Moving Charge and Magnetism Chapter-Wise Test 3

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A wire of length \( 1.3 \, \text{m} \) carrying \( 6 \, \text{A} \) is at \( 45^\circ \) to a magnetic field of \( 0.8 \, \text{T} \). What is the force on the wire?

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

\( F = 6 \times 1.3 \times 0.8 \times \sin 45^\circ = 6.24 \times 0.707 = 4.4117 \approx 4.41 \, \text{N} \).

3.12 N
6.24 N
2.21 N
4.41 N
4

Two parallel wires \( 0.04 \, \text{m} \) apart carry \( 10 \, \text{A} \) and \( 4 \, \text{A} \) in opposite directions. What is the force per unit length? (\( \mu_0 = 4 \pi \times 10^{-7} \, \text{T m/A} \))

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

\( f = \frac{4 \pi \times 10^{-7} \times 10 \times 4}{2 \pi \times 0.04} = \frac{160 \times 10^{-7}}{0.08} = 2 \times 10^{-5} \, \text{N/m} \).

1 × 10⁻⁵ N/m
4 × 10⁻⁵ N/m
2 × 10⁻⁵ N/m
3 × 10⁻⁵ N/m
3

A solenoid with 1750 turns per meter carries a current of \( 1.8 \, \text{A} \). What is the magnetic field inside it? (\( \mu_0 = 4 \pi \times 10^{-7} \, \text{T m/A} \))

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

\( B = 4 \pi \times 10^{-7} \times 1750 \times 1.8 = 12.6 \pi \times 10^{-4} \approx 3.96 \times 10^{-3} \, \text{T} \).

3.96 × 10⁻³ T
1.98 × 10⁻³ T
7.92 × 10⁻³ T
5.94 × 10⁻³ T
1

An electron moves at \( 6 \times 10^6 \, \text{m/s} \) perpendicular to a field of \( 0.4 \, \text{T} \). What is the magnetic force? (Charge = \( 1.6 \times 10^{-19} \, \text{C} \))

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

\( F = 1.6 \times 10^{-19} \times 6 \times 10^6 \times 0.4 = 3.84 \times 10^{-13} \, \text{N} \).

1.92 × 10⁻¹³ N
7.68 × 10⁻¹³ N
3.84 × 10⁻¹³ N
5.76 × 10⁻¹³ N
3

A square loop of side \( 0.1 \, \text{m} \) with 25 turns carries \( 2 \, \text{A} \) in a magnetic field of \( 0.8 \, \text{T} \). The plane of the loop makes \( 30^\circ \) with the field. What is the torque?

Torque \( \tau = N I A B \sin \theta \), where \( A = 0.1 \times 0.1 = 0.01 \, \text{m}^2 \).

\( \tau = 25 \times 2 \times 0.01 \times 0.8 \times \sin 30^\circ = 0.5 \times 0.8 \times 0.5 = 0.2 \, \text{N m} \).

0.1 N m
0.2 N m
0.4 N m
0.8 N m
2

A circular coil of radius \( 0.15 \, \text{m} \) with 20 turns carries \( 2.8 \, \text{A} \). What is the magnetic field at the center? (\( \mu_0 = 4 \pi \times 10^{-7} \, \text{T m/A} \))

\( B = \frac{\mu_0 N I}{2 R} \).

\( B = \frac{4 \pi \times 10^{-7} \times 20 \times 2.8}{2 \times 0.15} = \frac{22.4 \pi \times 10^{-6}}{0.3} = 7.4667 \pi \times 10^{-5} \approx 2.35 \times 10^{-4} \, \text{T} \).

1.18 × 10⁻⁴ T
4.70 × 10⁻⁴ T
1.76 × 10⁻⁴ T
2.35 × 10⁻⁴ T
4

A rectangular loop of area \( 0.08 \, \text{m}^2 \) with 10 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.

\( \tau = 10 \times 4 \times 0.08 \times 0.9 \times 1 = 2.88 \, \text{N m} \).

1.44 N m
5.76 N m
2.16 N m
2.88 N m
4

Two parallel wires \( 0.05 \, \text{m} \) apart carry currents of \( 8 \, \text{A} \) and \( 3 \, \text{A} \) 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 \( f = \frac{\mu_0 I_1 I_2}{2 \pi d} \).

\( f = \frac{4 \pi \times 10^{-7} \times 8 \times 3}{2 \pi \times 0.05} = \frac{96 \times 10^{-7}}{0.1} = 9.6 \times 10^{-6} \, \text{N/m} \).

4.8 × 10⁻⁶ N/m
9.6 × 10⁻⁶ N/m
1.92 × 10⁻⁵ N/m
7.2 × 10⁻⁶ N/m
2

A circular coil of 60 turns and radius \( 8 \, \text{cm} \) carries a current of \( 0.9 \, \text{A} \). What is the magnetic field at its center? (\( \mu_0 = 4 \pi \times 10^{-7} \, \text{T m/A} \))

Magnetic field \( B = \frac{\mu_0 N I}{2 R} \).

\( B = \frac{4 \pi \times 10^{-7} \times 60 \times 0.9}{2 \times 0.08} = \frac{21.6 \pi \times 10^{-6}}{0.16} = 1.35 \pi \times 10^{-4} \approx 4.24 \times 10^{-4} \, \text{T} \).

4.24 × 10⁻⁴ T
2.12 × 10⁻⁴ T
8.48 × 10⁻⁴ T
6.36 × 10⁻⁴ T
1

What is the primary source of the magnetic field in a solenoid?

The magnetic field in a solenoid is primarily due to the current flowing through its coils, which creates a uniform field inside when the turns are closely spaced.

External magnetic field
Current in the coils
Motion of free charges
Magnetic material inside
2

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