm{m\,s^{-1}}.

Solution

Given

q = 2\,\mu\mathrm{C}=2\times10^{-6}\,\mathrm{C}
\vec B = 2\,\hat j\,\mathrm{T}
\vec v = (2\hat i+3\hat j)\times10^{6}\,\mathrm{m\,s^{-1}}

To find

\vec F

Use \vec F=q(\vec v\times\vec B). Only the \hat i component of \vec v is not parallel to \hat j: the \hat j part of \vec v gives zero cross product with \hat j.
\vec v\times\vec B=(2\hat i+3\hat j)\times10^{6}\times2\hat j
Use \hat i\times\hat j=\hat k.
\vec v\times\vec B=(2\times2)\times10^{6}\,(\hat i\times\hat j)=4\times10^{6}\,\hat k
Multiply by the charge.
\vec F=(2\times10^{-6})(4\times10^{6})\hat k=8\,\hat k\,\mathrm{N}

\vec F=8\,\hat k\,\mathrm{N}\ \text{(magnitude } 8\,\mathrm{N}\text{, along } +z)

Book answer. 8\,\mathrm{N}, +\text{ve } z-axis

A coil with 25 turns of wire is wrapped on a frame with a square cross section 1.80\,\mathrm{cm} on a side. Each turn has the same area, equal to that of the frame, and the total resistance of the coil is 0.350\,\Omega. An applied uniform magnetic field is perpendicular to the plane of the coil, as in the figure. (a) Applying Faraday's law find the induced emf in the coil when the magnetic field is changing uniformly from zero to 0.50\,\mathrm{T} in 0.80\,\mathrm{s}. (b) Find the magnitude of induced current. (c) Applying Lenz's law find the direction of the induced current in the coil while the field is changing.

Solution

Given

N = 25
a = 1.80\,\mathrm{cm}=0.0180\,\mathrm{m} side of the square cross section
R = 0.350\,\Omega
\Delta B = 0.50\,\mathrm{T}-0\,\mathrm{T}
\Delta t = 0.80\,\mathrm{s}

To find

\varepsilon
I
\text{direction of } I

The field is perpendicular to the plane of the coil, so the flux through each turn is BA, with A=a^2.

(a) Find the induced emf in the coil when the field is changing uniformly from zero to 0.50\,\mathrm{T} in 0.80\,\mathrm{s}.

Area of one turn: A=a^2=(0.0180)^2=3.24\times10^{-4}\,\mathrm{m^2}.
A=3.24\times10^{-4}\,\mathrm{m^2}
Faraday's law for N turns with \Delta\Phi=A\,\Delta B per turn.
\varepsilon=N\frac{A\,\Delta B}{\Delta t}
Substitute the values.
\varepsilon=25\times\frac{(3.24\times10^{-4})(0.50)}{0.80}
Evaluate.
\varepsilon=5.06\times10^{-3}\,\mathrm{V}

\varepsilon=5.06\times10^{-3}\,\mathrm{V}

(b) Find the magnitude of the induced current.

The current is the emf divided by the total resistance: I=\varepsilon/R.
I=\frac{\varepsilon}{R}
Substitute the emf from part (a).
I=\frac{5.06\times10^{-3}}{0.350}
Evaluate.
I=1.45\times10^{-2}\,\mathrm{A}

I=1.45\times10^{-2}\,\mathrm{A}

The field points upward (+z) and is increasing, so the upward flux through the coil increases.
By Lenz's law the induced current must create a field that opposes the increase, so the induced field points downward (-z) inside the coil.
By the right-hand rule, a current that gives a downward field inside the coil flows clockwise when viewed from above.

\text{Clockwise, as seen from above}

\varepsilon=5.06\times10^{-3}\,\mathrm{V},\quad I=1.45\times10^{-2}\,\mathrm{A},\quad \text{clockwise}

Book answer. 5.06\times10^{-3}\,\mathrm{V}, 1.45\times10^{-2}\,\mathrm{A}, clockwise

FBISE 2025 and 2026 questions

Questions from the FBISE HSSC-I 2026 (1st Annual) and FBISE HSSC-I 2025 (1st Annual) papers that belong to this unit, answered the way an examiner expects.

