NCERT CLASS 12 PHYSICS

CHAPTER 4 – MOVING CHARGES AND MAGNETISM

PART A – IN-TEXT / CONCEPT-BASED QUESTIONS AND ANSWERS

1. What is the relationship between electricity and magnetism?

Electricity and magnetism are closely related phenomena. A stationary electric charge produces an electric field, whereas a moving electric charge or an electric current produces a magnetic field.

This relationship was demonstrated by Hans Christian Oersted through his experiment involving a current-carrying wire and a magnetic compass.

2. What is Oersted's experiment?

Oersted observed that a magnetic compass needle placed near a current-carrying wire gets deflected.

When the direction of current is reversed, the direction of deflection also reverses.

This experiment established that an electric current produces a magnetic field around a conductor.

3. What is a magnetic field?

A magnetic field is the region around a magnet, a current-carrying conductor or a moving electric charge in which magnetic effects can be experienced.

It is represented by the symbol B.

Its SI unit is tesla (T).

4. Define one tesla.

One tesla is the magnetic field in which a charge of one coulomb moving with a velocity of one metre per second perpendicular to the field experiences a magnetic force of one newton.

1 T = 1 N A⁻¹ m⁻¹

5. What is the magnetic force on a moving charge?

The magnetic force acting on a charge moving in a magnetic field is given by:

F = q(v × B)

Magnitude:

F = |q|vB sinθ

Where:

q = Electric charge

v = Velocity of the charge

B = Magnetic field

θ = Angle between velocity and magnetic field

6. When is the magnetic force on a moving charge maximum and minimum?

The magnetic force is maximum when the charge moves perpendicular to the magnetic field.

Fmax = |q|vB

The magnetic force is zero when the charge moves parallel or antiparallel to the magnetic field.

F = 0 when θ = 0° or 180°.

It is also zero when the charge is stationary.

7. Why does a magnetic field not change the kinetic energy of a moving charged particle?

The magnetic force always acts perpendicular to the instantaneous velocity of a charged particle.

Therefore, the magnetic force does no work on the particle.

As a result, its speed and kinetic energy remain unchanged. Only the direction of motion can change.

8. What is Lorentz force?

Lorentz force is the total force experienced by a charged particle moving in the presence of both electric and magnetic fields.

It is given by:

F = q(E + v × B)

Where:

E = Electric field

B = Magnetic field

v = Velocity of the charged particle

q = Charge of the particle

9. State the right-hand rule for determining the direction of magnetic force.

For a positive charge, point the fingers of the right hand in the direction of velocity and curl them towards the direction of the magnetic field.

The thumb gives the direction of magnetic force.

For a negative charge, the force acts in the opposite direction.

10. What is the force experienced by a current-carrying conductor placed in a magnetic field?

A current-carrying conductor placed in a magnetic field experiences a magnetic force.

The magnitude of the force is:

F = BIl sinθ

Where:

B = Magnetic field

I = Current

l = Length of the conductor

θ = Angle between current direction and magnetic field

11. When is the force on a current-carrying conductor maximum?

The force is maximum when the conductor is placed perpendicular to the magnetic field.

Fmax = BIl

12. What is the force on a current-carrying conductor when it is parallel to the magnetic field?

When the conductor is parallel to the magnetic field, the angle between current and magnetic field is zero.

F = BIl sin0°

F = 0

Therefore, no magnetic force acts on the conductor.

13. What is the motion of a charged particle when it enters a uniform magnetic field perpendicular to its velocity?

When a charged particle enters a uniform magnetic field perpendicular to its velocity, it moves in a circular path.

The magnetic force provides the necessary centripetal force.

The radius of the circular path is:

r = mv/|q|B

14. Derive the expression for the radius of the circular path of a charged particle in a magnetic field.

The magnetic force provides the centripetal force.

Therefore,

|q|vB = mv²/r

Cancelling v:

|q|B = mv/r

Therefore,

r = mv/|q|B

15. What is the cyclotron frequency?

Cyclotron frequency is the frequency with which a charged particle revolves in a circular path under the influence of a uniform magnetic field perpendicular to its velocity.

It is given by:

ν = |q|B/2πm

Its SI unit is hertz (Hz).

16. Does cyclotron frequency depend on the speed of a charged particle?

No. For a non-relativistic charged particle, cyclotron frequency is independent of its speed and radius of circular motion.

It depends on the charge-to-mass ratio of the particle and the magnetic field.

ν = |q|B/2πm

17. What is helical motion?

When a charged particle enters a uniform magnetic field at an angle other than 0° or 90°, it follows a helical path.

The component of velocity perpendicular to the magnetic field produces circular motion, while the parallel component produces uniform motion along the field.

The combination of these two motions produces a helix.

18. What is the pitch of a helix?

The pitch of a helix is the distance travelled by a charged particle along the direction of the magnetic field during one complete revolution.

p = v∥T

Where:

v∥ = Velocity component parallel to the magnetic field

T = Time period of revolution

19. State Biot–Savart law.

Biot–Savart law states that the magnetic field produced at a point due to a small current element is directly proportional to the current, the length of the current element and the sine of the angle between the current element and the position vector.

It is inversely proportional to the square of the distance between the current element and the observation point.

Mathematically,

dB = (μ₀/4π) × (I dl sinθ/r²)

20. What is the SI value of permeability of free space?

The permeability of free space is represented by μ₀.

