Moving Charge and Magnetism Chapter-Wise Test 12

Correct answer Carries: 4.

Wrong Answer Carries: -1.

Two parallel wires \( 0.01 \, \text{m} \) apart carry \( 5 \, \text{A} \) and \( 6 \, \text{A} \) in opposite directions. What is the force per unit length? (\( \mu_0 = 4 \pi \times 10^{-7} \, \text{T m/A} \))

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

\( f = \frac{4 \pi \times 10^{-7} \times 5 \times 6}{2 \pi \times 0.01} = \frac{120 \times 10^{-7}}{0.02} = 6 \times 10^{-5} \, \text{N/m} \).

3 × 10⁻⁵ N/m
1.2 × 10⁻⁴ N/m
6 × 10⁻⁵ N/m
9 × 10⁻⁵ N/m
3

Two parallel wires \( 0.1 \, \text{m} \) apart carry \( 8 \, \text{A} \) and \( 6 \, \text{A} \) in the same direction. What is the force per unit length? (\( \mu_0 = 4 \pi \times 10^{-7} \, \text{T m/A} \))

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

\( f = \frac{4 \pi \times 10^{-7} \times 8 \times 6}{2 \pi \times 0.1} = \frac{192 \times 10^{-7}}{0.2} = 9.6 \times 10^{-5} \, \text{N/m} \).

4.8 × 10⁻⁵ N/m
1.92 × 10⁻⁴ N/m
7.2 × 10⁻⁵ N/m
9.6 × 10⁻⁵ N/m
4

A solenoid has 850 turns per meter and carries a current of \( 1.6 \, \text{A} \). What is the magnetic field inside it? (\( \mu_0 = 4 \pi \times 10^{-7} \, \text{T m/A} \))

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

\( B = 4 \pi \times 10^{-7} \times 850 \times 1.6 = 5.44 \pi \times 10^{-4} \approx 1.71 \times 10^{-3} \, \text{T} \).

8.55 × 10⁻⁴ T
1.71 × 10⁻³ T
3.42 × 10⁻³ T
2.56 × 10⁻³ T
2

A square loop of side \( 0.12 \, \text{m} \) with 35 turns carries \( 1.8 \, \text{A} \) in a magnetic field of \( 0.5 \, \text{T} \). The plane of the loop is at \( 30^\circ \) to the field. What is the torque?

Torque \( \tau = N I A B \sin \theta \), where \( A = 0.12 \times 0.12 = 0.0144 \, \text{m}^2 \).

\( \tau = 35 \times 1.8 \times 0.0144 \times 0.5 \times \sin 30^\circ = 0.9072 \times 0.5 = 0.4536 \approx 0.45 \, \text{N m} \).

0.23 N m
0.45 N m
0.91 N m
1.36 N m
2

What is the condition for zero torque on a current-carrying loop in a magnetic field?

Torque \( \tau = N I A B \sin \theta \) is zero when \( \sin \theta = 0 \), i.e., \( \theta = 0^\circ \) or \( 180^\circ \), meaning the plane of the loop is parallel to the magnetic field.

Current is maximum
Field is perpendicular to the loop
Plane of the loop is parallel to the field
Area of the loop is zero
3

A square loop of side \( 0.25 \, \text{m} \) with 20 turns carries \( 2.5 \, \text{A} \) in a magnetic field of \( 0.3 \, \text{T} \). The plane of the loop is at \( 45^\circ \) to the field. What is the torque?

Torque \( \tau = N I A B \sin \theta \), where \( A = 0.25 \times 0.25 = 0.0625 \, \text{m}^2 \).

\( \tau = 20 \times 2.5 \times 0.0625 \times 0.3 \times \sin 45^\circ = 0.9375 \times 0.707 = 0.6633 \approx 0.66 \, \text{N m} \).

0.33 N m
0.66 N m
1.32 N m
0.99 N m
2

A proton moves at \( 4 \times 10^7 \, \text{m/s} \) perpendicular to a field of \( 0.05 \, \text{T} \). What is the radius of its path? (Mass = \( 1.67 \times 10^{-27} \, \text{kg} \), charge = \( 1.6 \times 10^{-19} \, \text{C} \))

\( r = \frac{mv}{qB} \).

\( r = \frac{1.67 \times 10^{-27} \times 4 \times 10^7}{1.6 \times 10^{-19} \times 0.05} = \frac{6.68 \times 10^{-20}}{8 \times 10^{-21}} = 8.35 \, \text{m} \).

4.18 m
16.7 m
8.35 m
12.5 m
3

An electron moves at \( 5 \times 10^6 \, \text{m/s} \) perpendicular to a magnetic field of \( 0.2 \, \text{T} \). What is the magnetic force on it? (Charge of electron = \( 1.6 \times 10^{-19} \, \text{C} \))

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

\( F = 1.6 \times 10^{-19} \times 5 \times 10^6 \times 0.2 = 1.6 \times 10^{-13} \, \text{N} \).

8 × 10⁻¹⁴ N
1.6 × 10⁻¹³ N
3.2 × 10⁻¹³ N
4.8 × 10⁻¹³ N
2

A rectangular loop of area \( 0.02 \, \text{m}^2 \) with 10 turns carries \( 3 \, \text{A} \) in a field of \( 0.5 \, \text{T} \) at \( 60^\circ \) to the plane. What is the torque?

\( \tau = N I A B \sin \theta \), where \( \theta = 60^\circ \) to the plane means \( \sin 30^\circ \) with the normal.

\( \tau = 10 \times 3 \times 0.02 \times 0.5 \times \sin 60^\circ = 0.3 \times 0.866 = 0.2598 \approx 0.26 \, \text{N m} \).

0.15 N m
0.3 N m
0.26 N m
0.52 N m
3

A straight wire of length \( 1.2 \, \text{m} \) carries a current of \( 6 \, \text{A} \) perpendicular to a uniform magnetic field of \( 0.25 \, \text{T} \). What is the force acting on the wire?

Force \( F = I l B \sin \theta \), where \( \theta = 90^\circ \), so \( \sin \theta = 1 \).

\( F = 6 \times 1.2 \times 0.25 = 1.8 \, \text{N} \).

1.8 N
2.4 N
1.2 N
3.0 N
1

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