2026 Annual I, Set B, MCQ 101 mark

What is the motion of an electron moving at an angle of 45^{\circ} to a uniform magnetic field?

ALinear motion
BCircular motion
CParabolic motion
DHelical motionAnswer

Why

Resolve the velocity into a component v\cos45^\circ along \vec B and a component v\sin45^\circ perpendicular to it. The magnetic force is zero along \vec B, so the parallel component stays constant and carries the electron forward. The perpendicular component makes it move in a circle. The two together give a spiral path, a helix.

Circular motion (B) occurs only when the velocity is exactly perpendicular to the field (90^\circ), and straight-line motion (A) only when it is parallel or antiparallel (0^\circ or 180^\circ). A magnetic force never gives a parabolic path (C).

2025 Annual I, Set B, MCQ 131 mark

Two closely placed parallel wires are carrying current in opposite direction, the wires will:

ARepel each otherAnswer
BAttract each other
CNot affect each other
DRotate each other

Why

Each wire lies in the field of the other. Between two parallel wires carrying currents in opposite directions the two fields point the same way, so the field between them is strengthened and the wires are pushed apart: they repel. The force per unit length is \dfrac{\mu_0I_1I_2}{2\pi r}.

Option B (attract) is the result for currents in the same direction. Option C is wrong because the force is not zero, and D is wrong because the forces lie along the line joining the wires and give no turning.

2025 Annual I, Set B, MCQ 141 mark

One weber is equal to:

A\mathrm{NA^{-1}}
B\mathrm{Nm\,A^{-1}}Answer
C\mathrm{Nm^{-1}A}
D\mathrm{Nm^{-1}A^{-1}}

Why

Magnetic flux is \Phi=BA, and B=\dfrac{F}{IL} has the unit \mathrm{N\,A^{-1}\,m^{-1}}. Multiplying by an area in \mathrm{m^{2}} gives 1\,\mathrm{Wb}=1\,\mathrm{N\,A^{-1}\,m^{-1}}\times\mathrm{m^{2}}=1\,\mathrm{N\,m\,A^{-1}}, option B.

Option A (\mathrm{N\,A^{-1}}) has lost the metre, so it is not a flux unit. Options C and D do not follow from \Phi=BA: C has \mathrm{A} in the numerator and D has \mathrm{m^{-1}}, where a flux needs \mathrm{m^{+1}}. D is close to the unit of the tesla, \mathrm{N\,A^{-1}\,m^{-1}}, which is a flux density, not a flux.

2026 Annual I, Set B, Q.2 (iii) OR3 marks

Show that \varepsilon and \dfrac{\Delta\phi}{\Delta t} have the same units.

Solution

By Faraday's law the magnitude of the induced emf is \varepsilon=N\dfrac{\Delta\phi}{\Delta t}, where N is a pure number. So \varepsilon has the same units as \dfrac{\Delta\phi}{\Delta t}, and we show that the units of \Delta\phi/\Delta t reduce to the volt.

\left[\dfrac{\Delta\phi}{\Delta t}\right]=\dfrac{\mathrm{weber}}{\mathrm{second}}=\dfrac{\mathrm{N\,m\,A^{-1}}}{\mathrm{s}}=\dfrac{\mathrm{N\,m}}{\mathrm{A\,s}}=\dfrac{\mathrm{N\,m}}{\mathrm{C}}=\dfrac{\mathrm{J}}{\mathrm{C}}=\mathrm{V}

Here 1\,\mathrm{Wb}=1\,\mathrm{N\,m\,A^{-1}}, 1\,\mathrm{A\,s}=1\,\mathrm{C} and 1\,\mathrm{J\,C^{-1}}=1\,\mathrm{V}. Both \varepsilon and \Delta\phi/\Delta t are measured in volts, so they have the same units.

2025 Annual I, Set B, Q.2 (vii) OR2+1 = 3 marks

State Faraday's law of electromagnetic induction. Give its mathematical form.

Solution

Statement (2 marks). The average emf induced in a conducting coil of N turns is directly proportional to the rate of change of magnetic flux through the coil.

Mathematical form (1 mark).

\varepsilon=N\dfrac{\Delta\Phi}{\Delta t}

Here \varepsilon is the magnitude of the induced emf, N the number of turns and \Delta\Phi/\Delta t the rate of change of flux. With the sign, \varepsilon=-N\dfrac{\Delta\Phi}{\Delta t}, the minus sign shows that the direction of the induced current is given by Lenz's law.