μ₀/4π = 10⁻⁷ T m A⁻¹

Therefore,

μ₀ = 4π × 10⁻⁷ T m A⁻¹

21. State the right-hand thumb rule.

According to the right-hand thumb rule, if a straight current-carrying conductor is held in the right hand such that the thumb points in the direction of current, then the curled fingers indicate the direction of the magnetic field lines around the conductor.

22. What is the magnetic field at the centre of a circular current-carrying coil?

The magnetic field at the centre of a circular coil of radius R carrying current I is:

B = μ₀I/2R

For a coil having N turns:

B = μ₀NI/2R

23. What is Ampere's circuital law?

Ampere's circuital law states that the line integral of the magnetic field around any closed path is equal to μ₀ times the total current enclosed by that path.

∮ B · dl = μ₀I

Where:

μ₀ = Permeability of free space

I = Total current enclosed by the closed path

24. What is the magnetic field due to a long straight current-carrying conductor?

The magnetic field at a perpendicular distance r from a long straight conductor carrying current I is:

B = μ₀I/2πr

The magnetic field decreases as the distance from the conductor increases.

25. What is a solenoid?

A solenoid is a long cylindrical coil consisting of a large number of closely wound turns of insulated copper wire.

When current passes through it, a magnetic field is produced inside the solenoid.

26. What is the magnetic field inside a long solenoid?

The magnetic field inside a long solenoid is approximately uniform and is given by:

B = μ₀nI

Where:

n = Number of turns per unit length

I = Current flowing through the solenoid

27. Why is the magnetic field outside a long solenoid nearly zero?

The magnetic fields produced by the different turns of a long solenoid nearly cancel each other outside it.

Therefore, the magnetic field outside an ideal long solenoid is approximately zero.

28. What is a toroid?

A toroid is a solenoid bent into a circular ring so that its ends meet.

It consists of a large number of turns wound around a circular core.

29. What is the magnetic field inside a toroid?

For an ideal toroid, the magnetic field at a distance r from its centre is:

B = μ₀NI/2πr

Where:

N = Total number of turns

I = Current

r = Distance from the centre of the toroid

30. What is the force between two parallel current-carrying conductors?

Two parallel current-carrying conductors exert magnetic forces on each other.

If currents flow in the same direction, the conductors attract each other.

If currents flow in opposite directions, the conductors repel each other.

31. Derive the expression for force per unit length between two long parallel current-carrying conductors.

Consider two long parallel conductors separated by a distance d.

Let their currents be I₁ and I₂.

The magnetic field produced by the first conductor at the position of the second conductor is:

B₁ = μ₀I₁/2πd

The force on a length l of the second conductor is:

F = I₂lB₁

Substituting:

F = I₂l × μ₀I₁/2πd

Therefore,

F = μ₀I₁I₂l/2πd

Dividing by l:

F/l = μ₀I₁I₂/2πd

32. What happens when two parallel wires carry currents in the same direction?

When two parallel wires carry currents in the same direction, they attract each other.

33. What happens when two parallel wires carry currents in opposite directions?

When two parallel wires carry currents in opposite directions, they repel each other.

34. What is a current loop?

A current loop is a closed conducting path through which electric current flows.

A current-carrying loop placed in a magnetic field experiences a torque if its magnetic moment is not parallel to the magnetic field.

35. What is the torque acting on a current-carrying rectangular loop in a uniform magnetic field?

The torque acting on a current-carrying coil is:

τ = NIAB sinθ

Where:

N = Number of turns

I = Current

A = Area of the coil

B = Magnetic field

θ = Angle between the normal to the coil and the magnetic field

36. What is the magnetic dipole moment of a current loop?

The magnetic dipole moment of a current loop is the product of current and the area of the loop.

m = IA

For a coil having N turns:

m = NIA

Its SI unit is A m².

37. What is the direction of magnetic dipole moment of a current loop?

The direction of the magnetic dipole moment is perpendicular to the plane of the current loop.

It is determined by the right-hand thumb rule.

38. What is a moving-coil galvanometer?

A moving-coil galvanometer is an instrument used to detect and measure small electric currents in a circuit.

It works on the principle that a current-carrying coil placed in a magnetic field experiences a torque.

39. State the principle of a moving-coil galvanometer.

A current-carrying coil placed in a radial magnetic field experiences a torque.

This torque causes the coil to rotate.

The deflection of the coil is proportional to the current flowing through it.

40. What is current sensitivity of a galvanometer?

Current sensitivity is defined as the deflection produced in a galvanometer per unit current flowing through it.

Current sensitivity = θ/I

For a moving-coil galvanometer:

θ/I = NBA/k

Where:

N = Number of turns

B = Magnetic field

A = Area of the coil

k = Torsional constant

41. What is voltage sensitivity of a galvanometer?

Voltage sensitivity is defined as the deflection produced per unit potential difference applied across the galvanometer.

Voltage sensitivity = θ/V

For a galvanometer of resistance G:

θ/V = NBA/kG

42. How can a galvanometer be converted into an ammeter?

A galvanometer is converted into an ammeter by connecting a very low resistance called a shunt resistance in parallel with it.

The shunt allows most of the current to pass through the low-resistance path.

The resistance of an ideal ammeter is very small.

