Moving Charge and Magnetism Chapter-Wise Test 11

Correct answer Carries: 4.

Wrong Answer Carries: -1.

Two parallel wires \( 0.11 \, \text{m} \) apart carry currents of \( 6 \, \text{A} \) 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 \( f = \frac{\mu_0 I_1 I_2}{2 \pi d} \).

\( f = \frac{4 \pi \times 10^{-7} \times 6 \times 2}{2 \pi \times 0.11} = \frac{48 \times 10^{-7}}{0.22} = 2.18 \times 10^{-6} \, \text{N/m} \).

1.09 × 10⁻⁶ N/m
2.18 × 10⁻⁶ N/m
4.36 × 10⁻⁶ N/m
3.27 × 10⁻⁶ N/m
2

A rectangular loop of area \( 0.07 \, \text{m}^2 \) with 10 turns carries \( 5 \, \text{A} \) in a field of \( 0.8 \, \text{T} \) 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.

\( \tau = 10 \times 5 \times 0.07 \times 0.8 \times 1 = 2.8 \, \text{N m} \).

1.4 N m
5.6 N m
2.1 N m
2.8 N m
4

A circular loop of radius \( 0.2 \, \text{m} \) with 10 turns carries \( 1.5 \, \text{A} \). What is the magnetic field at its center? (\( \mu_0 = 4 \pi \times 10^{-7} \, \text{T m/A} \))

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

\( B = \frac{4 \pi \times 10^{-7} \times 10 \times 1.5}{2 \times 0.2} = \frac{6 \pi \times 10^{-6}}{0.4} = 1.5 \pi \times 10^{-5} \approx 4.71 \times 10^{-5} \, \text{T} \).

2.36 × 10⁻⁵ T
4.71 × 10⁻⁵ T
7.85 × 10⁻⁵ T
9.42 × 10⁻⁵ T
2

A proton moves at \( 2.5 \times 10^7 \, \text{m/s} \) perpendicular to a field of \( 0.15 \, \text{T} \). What is the magnetic force? (Charge = \( 1.6 \times 10^{-19} \, \text{C} \))

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

\( F = 1.6 \times 10^{-19} \times 2.5 \times 10^7 \times 0.15 = 6 \times 10^{-13} \, \text{N} \).

3 × 10⁻¹³ N
1.2 × 10⁻¹² N
6 × 10⁻¹³ N
9 × 10⁻¹³ N
3

A circular coil of radius \( 0.11 \, \text{m} \) with 25 turns carries \( 3 \, \text{A} \). What is the magnetic field at the center? (\( \mu_0 = 4 \pi \times 10^{-7} \, \text{T m/A} \))

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

\( B = \frac{4 \pi \times 10^{-7} \times 25 \times 3}{2 \times 0.11} = \frac{30 \pi \times 10^{-6}}{0.22} = 1.3636 \pi \times 10^{-4} \approx 4.28 \times 10^{-4} \, \text{T} \).

2.14 × 10⁻⁴ T
8.56 × 10⁻⁴ T
3.21 × 10⁻⁴ T
4.28 × 10⁻⁴ T
4

A solenoid with 500 turns per meter carries \( 2 \, \text{A} \). What is the magnetic field inside it? (\( \mu_0 = 4 \pi \times 10^{-7} \, \text{T m/A} \))

\( B = \mu_0 n I \).

\( B = 4 \pi \times 10^{-7} \times 500 \times 2 = 4 \pi \times 10^{-4} \approx 1.26 \times 10^{-3} \, \text{T} \).

6.28 × 10⁻⁴ T
2.51 × 10⁻³ T
1.26 × 10⁻³ T
3.14 × 10⁻³ T
3

A proton moves at \( 3.5 \times 10^7 \, \text{m/s} \) perpendicular to a field of \( 0.08 \, \text{T} \). What is the radius of its path? (Mass = \( 1.67 \times 10^{-27} \, \text{kg} \), charge = \( 1.6 \times 10^{-19} \, \text{C} \))

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

\( r = \frac{1.67 \times 10^{-27} \times 3.5 \times 10^7}{1.6 \times 10^{-19} \times 0.08} = \frac{5.845 \times 10^{-20}}{1.28 \times 10^{-20}} = 4.5664 \approx 4.57 \, \text{m} \).

2.28 m
9.14 m
4.57 m
6.85 m
3

A solenoid with 1000 turns per meter carries \( 1.5 \, \text{A} \). What is the magnetic field inside? (\( \mu_0 = 4 \pi \times 10^{-7} \, \text{T m/A} \))

\( B = \mu_0 n I \).

\( B = 4 \pi \times 10^{-7} \times 1000 \times 1.5 = 6 \pi \times 10^{-4} \approx 1.885 \times 10^{-3} \, \text{T} \).

9.42 × 10⁻⁴ T
3.77 × 10⁻³ T
1.26 × 10⁻³ T
1.89 × 10⁻³ T
4

Two parallel wires \( 0.07 \, \text{m} \) apart carry \( 8 \, \text{A} \) and \( 6 \, \text{A} \) in the same direction. What is the force per unit length? (\( \mu_0 = 4 \pi \times 10^{-7} \, \text{T m/A} \))

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

\( f = \frac{4 \pi \times 10^{-7} \times 8 \times 6}{2 \pi \times 0.07} = \frac{192 \times 10^{-7}}{0.14} = 1.3714 \times 10^{-5} \approx 1.37 \times 10^{-5} \, \text{N/m} \).

6.86 × 10⁻⁶ N/m
2.74 × 10⁻⁵ N/m
1.03 × 10⁻⁵ N/m
1.37 × 10⁻⁵ N/m
4

Two parallel wires \( 0.06 \, \text{m} \) apart carry \( 9 \, \text{A} \) and \( 5 \, \text{A} \) in the same direction. What is the force per unit length? (\( \mu_0 = 4 \pi \times 10^{-7} \, \text{T m/A} \))

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

\( f = \frac{4 \pi \times 10^{-7} \times 9 \times 5}{2 \pi \times 0.06} = \frac{180 \times 10^{-7}}{0.12} = 1.5 \times 10^{-5} \, \text{N/m} \).

7.5 × 10⁻⁶ N/m
3 × 10⁻⁵ N/m
1.125 × 10⁻⁵ N/m
1.5 × 10⁻⁵ N/m
4

Performance Summary

Score:

Category Details
Total Attempts:
Total Skipped:
Total Wrong Answers:
Total Correct Answers:
Time Taken:
Average Time Taken per Question:
Accuracy:
0