2025 Annual I, Set B, Q.2 (viii)1.5+1.5 = 3 marks

Does the induced emf in a circuit depend on the resistance of the circuit? Does the induced current depend on the resistance of the circuit? Justify.

Solution

Induced emf (1.5 marks). No. By Faraday's law \varepsilon=N\dfrac{\Delta\Phi}{\Delta t}, so the emf depends only on the number of turns and the rate of change of flux. The resistance of the circuit does not appear.

Induced current (1.5 marks). Yes. The induced current in the circuit is found from Ohm's law:

I=\dfrac{\varepsilon}{R}

For the same induced emf, a larger resistance gives a smaller current. For example, if the circuit is open (R\to\infty) an emf is still induced but no current flows.

2025 Annual I, Set B, Q.2 (ix)3 marks

The Kinetic energy of a charged particle moving in a uniform magnetic field does not change. Why? Explain briefly.

Solution

The magnetic force on a moving charge is always perpendicular to its velocity, \vec F=q(\vec v\times\vec B). A force perpendicular to the displacement does no work, because W=\vec F\cdot\vec d=0.

Since no work is done, the change in kinetic energy is zero, \Delta K=W=0, so \tfrac12mv^{2} stays constant and the speed does not change. Only the direction of the velocity changes, which is why the particle moves in a circle.

2026 Annual I, Set B, Q.4 OR4+2 = 6 marks

Derive an expression for a magnetic force experienced by a moving charge in a uniform magnetic field. Also discuss the direction of that force.

Solution

Expression for the force (4 marks)

A charge at rest feels no magnetic force. A charge q moving with velocity \vec v in a uniform field \vec B feels a force perpendicular to both. It is found by treating the moving charge as a current in a conductor.

Derivation Force on a moving charge

Aim. Find the magnetic force on a charge q moving with speed v at an angle \theta to a uniform field B, by treating the moving charge as a current.

Assumptions.

A charge q passes a fixed point in a time t, so it forms a current I=q/t in the conductor it represents.
In time t the charge travels a distance L=vt in the field, at the angle \theta.

For a positive charge moving to the right in a field into the page, the force is upward, perpendicular to both \vec v and \vec B (right-hand rule II).

A moving charge is equivalent to a current I=\dfrac{q}{t} over the length L=vt it travels in the field.
I=\frac{q}{t},\qquad L=vt
Use the force on a current-carrying conductor, F=BIL\sin\theta, from the previous part.
F=BIL\sin\theta
Substitute the current and the length.
F=B\left(\frac{q}{t}\right)(vt)\sin\theta
The time t cancels, so the force does not depend on how long the charge takes to cross the field.
F=qvB\sin\theta

ResultF=qvB\sin\theta

The force is largest when the charge moves at 90^\circ to the field (F=qvB). It is zero when the charge moves parallel or antiparallel to the field, and then it goes in a straight line.

Direction of the force (2 marks)

The force \vec F=q(\vec v\times\vec B) is perpendicular to the plane containing \vec v and \vec B. Use the right-hand rule: curl the fingers of the right hand from \vec v towards \vec B through the smaller angle, and the thumb points along the force on a positive charge. (Equivalently, use Fleming's left-hand rule with the current replaced by the velocity of the positive charge.) For a negative charge the force is in the opposite direction.

The force is largest when \theta=90^\circ, F=qvB.
It is zero when \vec v is parallel or antiparallel to \vec B (\theta=0^\circ or 180^\circ).
Because \vec F\perp\vec v, the force changes only the direction of the motion, not the speed: the particle moves in a circle for \theta=90^\circ and in a helix for other angles.

2026 Annual I, Set D, MCQ 111 mark

Path of charged particle in magnetic field is always:

ARondon
BStraight
CParabolic
DCircularAnswer

Why

The magnetic force \vec F=q(\vec v\times\vec B) is always perpendicular to the velocity. It therefore changes only the direction of the velocity, never the speed, and acts as a centripetal force: qvB=\dfrac{mv^{2}}{r} gives the circle of radius r=\dfrac{mv}{qB}. The answer is D.