43. How can a galvanometer be converted into a voltmeter?

A galvanometer is converted into a voltmeter by connecting a high resistance in series with it.

The high resistance limits the current through the galvanometer.

The resistance of an ideal voltmeter is very high.

44. What is a shunt?

A shunt is a very low resistance connected in parallel with a galvanometer to convert it into an ammeter.

Its value is:

S = IgG/(I − Ig)

Where:

Ig = Full-scale deflection current

G = Galvanometer resistance

I = Desired maximum current range

45. What is the difference between an ammeter and a voltmeter?

Ammeter:

An ammeter measures electric current.
It is connected in series with the circuit.
It has very low resistance.
It is represented by the symbol A.

Voltmeter:

A voltmeter measures potential difference.
It is connected in parallel with the component.
It has very high resistance.
It is represented by the symbol V.

PART B – NCERT EXERCISE QUESTIONS AND ANSWERS

QUESTION 4.1

A circular coil of wire consisting of 100 turns, each of radius 8.0 cm, carries a current of 0.40 A. What is the magnitude of the magnetic field B at the centre of the coil?

Solution:

Given:

Number of turns, N = 100

Radius, R = 8.0 cm = 0.08 m

Current, I = 0.40 A

Permeability of free space:

μ₀ = 4π × 10⁻⁷ T m A⁻¹

The magnetic field at the centre of a circular coil is:

B = μ₀NI/2R

Substituting the values:

B = (4π × 10⁻⁷ × 100 × 0.40)/(2 × 0.08)

B = (160π × 10⁻⁷)/0.16

B = π × 10⁻⁴ T

B = 3.14 × 10⁻⁴ T

Answer: The magnetic field at the centre of the coil is 3.14 × 10⁻⁴ T.

QUESTION 4.2

A long straight wire carries a current of 35 A. What is the magnitude of the magnetic field B at a point 20 cm from the wire?

Solution:

Given:

Current, I = 35 A

Distance, r = 20 cm = 0.20 m

The magnetic field due to a long straight conductor is:

B = μ₀I/2πr

Substituting the values:

B = (4π × 10⁻⁷ × 35)/(2π × 0.20)

B = (2 × 10⁻⁷ × 35)/0.20

B = 70 × 10⁻⁷/0.20

B = 3.5 × 10⁻⁵ T

Answer: The magnetic field is 3.5 × 10⁻⁵ T.

QUESTION 4.3

A long straight wire in the horizontal plane carries a current of 50 A in the north-to-south direction. Give the magnitude and direction of the magnetic field B at a point 2.5 m east of the wire.

Solution:

Given:

Current, I = 50 A

Distance, r = 2.5 m

The magnetic field is:

B = μ₀I/2πr

Substituting:

B = (4π × 10⁻⁷ × 50)/(2π × 2.5)

B = (2 × 10⁻⁷ × 50)/2.5

B = 100 × 10⁻⁷/2.5

B = 4.0 × 10⁻⁶ T

Direction:

Using the right-hand thumb rule, point the thumb towards the south, in the direction of current.

At a point east of the wire, the curled fingers indicate that the magnetic field is vertically upwards.

Answer:

Magnitude = 4.0 × 10⁻⁶ T

Direction = Vertically upwards.

QUESTION 4.4

A horizontal overhead power line carries a current of 90 A in the east-to-west direction. What is the magnitude and direction of the magnetic field due to the current 1.5 m below the line?

Solution:

Given:

Current, I = 90 A

Distance, r = 1.5 m

The magnetic field due to a long straight wire is:

B = μ₀I/2πr

Substituting:

B = (4π × 10⁻⁷ × 90)/(2π × 1.5)

B = (2 × 10⁻⁷ × 90)/1.5

B = 180 × 10⁻⁷/1.5

B = 1.2 × 10⁻⁵ T

Direction:

Using the right-hand thumb rule, point the thumb towards the west.

At a point below the wire, the magnetic field is directed towards the south.

Answer:

Magnitude = 1.2 × 10⁻⁵ T

Direction = Horizontally towards the south.

QUESTION 4.5

What is the magnitude of magnetic force per unit length on a wire carrying a current of 8 A and making an angle of 30° with the direction of a uniform magnetic field of 0.15 T?

Solution:

Given:

Current, I = 8 A

Magnetic field, B = 0.15 T

Angle, θ = 30°

Force on a current-carrying conductor:

F = BIl sinθ

Force per unit length:

F/l = BI sinθ

Substituting:

F/l = 0.15 × 8 × sin30°

We know:

sin30° = 0.5

Therefore,

F/l = 0.15 × 8 × 0.5

F/l = 0.6 N/m

Answer: The magnetic force per unit length is 0.6 N/m.

QUESTION 4.6

A 3.0 cm wire carrying a current of 10 A is placed inside a solenoid perpendicular to its axis. The magnetic field inside the solenoid is 0.27 T. What is the magnetic force on the wire?

Solution:

Given:

Length, l = 3.0 cm = 0.03 m

Current, I = 10 A

Magnetic field, B = 0.27 T

Angle, θ = 90°

Force on the conductor:

F = BIl sinθ

Substituting:

F = 0.27 × 10 × 0.03 × sin90°

Since:

sin90° = 1

F = 0.27 × 10 × 0.03

F = 0.081 N

Answer: The magnetic force on the wire is 8.1 × 10⁻² N.

