Chapter 4 Moving Charges and Magnetism Class 12 Physics

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?

Concept

The magnetic field at the center of a circular coil with $N$ turns is given by $B = \frac{\mu_0 N I}{2r}$.

Given

$N = 100$
$r = 8.0 \text{ cm} = 0.08 \text{ m}$
$I = 0.40 \text{ A}$
$\mu_0 = 4\pi \times 10^{-7} \text{ T m A}^{-1}$

Calculation

$$B = \frac{4\pi \times 10^{-7} \times 100 \times 0.40}{2 \times 0.08} = \frac{4\pi \times 10^{-5} \times 0.40}{0.16}$$ $$B = \pi \times 10^{-4} \text{ T} \approx 3.14 \times 10^{-4} \text{ T}$$

Result: $3.14 \times 10^{-4} \text{ T}$

Question 4.2

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

Formula

For a long straight wire, $B = \frac{\mu_0 I}{2\pi r}$.

Calculation

$I = 35 \text{ A}, r = 0.2 \text{ m}$.

$$B = \frac{4\pi \times 10^{-7} \times 35}{2\pi \times 0.2} = \frac{2 \times 10^{-7} \times 35}{0.2} = 350 \times 10^{-7} \text{ T}$$ $$B = 3.5 \times 10^{-5} \text{ T}$$

Result: $3.5 \times 10^{-5} \text{ T}$

Question 4.3

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

1. Magnitude

$$B = \frac{\mu_0 I}{2\pi r} = \frac{4\pi \times 10^{-7} \times 50}{2\pi \times 2.5} = \frac{2 \times 10^{-7} \times 50}{2.5} = 4 \times 10^{-6} \text{ T}$$

2. Direction

Using the Right-Hand Thumb Rule: If current is North to South (thumb points South), fingers curl Upwards (Vertically Out of plane) at a point to the East.

Result: $4 \times 10^{-6} \text{ T}$, Vertically Upwards

Question 4.4

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

1. Magnitude

$$B = \frac{\mu_0 I}{2\pi r} = \frac{2 \times 10^{-7} \times 90}{1.5} = 1.2 \times 10^{-4} \text{ T}$$

2. Direction

Current is East to West. Point is below the wire. By Right-Hand Thumb Rule, the field points towards the South.

Result: $1.2 \times 10^{-4} \text{ T}$, Towards 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 $30^\circ$ with the direction of a uniform magnetic field of 0.15 T?

Formula

Force on a current carrying conductor: $F = I L B \sin\theta$.
Force per unit length: $f = \frac{F}{L} = I B \sin\theta$.

Calculation

$$f = 8 \times 0.15 \times \sin 30^\circ = 1.2 \times 0.5 = 0.6 \text{ N m}^{-1}$$

Result: $0.6 \text{ N/m}$

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 given to be 0.27 T. What is the magnetic force on the wire?

Given

$L = 0.03 \text{ m}, I = 10 \text{ A}, B = 0.27 \text{ T}, \theta = 90^\circ$.

[Image of Solenoid Magnetic Field]

Calculation

$$F = I L B \sin 90^\circ = 10 \times 0.03 \times 0.27 \times 1 = 8.1 \times 10^{-2} \text{ N}$$

Result: $8.1 \times 10^{-2} \text{ N}$

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.

Concept

Parallel currents attract. Force per unit length: $f = \frac{\mu_0 I_1 I_2}{2\pi d}$.

[Image of Force between parallel currents]

Calculation

$$F = f \times L = \frac{2 \times 10^{-7} \times 8 \times 5}{0.04} \times 0.1$$ $$F = \frac{80 \times 10^{-7}}{0.04} \times 0.1 = 2000 \times 10^{-7} \times 0.1 = 2 \times 10^{-5} \text{ N}$$

Result: $2 \times 10^{-5} \text{ N}$ (Attractive)

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.

1. Number Density (n)

Total turns $N = 5 \times 400 = 2000$. Length $L = 0.8 \text{ m}$.

$$n = \frac{N}{L} = \frac{2000}{0.8} = 2500 \text{ turns/m}$$

2. Magnetic Field

$$B = \mu_0 n I = 4\pi \times 10^{-7} \times 2500 \times 8.0$$ $$B = 4 \times 3.14 \times 20000 \times 10^{-7} = 2.5 \times 10^{-2} \text{ T}$$

Result: $2.5 \times 10^{-2} \text{ T}$

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 $30^\circ$ with the direction of a uniform horizontal magnetic field of magnitude 0.80 T. What is the magnitude of torque experienced by the coil?