The path is straight (B) only if the particle moves along the field, where the force is zero. It is never parabolic (C), because a force of constant magnitude perpendicular to the velocity does not give a parabola. A (random) is wrong because the path is fixed by v, B, q and m.

Note. Option A is printed 'Rondon' in the paper, an apparent misprint of 'Random'. The word 'always' is loose: the path is a circle when the velocity is perpendicular to the field, a straight line when it is parallel, and a helix otherwise. Of the four options, circular is the one the question intends.

2025 Annual I, Set D, MCQ 141 mark

The SI unit of magnetic induction is:

ANm
BGauss
CWeber
DTeslaAnswer

Why

The tesla is the SI unit of magnetic induction (flux density). It is defined from F=BIL\sin\theta: 1\,\mathrm{T}=1\,\mathrm{N\,A^{-1}\,m^{-1}}.

The gauss (B) is a unit of the same quantity, but in the CGS system, not the SI (1\,\mathrm{T}=10^{4}\,\mathrm{G}). The weber (C) is the unit of magnetic flux, and the newton metre (A) is the unit of work or torque.

2026 Annual I, Set D, Q.2 (iii) OR3 marks

How a seismometer uses electromagnetic induction to detect earthquakes? Explain briefly.

Solution

Construction. A seismometer has a coil wound on a frame that hangs from springs fixed to the casing. A permanent magnet is fixed to the casing, which rests on the ground, in such a way that the coil can oscillate freely relative to the magnet.

Figure 12.28: Seismometer. The magnet is fixed to the casing, which rests on the ground; the coil on its frame hangs from springs and can move relative to the magnet.

When an earthquake shakes the ground, the casing and magnet move with it, while the coil on its springs lags behind. There is relative motion between the magnet and the coil.
This changes the magnetic flux through the coil. By Faraday's law of electromagnetic induction an emf, and so a current, is induced in the coil.
\varepsilon=N\dfrac{\Delta\Phi}{\Delta t}
The induced current is sent to an external circuit, where it is recorded and a graph is plotted. In a strong earthquake the relative velocity is larger, the flux changes faster and the induced emf is larger.

2026 Annual I, Set D, Q.2 (ix)1+2 = 3 marks

What is magnetic flux? Also describe magnetic flux density and write its formula.

Solution

Magnetic flux (1 mark). Magnetic flux is the number of magnetic lines of force passing through a certain area. It is the scalar product of the uniform field \mathbf{B} and the vector area \mathbf{A} (directed along the normal to the surface):

\Phi_B=\mathbf{B}\cdot\mathbf{A}=BA\cos\theta

Here \theta is the angle between \mathbf{B} and \mathbf{A}. Its unit is the weber, 1\,\mathrm{Wb}=1\,\mathrm{N\,m\,A^{-1}}.

Magnetic flux density (2 marks). The magnetic flux density B is the maximum magnetic flux per unit area, that is, the flux through unit area held perpendicular to the field:

B=\dfrac{\Phi}{A}

Its unit is \mathrm{Wb\,m^{-2}}, which is the tesla (T). It is also defined as the force per unit current per unit length on a conductor at right angles to the field, B=\dfrac{F}{IL}; it is a vector quantity.

2025 Annual I, Set D, Q.2 (x) OR3 marks

What is meant by magnetic flux linkage? Explain briefly.

Solution

Definition. Magnetic flux linkage is the product of the magnetic flux through one turn of a coil and the number of turns N of the coil.

\text{flux linkage}=N\Phi=NBA\cos\theta

Here B is the flux density, A the area of one turn and \theta the angle between the area vector (the normal to the coil) and \mathbf{B}. When the coil faces the field, \cos\theta=1 and N\Phi=NBA.

Explanation. The flux through a single turn is usually very small. A coil or solenoid has N turns, each linked by the same flux, so the total linkage is N times larger. The unit is the weber turn (Wb turn). For example, 300 turns of area 0.80\,\mathrm{m^{2}} facing a field of 4\,\mathrm{mT} have N\Phi=300\times4\times10^{-3}\times0.80=0.96\,\mathrm{Wb\ turns}.

2025 Annual I, Set D, Q.66 marks

Discuss and derive an expression for magnetic force acting per unit length of conductor, when current is passing through two parallel straight conductors in upward direction.