QUESTION 4.7

Two long and parallel straight wires A and B carrying currents of 8.0 A and 5.0 A in the same direction are separated by a distance of 4.0 cm. Estimate the force on a 10 cm section of wire A.

Solution:

Given:

Current in wire A, I₁ = 8.0 A

Current in wire B, I₂ = 5.0 A

Distance between wires, d = 4.0 cm = 0.04 m

Length of wire, l = 10 cm = 0.10 m

The force between two parallel current-carrying wires is:

F = μ₀I₁I₂l/2πd

Substituting:

F = (4π × 10⁻⁷ × 8 × 5 × 0.10)/(2π × 0.04)

F = (2 × 10⁻⁷ × 8 × 5 × 0.10)/0.04

F = (8 × 10⁻⁷)/0.04

F = 2.0 × 10⁻⁵ N

Since both currents flow in the same direction, the wires attract each other.

Answer: The force on the 10 cm section of wire A is 2.0 × 10⁻⁵ N, directed towards wire B.

QUESTION 4.8

A closely wound solenoid 80 cm long has 5 layers of windings of 400 turns each. The diameter of the solenoid is 1.8 cm. If the current carried is 8.0 A, estimate the magnitude of B inside the solenoid near its centre.

Solution:

Given:

Length, l = 80 cm = 0.80 m

Number of layers = 5

Turns in each layer = 400

Current, I = 8.0 A

Total number of turns:

N = 5 × 400

N = 2000

Number of turns per unit length:

n = N/l

n = 2000/0.80

n = 2500 turns/m

The magnetic field inside a long solenoid is:

B = μ₀nI

Substituting:

B = 4π × 10⁻⁷ × 2500 × 8

B = 8π × 10⁻³ T

B = 2.51 × 10⁻² T

Answer: The magnetic field inside the solenoid is approximately 2.51 × 10⁻² T.

QUESTION 4.9

A square coil of side 10 cm consists of 20 turns and carries a current of 12 A. The coil is suspended vertically and the normal to the plane of the coil makes an angle of 30° with the direction of a uniform horizontal magnetic field of magnitude 0.80 T. What is the magnitude of torque experienced by the coil?

Solution:

Given:

Side of square coil = 10 cm = 0.10 m

Number of turns, N = 20

Current, I = 12 A

Magnetic field, B = 0.80 T

Angle, θ = 30°

Area of the square coil:

A = side²

A = (0.10)²

A = 0.01 m²

Torque on a current-carrying coil:

τ = NIAB sinθ

Substituting:

τ = 20 × 12 × 0.01 × 0.80 × sin30°

τ = 20 × 12 × 0.01 × 0.80 × 0.5

τ = 0.96 N m

Answer: The magnitude of torque experienced by the coil is 0.96 N m.

QUESTION 4.10

Two moving-coil meters M₁ and M₂ have the following characteristics:

For meter M₁:

Resistance, R₁ = 10 Ω

Number of turns, N₁ = 30

Area of coil, A₁ = 3.6 × 10⁻³ m²

Magnetic field, B₁ = 0.25 T

For meter M₂:

Resistance, R₂ = 14 Ω

Number of turns, N₂ = 42

Area of coil, A₂ = 1.8 × 10⁻³ m²

Magnetic field, B₂ = 0.50 T

The spring constants of the two meters are identical. Determine the ratio of:

(a) Current sensitivity of M₂ to M₁.

(b) Voltage sensitivity of M₂ to M₁.

Solution:

Given:

k₁ = k₂

Current sensitivity is:

Si = θ/I = NAB/k

Therefore,

Si₂/Si₁ = (N₂A₂B₂)/(N₁A₁B₁)

Substituting:

Si₂/Si₁ = (42 × 1.8 × 10⁻³ × 0.50)/(30 × 3.6 × 10⁻³ × 0.25)

Si₂/Si₁ = 37.8/27

Si₂/Si₁ = 1.4

Therefore, the current sensitivity of M₂ is 1.4 times that of M₁.

(b) Ratio of voltage sensitivity

Voltage sensitivity is:

Sv = θ/V

Since V = IG,

Sv = NAB/kG

Therefore,

Sv₂/Sv₁ = (N₂A₂B₂/G₂)/(N₁A₁B₁/G₁)

Sv₂/Sv₁ = (Si₂/Si₁) × (G₁/G₂)

Substituting:

Sv₂/Sv₁ = 1.4 × 10/14

Sv₂/Sv₁ = 1.4 × 0.714

Sv₂/Sv₁ = 1

Answer:

(a) Ratio of current sensitivity = 1.4

(b) Ratio of voltage sensitivity = 1

QUESTION 4.11

In a chamber, a magnetic field of 6.5 G is maintained. An electron is shot into the field with a speed of 4.8 × 10⁶ m/s perpendicular to the field. Explain why the path of the electron is a circle and determine its radius. (e = 1.6 × 10⁻¹⁹ C, m = 9.1 × 10⁻³¹ kg)

Solution:

Given:

Magnetic field, B = 6.5 G

We know:

1 G = 10⁻⁴ T

Therefore,

B = 6.5 × 10⁻⁴ T

Velocity, v = 4.8 × 10⁶ m/s

Charge of electron, e = 1.6 × 10⁻¹⁹ C

Mass of electron, m = 9.1 × 10⁻³¹ kg

Why is the path circular?