Given

$N = 20, I = 12 \text{ A}, B = 0.80 \text{ T}, \theta = 30^\circ$.
Area $A = 0.1 \times 0.1 = 0.01 \text{ m}^2$.

Calculation

$$\tau = N I A B \sin\theta$$ $$\tau = 20 \times 12 \times 0.01 \times 0.80 \times \sin 30^\circ$$ $$\tau = 2.4 \times 0.8 \times 0.5 = 0.96 \text{ N m}$$

Torque: 0.96 N m

Question 4.10

Two moving coil meters, M1 and M2, have the following particulars:
$R_1 = 10 \Omega, N_1 = 30, A_1 = 3.6 \times 10^{-3} \text{ m}^2, B_1 = 0.25 \text{ T}$
$R_2 = 14 \Omega, N_2 = 42, A_2 = 1.8 \times 10^{-3} \text{ m}^2, B_2 = 0.50 \text{ T}$
Determine the ratio of (a) current sensitivity and (b) voltage sensitivity of M2 and M1.

(a) Current Sensitivity

$S_I = \frac{NBA}{k}$. Assuming spring constants $k$ are equal:

$$\text{Ratio} = \frac{S_{I2}}{S_{I1}} = \frac{N_2 B_2 A_2}{N_1 B_1 A_1} = \frac{42 \times 0.50 \times 1.8 \times 10^{-3}}{30 \times 0.25 \times 3.6 \times 10^{-3}} = \frac{42}{30} \times 2 \times 0.5 = 1.4$$

(b) Voltage Sensitivity

$S_V = \frac{S_I}{R}$.

$$\text{Ratio} = \frac{S_{V2}}{S_{V1}} = \frac{S_{I2}}{S_{I1}} \times \frac{R_1}{R_2} = 1.4 \times \frac{10}{14} = 1$$

Current Ratio = 1.4, Voltage Ratio = 1

Question 4.11

In a chamber, a uniform magnetic field of 6.5 G is maintained. An electron is shot into the field with a speed of $4.8 \times 10^6 \text{ m/s}$ normal to the field. Determine the radius of the circular orbit. ($e = 1.6 \times 10^{-19} \text{ C}, m_e = 9.1 \times 10^{-31} \text{ kg}$).

Concept

Lorentz force provides centripetal force: $qvB = \frac{mv^2}{r} \implies r = \frac{mv}{qB}$.

Calculation

$B = 6.5 \text{ G} = 6.5 \times 10^{-4} \text{ T}$.

$$r = \frac{9.1 \times 10^{-31} \times 4.8 \times 10^6}{1.6 \times 10^{-19} \times 6.5 \times 10^{-4}} = \frac{43.68 \times 10^{-25}}{10.4 \times 10^{-23}} \approx 4.2 \times 10^{-2} \text{ m}$$

Radius = 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?

Frequency

$$\nu = \frac{v}{2\pi r} = \frac{v}{2\pi (mv/qB)} = \frac{qB}{2\pi m}$$ $$\nu = \frac{1.6 \times 10^{-19} \times 6.5 \times 10^{-4}}{2 \times 3.14 \times 9.1 \times 10^{-31}} \approx 18.18 \times 10^6 \text{ Hz}$$

Dependence

No, frequency $\nu = \frac{qB}{2\pi m}$ is independent of speed $v$.

Freq $\approx 18 \text{ MHz}$. Independent of speed.

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 field lines make an angle of $60^\circ$ with the normal of the coil. Calculate the magnitude of the counter torque that must be applied to prevent the coil from turning.
(b) Would your answer change if the circular coil in (a) were replaced by a planar coil of some irregular shape that encloses the same area?

(a) Counter Torque

Area $A = \pi r^2 = \pi (0.08)^2 \approx 0.0201 \text{ m}^2$.

$$\tau = N I A B \sin\theta$$ $$\tau = 30 \times 6.0 \times 0.0201 \times 1.0 \times \sin 60^\circ$$ $$\tau = 3.618 \times 0.866 \approx 3.13 \text{ N m}$$

(b) Irregular Shape

No. The torque depends only on the Area $A$ enclosed, not the shape of the loop.

Torque = 3.13 N m. Answer unchanged for irregular shape.

NCERT Solutions