Solution

Discussion

Each of two parallel current-carrying wires produces its own magnetic field, and each wire lies in the field of the other. A wire in a magnetic field experiences a force at right angles to both the wire and the field, so there is a force between the wires. With both currents upward (the same direction) the field between the wires is weakened and the outer field is stronger, so the wires are pulled together: the force is attractive. Currents in opposite directions give repulsion.

Derivation of the force per unit length

Derivation Force per unit length between two parallel wires

Aim. Find the force per unit length on each of two long parallel wires, carrying currents I_1 and I_2 and separated by a distance r.

Assumptions.

Both wires are long and straight, and of length L in the region of interest.
The wires are in vacuum (or a medium with permeability \mu_0), and r\gg the wire diameter.

Figure 12.19 (c): Two wires carrying currents in the same direction attract each other. F_{12} is the force on wire 2 due to wire 1, and F_{21} the force on wire 1 due to wire 2.

Wire 1 produces a field at the position of wire 2: B_1=\dfrac{\mu_0I_1}{2\pi r}.
B_1=\frac{\mu_0 I_1}{2\pi r}
Wire 2 is perpendicular to B_1, so the force on the length L of wire 2 is F_{12}=B_1I_2L\sin90^\circ.
F_{12}=B_1I_2L
Substitute B_1 into the force.
F_{12}=\frac{\mu_0I_1}{2\pi r}I_2L
Divide by L to get the force per unit length on wire 2.
\frac{F_{12}}{L}=\frac{\mu_0I_1I_2}{2\pi r}
By the same method, wire 2 produces B_2=\dfrac{\mu_0I_2}{2\pi r} at wire 1, so F_{21}=B_2I_1L and \dfrac{F_{21}}{L}=\dfrac{\mu_0I_1I_2}{2\pi r}.
\frac{F_{21}}{L}=\frac{\mu_0I_1I_2}{2\pi r}

Result\frac{F}{L}=\frac{\mu_0I_1I_2}{2\pi r}

The two forces are equal in magnitude and opposite in direction, an action and reaction pair. This result defines the ampere: I_1=I_2=1\,\mathrm{A} at r=1\,\mathrm{m} gives 2\times10^{-7}\,\mathrm{N\,m^{-1}}.

Exam focus

Derive F=qvB\sin\theta for a charge moving in a magnetic field and discuss the direction of the force. (long)
2026 Set B Q.4 OR (6 marks, 4+2); comprehensive question 3.
Go to answer
Force per unit length between two parallel current-carrying wires (derivation and direction). (long)
2025 Set D Q.6 (6 marks) and 2025 Set B MCQ 13 (opposite currents repel); comprehensive question 6.
Go to answer
State Faraday's law; does the induced emf or current depend on the resistance? (short)
2025 Set B Q.2 (vii) OR and (viii), and 2026 Set B Q.2 (iii) OR (units of \varepsilon and \Delta\Phi/\Delta t); comprehensive question 7 and short question 13.
Go to answer
Magnetic flux, flux density and flux linkage with their formulas and units. (short)
2026 Set D Q.2 (ix), 2025 Set D Q.2 (x) OR and 2025 Set B MCQ 14 (weber equals \mathrm{N\,m\,A^{-1}}); comprehensive question 4.
Go to answer
Why the kinetic energy of a charged particle in a magnetic field does not change; helical and circular paths. (short)
2025 Set B Q.2 (ix), 2026 Set B MCQ 10 (helix at 45^\circ) and 2026 Set D MCQ 11 (circular path); short question 7 and MCQ 4 of the book.
Go to answer
How a seismometer uses electromagnetic induction. (short)
2026 Set D Q.2 (iii) OR.
Go to answer
SI unit of magnetic induction (tesla) and its definition from B=F/IL. (mcq)
2025 Set D MCQ 14; the first MCQ of the book.
Go to answer
Lenz's law and the law of conservation of energy. (long)
Not in the 2025 or 2026 papers, but comprehensive question 8 and the unit SLOs.
Go to answer
Numericals: force on a wire, force on a moving charge, flux linkage and induced emf. (numerical)
Not asked in the 2025 or 2026 papers; numericals 5 to 10 of the book.
Go to answer