The electron enters the magnetic field perpendicular to its velocity.

Therefore, the magnetic force acts perpendicular to its velocity at every instant.

This force provides the necessary centripetal force, causing the electron to move in a circular path.

The radius is given by:

r = mv/eB

Substituting:

r = (9.1 × 10⁻³¹ × 4.8 × 10⁶)/(1.6 × 10⁻¹⁹ × 6.5 × 10⁻⁴)

r = (43.68 × 10⁻²⁵)/(10.4 × 10⁻²³)

r = 0.042 m

r = 4.2 cm

Answer: The radius of the circular path is approximately 4.2 cm.

QUESTION 4.12

In Exercise 4.11, obtain the frequency of revolution of the electron. Does the answer depend on the speed of the electron?

Solution:

Given:

Charge of electron, e = 1.6 × 10⁻¹⁹ C

Mass of electron, m = 9.1 × 10⁻³¹ kg

Magnetic field, B = 6.5 × 10⁻⁴ T

The frequency of revolution is:

ν = eB/2πm

Substituting:

ν = (1.6 × 10⁻¹⁹ × 6.5 × 10⁻⁴)/(2π × 9.1 × 10⁻³¹)

ν = 10.4 × 10⁻²³/(57.18 × 10⁻³¹)

ν ≈ 1.82 × 10⁷ Hz

Does the frequency depend on the speed of the electron?

No. The frequency of revolution is independent of the speed of the electron.

It depends only on the charge-to-mass ratio of the particle and the magnetic field.

Answer:

Frequency = 1.82 × 10⁷ Hz

The frequency does not depend on the speed of the electron.

QUESTION 4.13

(a) A circular coil of 30 turns and radius 8.0 cm carrying a current of 6.0 A is suspended vertically in a uniform horizontal magnetic field of magnitude 1.0 T. The magnetic field lines make an angle of 60° with the normal to the coil. What is the magnitude of the counter torque needed to prevent the coil from turning?

(b) Would your answer change if the circular coil in part (a) were replaced by a planar coil of some irregular shape that encloses the same area?

Solution:

(a) Calculation of counter torque

Given:

Number of turns, N = 30

Radius, R = 8.0 cm = 0.08 m

Current, I = 6.0 A

Magnetic field, B = 1.0 T

Angle, θ = 60°

Area of the circular coil:

A = πR²

A = π × (0.08)²

A = π × 0.0064

A ≈ 0.0201 m²

Torque on the coil:

τ = NIAB sinθ

Substituting:

τ = 30 × 6.0 × 0.0201 × 1.0 × sin60°

We know:

sin60° = 0.866

Therefore,

τ = 30 × 6.0 × 0.0201 × 0.866

τ ≈ 3.13 N m

The external counter torque must have the same magnitude and act in the opposite direction to prevent rotation.

Answer (a): The required counter torque is approximately 3.13 N m.

(b) Effect of changing the shape of the coil

No, the answer would not change.

The torque on a planar current loop in a uniform magnetic field depends on its area, current, number of turns and the angle between its normal and the magnetic field.

It does not depend on the particular shape of the planar loop, provided the enclosed area and other quantities remain the same.

Answer (b): No. The counter torque remains unchanged.

PART C – IMPORTANT DERIVATIONS

1. Derive the expression for magnetic force on a current-carrying conductor.

Answer:

Consider a straight conductor of length l and area of cross-section A placed in a uniform magnetic field B.

Let:

n = Number of free charge carriers per unit volume

q = Charge of each carrier

vd = Drift velocity

The total number of charge carriers in the conductor is:

N = nAl

Force on each charge carrier:

f = q(vd × B)

Total magnetic force:

F = Nq(vd × B)

Substituting N:

F = nAlq(vd × B)

We know that:

I = nqAvd

Therefore,

F = I(l × B)

The magnitude is:

F = BIl sinθ

Hence proved.

2. Derive the expression for the radius of the circular path of a charged particle in a uniform magnetic field.

Answer:

Consider a charged particle of mass m and charge q moving with velocity v perpendicular to a uniform magnetic field B.

The magnetic force is:

F = |q|vB

This force provides the centripetal force required for circular motion.

Therefore,

|q|vB = mv²/r

Cancelling v:

|q|B = mv/r

Therefore,

r = mv/|q|B

Hence proved.

3. Derive the expression for the cyclotron frequency.

Answer:

The radius of the circular path is:

r = mv/|q|B

The time period of circular motion is:

T = 2πr/v

Substituting the value of r:

T = 2π/v × mv/|q|B

T = 2πm/|q|B

Frequency is the reciprocal of time period.

ν = 1/T

Therefore,

ν = |q|B/2πm

Hence proved.

4. Derive the expression for the pitch of a helical path.

Answer:

When a charged particle enters a magnetic field at an angle θ, its velocity has two components.

Perpendicular component:

v⊥ = v sinθ

Parallel component:

v∥ = v cosθ

The perpendicular component produces circular motion, while the parallel component remains unchanged.

The pitch is the distance travelled along the field direction during one complete revolution.

Therefore,

p = v∥T

Substituting:

p = v cosθ × 2πm/|q|B

Therefore,

p = 2πmv cosθ/|q|B

Hence proved.

