Moving Charge and Magnetism Chapter-Wise Test 10

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A solenoid with 1300 turns per meter carries \( 1.2 \, \text{A} \). What is the magnetic field inside? (\( \mu_0 = 4 \pi \times 10^{-7} \, \text{T m/A} \))

\( B = \mu_0 n I \).

\( B = 4 \pi \times 10^{-7} \times 1300 \times 1.2 = 6.24 \pi \times 10^{-4} \approx 1.96 \times 10^{-3} \, \text{T} \).

9.8 × 10⁻⁴ T
3.92 × 10⁻³ T
1.63 × 10⁻³ T
1.96 × 10⁻³ T
4

A proton moves at \( 7.5 \times 10^7 \, \text{m/s} \) perpendicular to a field of \( 0.12 \, \text{T} \). What is the magnetic force? (Charge = \( 1.6 \times 10^{-19} \, \text{C} \))

Force \( F = q v B \sin \theta \), \( \theta = 90^\circ \), so \( \sin \theta = 1 \).

\( F = 1.6 \times 10^{-19} \times 7.5 \times 10^7 \times 0.12 = 1.44 \times 10^{-12} \, \text{N} \).

7.2 × 10⁻¹³ N
2.88 × 10⁻¹² N
1.44 × 10⁻¹² N
2.16 × 10⁻¹² N
3

A rectangular loop of area \( 0.03 \, \text{m}^2 \) with 20 turns carries \( 4 \, \text{A} \) in a field of \( 0.9 \, \text{T} \) perpendicular to the plane. What is the torque?

\( \tau = N I A B \sin \theta \), \( \theta = 90^\circ \) to plane means \( \sin 0^\circ = 1 \) with normal.

\( \tau = 20 \times 4 \times 0.03 \times 0.9 \times 1 = 2.16 \, \text{N m} \).

1.08 N m
4.32 N m
1.8 N m
2.16 N m
4

Two parallel wires \( 0.06 \, \text{m} \) apart carry \( 7 \, \text{A} \) and \( 5 \, \text{A} \) in the same direction. What is the force per unit length? (\( \mu_0 = 4 \pi \times 10^{-7} \, \text{T m/A} \))

\( f = \frac{\mu_0 I_1 I_2}{2 \pi d} \).

\( f = \frac{4 \pi \times 10^{-7} \times 7 \times 5}{2 \pi \times 0.06} = \frac{140 \times 10^{-7}}{0.12} = 1.166 \times 10^{-5} \approx 1.17 \times 10^{-5} \, \text{N/m} \).

5.83 × 10⁻⁶ N/m
2.33 × 10⁻⁵ N/m
8.75 × 10⁻⁶ N/m
1.17 × 10⁻⁵ N/m
4

A solenoid with 1800 turns per meter carries a current of \( 1.5 \, \text{A} \). What is the magnetic field inside it? (\( \mu_0 = 4 \pi \times 10^{-7} \, \text{T m/A} \))

Magnetic field \( B = \mu_0 n I \).

\( B = 4 \pi \times 10^{-7} \times 1800 \times 1.5 = 10.8 \pi \times 10^{-4} \approx 3.39 \times 10^{-3} \, \text{T} \).

3.39 × 10⁻³ T
1.70 × 10⁻³ T
6.78 × 10⁻³ T
2.26 × 10⁻³ T
1

Two parallel wires \( 0.12 \, \text{m} \) apart carry \( 7 \, \text{A} \) and \( 3 \, \text{A} \) in the same direction. What is the force per unit length? (\( \mu_0 = 4 \pi \times 10^{-7} \, \text{T m/A} \))

\( f = \frac{\mu_0 I_1 I_2}{2 \pi d} \).

\( f = \frac{4 \pi \times 10^{-7} \times 7 \times 3}{2 \pi \times 0.12} = \frac{84 \times 10^{-7}}{0.24} = 3.5 \times 10^{-6} \, \text{N/m} \).

1.75 × 10⁻⁶ N/m
7 × 10⁻⁶ N/m
2.63 × 10⁻⁶ N/m
3.5 × 10⁻⁶ N/m
4

A wire of length \( 2.3 \, \text{m} \) carrying \( 4 \, \text{A} \) is at \( 60^\circ \) to a magnetic field of \( 0.6 \, \text{T} \). What is the force on the wire?

\( F = I l B \sin \theta \).

\( F = 4 \times 2.3 \times 0.6 \times \sin 60^\circ = 5.52 \times 0.866 = 4.7803 \approx 4.78 \, \text{N} \).

2.76 N
5.52 N
4.14 N
4.78 N
4

Two parallel wires \( 0.07 \, \text{m} \) apart carry currents of \( 9 \, \text{A} \) and \( 2 \, \text{A} \) in the same direction. What is the force per unit length between them? (\( \mu_0 = 4 \pi \times 10^{-7} \, \text{T m/A} \))

Force per unit length \( f = \frac{\mu_0 I_1 I_2}{2 \pi d} \).

\( f = \frac{4 \pi \times 10^{-7} \times 9 \times 2}{2 \pi \times 0.07} = \frac{72 \times 10^{-7}}{0.14} = 5.14 \times 10^{-6} \, \text{N/m} \).

2.57 × 10⁻⁶ N/m
5.14 × 10⁻⁶ N/m
1.03 × 10⁻⁵ N/m
7.71 × 10⁻⁶ N/m
2

A rectangular loop of area \( 0.09 \, \text{m}^2 \) with 18 turns carries \( 2 \, \text{A} \) in a field of \( 0.6 \, \text{T} \) perpendicular to the plane. What is the torque?

\( \tau = N I A B \sin \theta \), \( \theta = 90^\circ \) to plane means \( \sin 0^\circ = 1 \) with normal.

\( \tau = 18 \times 2 \times 0.09 \times 0.6 \times 1 = 1.944 \approx 1.94 \, \text{N m} \).

0.97 N m
3.88 N m
1.62 N m
1.94 N m
4

In a moving coil galvanometer, what provides the restoring torque to balance the magnetic torque?

In a moving coil galvanometer, the spring provides a restoring torque proportional to the deflection, balancing the magnetic torque caused by the current in the coil.

Magnetic field
Current in the coil
Spring
External weight
3

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