5. State and explain Biot–Savart law.

Answer:

Biot–Savart law gives the magnetic field produced at a point due to a small current element.

According to this law, the magnetic field due to a current element is:

Directly proportional to the current I.
Directly proportional to the length dl of the current element.
Directly proportional to sinθ.
Inversely proportional to the square of the distance r.

Mathematically,

dB = (μ₀/4π)(I dl sinθ/r²)

In vector form:

dB = (μ₀/4π) [I(dl × r̂)/r²]

Where r̂ is the unit vector directed from the current element to the observation point.

6. Derive the expression for the magnetic field at the centre of a circular current-carrying coil.

Answer:

Consider a circular coil of radius R carrying current I.

According to Biot–Savart law:

dB = (μ₀/4π)(I dl sinθ/R²)

At the centre of the circular coil, every current element is perpendicular to the radius vector.

Therefore:

θ = 90°

sin90° = 1

Thus,

dB = (μ₀/4π)(I dl/R²)

All the magnetic field contributions are in the same direction at the centre.

Therefore,

B = ∮dB

B = (μ₀I/4πR²)∮dl

The total length of the circular coil is:

∮dl = 2πR

Therefore,

B = (μ₀I/4πR²) × 2πR

B = μ₀I/2R

For a coil of N turns:

B = μ₀NI/2R

Hence proved.

7. Derive the expression for the magnetic field along the axis of a circular current loop.

Answer:

Consider a circular loop of radius R carrying current I.

Let P be a point on its axis at a distance x from the centre.

The distance between any current element and point P is:

r = √(R² + x²)

According to Biot–Savart law:

dB = (μ₀/4π)(I dl/r²)

The components of magnetic field perpendicular to the axis cancel each other due to symmetry.

Only the axial components add up.

The axial component is:

dBx = dB × R/r

Therefore,

dBx = (μ₀/4π)(I dl/r²) × R/r

dBx = (μ₀IR/4πr³)dl

Integrating around the complete circular loop:

Bx = (μ₀IR/4πr³)∮dl

Since:

∮dl = 2πR

Bx = (μ₀IR/4πr³) × 2πR

Bx = μ₀IR²/2r³

Since:

r = √(R² + x²)

Therefore,

B = μ₀IR²/[2(R² + x²)³ᐟ²]

For N turns:

B = μ₀NIR²/[2(R² + x²)³ᐟ²]

Hence proved.

8. State and explain Ampere's circuital law.

Answer:

Ampere's circuital law states that the line integral of the magnetic field around a closed path is equal to μ₀ times the total current enclosed by the path.

Mathematically,

∮B · dl = μ₀I

Where:

B = Magnetic field

dl = Small length element of the closed path

μ₀ = Permeability of free space

I = Total current enclosed by the path

This law is useful for calculating magnetic fields in situations having suitable symmetry.

9. Derive the expression for the magnetic field due to a long straight current-carrying wire using Ampere's law.

Answer:

Consider a long straight conductor carrying current I.

Choose a circular Amperian loop of radius r around the conductor.

By symmetry, the magnetic field has the same magnitude at every point on the circular path and is tangential to it.

According to Ampere's circuital law:

∮B · dl = μ₀I

Since B is constant along the circular path:

B∮dl = μ₀I

The circumference of the circular path is:

∮dl = 2πr

Therefore,

B × 2πr = μ₀I

Hence,

B = μ₀I/2πr

Hence proved.

10. Derive the expression for the magnetic field inside a long solenoid.

Answer:

Consider a long solenoid having:

N = Total number of turns

l = Length of solenoid

I = Current through the solenoid

Number of turns per unit length:

n = N/l

According to Ampere's circuital law:

∮B · dl = μ₀I

Choose a rectangular Amperian path partly inside and partly outside the solenoid.

For a long solenoid, the magnetic field outside is approximately zero.

The field inside is uniform and parallel to the axis.

If the length of the path inside the solenoid is h:

Bh = μ₀Ienclosed

The number of turns enclosed is:

Therefore, the enclosed current is:

Ienclosed = nhI

Substituting:

Bh = μ₀nhI

Cancelling h:

B = μ₀nI

Since n = N/l:

B = μ₀NI/l

Hence proved.

11. Derive the expression for force per unit length between two long parallel current-carrying conductors.

Answer:

Consider two long parallel conductors separated by distance d.

Let the currents through them be I₁ and I₂.

The magnetic field produced by the first conductor at the position of the second conductor is:

B₁ = μ₀I₁/2πd

The force on a length l of the second conductor is:

F = I₂lB₁

Substituting:

F = I₂l × μ₀I₁/2πd

Therefore,

F = μ₀I₁I₂l/2πd

Dividing by l:

F/l = μ₀I₁I₂/2πd

Hence proved.

12. Derive the expression for torque on a current-carrying rectangular coil in a uniform magnetic field.

Answer:

Consider a rectangular coil having:

Length = l

Breadth = b

Area = A = lb

Number of turns = N

Current = I

Magnetic field = B

Let θ be the angle between the normal to the coil and the magnetic field.

The two opposite sides of the coil experience equal and opposite forces.

These forces form a couple and produce torque.

The force on each effective side is:

F = BIl

The perpendicular distance between the lines of action of the forces is:

d = b sinθ

Torque on one turn:

τ = F × d

τ = BIl × b sinθ

Since A = lb:

τ = IAB sinθ

For N turns:

τ = NIAB sinθ

Hence proved.

13. Derive the expression for magnetic dipole moment of a current loop.

Answer:

A current-carrying loop behaves like a magnetic dipole.

The magnetic dipole moment is defined as the product of current and area of the loop.

For one turn:

m = IA

For a coil of N turns:

m = NIA

Its direction is perpendicular to the plane of the coil and is determined by the right-hand thumb rule.

Its SI unit is A m².

14. Explain the principle and working of a moving-coil galvanometer.

Answer:

Principle:

A current-carrying coil placed in a magnetic field experiences a torque.

This torque causes the coil to rotate.

Working:

A rectangular coil is suspended between the concave pole pieces of a permanent magnet.

The coil is placed in a radial magnetic field.

When current flows through the coil, it experiences a deflecting torque.

The suspension fibre produces a restoring torque.

At equilibrium:

Deflecting torque = Restoring torque

NIAB = kθ

Therefore,

I = kθ/NAB

Thus,

θ/I = NAB/k

This shows that the deflection is directly proportional to the current.

Hence, a moving-coil galvanometer can detect and measure small currents.

15. Derive the expression for converting a galvanometer into an ammeter.

Answer:

Let:

G = Resistance of galvanometer

Ig = Full-scale deflection current

I = Desired maximum current of the ammeter

S = Shunt resistance

To convert a galvanometer into an ammeter, a small resistance S is connected in parallel with it.

Since both are connected in parallel, the potential difference across them is equal.

Therefore,

IgG = (I − Ig)S

Rearranging:

S = IgG/(I − Ig)

Thus, a low resistance connected in parallel converts a galvanometer into an ammeter.

16. Derive the expression for converting a galvanometer into a voltmeter.

Answer:

Let:

G = Resistance of galvanometer

Ig = Full-scale deflection current

V = Desired voltage range

R = High resistance connected in series

The galvanometer is connected in series with a high resistance.

The total resistance is:

R + G

Using Ohm's law:

V = Ig(R + G)

Therefore,

V/Ig = R + G

Hence,

R = V/Ig − G

Thus, a high resistance connected in series converts a galvanometer into a voltmeter.

PART D – ADDITIONAL IMPORTANT NUMERICAL QUESTIONS

1. A charge of 2 × 10⁻⁶ C moves with a velocity of 3 × 10⁴ m/s perpendicular to a magnetic field of 0.5 T. Calculate the magnetic force acting on it.

Solution:

Given:

q = 2 × 10⁻⁶ C

v = 3 × 10⁴ m/s

B = 0.5 T

θ = 90°

Formula:

F = qvB sinθ

Substituting:

F = 2 × 10⁻⁶ × 3 × 10⁴ × 0.5 × 1

F = 3 × 10⁻² N

Answer: The magnetic force is 0.03 N.

2. An electron moves with a velocity of 2 × 10⁶ m/s perpendicular to a magnetic field of 0.01 T. Calculate the magnetic force acting on it.

Solution:

Given:

e = 1.6 × 10⁻¹⁹ C

v = 2 × 10⁶ m/s

B = 0.01 T

θ = 90°

Formula:

F = evB

Substituting:

F = 1.6 × 10⁻¹⁹ × 2 × 10⁶ × 0.01

F = 3.2 × 10⁻¹⁵ N

Answer: The magnitude of magnetic force is 3.2 × 10⁻¹⁵ N.

3. A proton moves with a velocity of 5 × 10⁵ m/s perpendicular to a magnetic field of 0.2 T. Calculate the radius of its circular path. (Mass of proton = 1.67 × 10⁻²⁷ kg; charge = 1.6 × 10⁻¹⁹ C.)

Solution:

Given:

m = 1.67 × 10⁻²⁷ kg

v = 5 × 10⁵ m/s

q = 1.6 × 10⁻¹⁹ C

B = 0.2 T

Formula:

r = mv/qB

Substituting:

r = (1.67 × 10⁻²⁷ × 5 × 10⁵)/(1.6 × 10⁻¹⁹ × 0.2)

r = 8.35 × 10⁻²²/3.2 × 10⁻²⁰

r = 0.0261 m

r = 2.61 cm

Answer: The radius of the circular path is approximately 2.61 cm.

4. A circular coil has 50 turns and a radius of 10 cm. It carries a current of 2 A. Calculate the magnetic field at its centre.

Solution:

Given:

N = 50

R = 10 cm = 0.10 m

I = 2 A

Formula:

B = μ₀NI/2R

Substituting:

B = (4π × 10⁻⁷ × 50 × 2)/(2 × 0.10)

B = 4π × 10⁻⁴ T

B = 1.26 × 10⁻³ T

Answer: The magnetic field at the centre is 1.26 × 10⁻³ T.

5. A long straight conductor carries a current of 10 A. Calculate the magnetic field at a distance of 5 cm from it.

Solution:

Given:

I = 10 A

r = 5 cm = 0.05 m

Formula:

B = μ₀I/2πr

Substituting:

B = (4π × 10⁻⁷ × 10)/(2π × 0.05)

B = (2 × 10⁻⁷ × 10)/0.05

B = 4 × 10⁻⁵ T

Answer: The magnetic field is 4 × 10⁻⁵ T.

6. A solenoid has 1000 turns per metre and carries a current of 3 A. Calculate the magnetic field inside it.

Solution:

Given:

n = 1000 turns/m

I = 3 A

Formula:

B = μ₀nI

Substituting:

B = 4π × 10⁻⁷ × 1000 × 3

B = 12π × 10⁻⁴ T

B = 3.77 × 10⁻³ T

Answer: The magnetic field inside the solenoid is 3.77 × 10⁻³ T.

7. Two long parallel wires separated by 10 cm carry currents of 5 A and 10 A in the same direction. Calculate the force per unit length between them.

Solution:

Given:

I₁ = 5 A

I₂ = 10 A

d = 10 cm = 0.10 m

Formula:

F/l = μ₀I₁I₂/2πd

Substituting:

F/l = (4π × 10⁻⁷ × 5 × 10)/(2π × 0.10)

F/l = (2 × 10⁻⁷ × 50)/0.10

F/l = 1 × 10⁻⁴ N/m

Answer: The force per unit length is 1 × 10⁻⁴ N/m. The wires attract each other because their currents flow in the same direction.

8. A coil of 100 turns and area 0.02 m² carries a current of 2 A. It is placed in a magnetic field of 0.5 T. Calculate the maximum torque acting on it.

Solution:

Given:

N = 100

A = 0.02 m²

I = 2 A

B = 0.5 T

Maximum torque occurs when:

θ = 90°

Formula:

τmax = NIAB

Substituting:

τmax = 100 × 2 × 0.02 × 0.5

τmax = 2 N m

Answer: The maximum torque is 2 N m.

PART E – IMPORTANT FORMULAS

Magnetic force on a moving charge:

F = qvB sinθ

Lorentz force:

F = q(E + v × B)

Magnetic force on a current-carrying conductor:

F = BIl sinθ

Radius of circular path:

r = mv/|q|B

Time period of circular motion:

T = 2πm/|q|B

Cyclotron frequency:

ν = |q|B/2πm

Angular frequency:

ω = |q|B/m

Pitch of helical path:

p = 2πmv cosθ/|q|B

Biot–Savart law:

dB = (μ₀/4π)(I dl sinθ/r²)

Magnetic field at the centre of a circular coil:

B = μ₀I/2R

Magnetic field at the centre of an N-turn coil:

B = μ₀NI/2R

Magnetic field on the axis of a circular loop:

B = μ₀NIR²/[2(R² + x²)³ᐟ²]

Ampere's circuital law:

∮B · dl = μ₀I

Magnetic field due to a long straight wire:

B = μ₀I/2πr

Magnetic field inside a long solenoid:

B = μ₀nI

Magnetic field inside a toroid:

B = μ₀NI/2πr

Force per unit length between parallel wires:

F/l = μ₀I₁I₂/2πd

Torque on a current-carrying coil:

τ = NIAB sinθ

Magnetic dipole moment:

m = NIA

Current sensitivity of a galvanometer:

Si = θ/I = NBA/k

Voltage sensitivity of a galvanometer:

Sv = θ/V = NBA/kG

Shunt resistance for converting galvanometer into ammeter:

S = IgG/(I − Ig)

Series resistance for converting galvanometer into voltmeter:

R = V/Ig − G

PART F – IMPORTANT SI UNITS

Physical Quantity	SI Unit
Magnetic field	Tesla (T)
Magnetic force	Newton (N)
Electric charge	Coulomb (C)
Current	Ampere (A)
Velocity	Metre per second (m/s)
Radius	Metre (m)
Frequency	Hertz (Hz)
Time period	Second (s)
Permeability of free space	T m A⁻¹
Magnetic dipole moment	A m²
Torque	Newton metre (N m)
Force per unit length	N/m
Area	Square metre (m²)
Resistance	Ohm (Ω)
Torsional constant	N m rad⁻¹
Current sensitivity	rad/A
Voltage sensitivity	rad/V

PART G – QUICK REVISION POINTS

Moving electric charges produce magnetic fields.
Oersted demonstrated the magnetic effect of electric current.
The SI unit of magnetic field is tesla.
Magnetic force on a moving charge is F = q(v × B).
Magnetic force is maximum when velocity is perpendicular to the magnetic field.
Magnetic force is zero when velocity is parallel or antiparallel to the magnetic field.
Magnetic force does no work on a moving charged particle.
A charged particle moving perpendicular to a uniform magnetic field follows a circular path.
A charged particle entering a magnetic field at an angle follows a helical path.
Cyclotron frequency is independent of the speed of a non-relativistic particle.
Biot–Savart law is used to calculate magnetic fields due to current elements.
The magnetic field at the centre of a circular coil is B = μ₀NI/2R.
Ampere's circuital law is ∮B · dl = μ₀I.
The magnetic field around a long straight wire is B = μ₀I/2πr.
The magnetic field inside a long solenoid is B = μ₀nI.
Parallel currents attract each other.
Antiparallel currents repel each other.
A current-carrying coil behaves like a magnetic dipole.
The torque on a current-carrying coil is τ = NIAB sinθ.
Magnetic dipole moment is m = NIA.
A moving-coil galvanometer detects small electric currents.
A low resistance connected in parallel converts a galvanometer into an ammeter.
A high resistance connected in series converts a galvanometer into a voltmeter.
Current sensitivity is the deflection per unit current.
Voltage sensitivity is the deflection per unit voltage.