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  4. Moving Charge and Magnetism Chapter-Wise Test 2

Moving Charge and Magnetism Chapter-Wise Test 2

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A long straight wire carries a current of \( 10 \, \text{A} \). What is the magnetic field at a point \( 0.1 \, \text{m} \) from the wire? (\( \mu_0 = 4 \pi \times 10^{-7} \, \text{T m/A} \))

Magnetic field \( B = \frac{\mu_0 I}{2 \pi r} \).

\( B = \frac{4 \pi \times 10^{-7} \times 10}{2 \pi \times 0.1} = \frac{4 \times 10^{-6}}{0.2} = 2 \times 10^{-5} \, \text{T} \).

2 × 10⁻⁵ T
1 × 10⁻⁵ T
4 × 10⁻⁵ T
5 × 10⁻⁵ T
1

A long straight wire carries a current of \( 40 \, \text{A} \). What is the magnetic field at a distance of \( 0.5 \, \text{m} \) from it? (\( \mu_0 = 4 \pi \times 10^{-7} \, \text{T m/A} \))

Magnetic field \( B = \frac{\mu_0 I}{2 \pi r} \).

\( B = \frac{4 \pi \times 10^{-7} \times 40}{2 \pi \times 0.5} = \frac{160 \times 10^{-7}}{1} = 1.6 \times 10^{-5} \, \text{T} \).

1.6 × 10⁻⁵ T
3.2 × 10⁻⁵ T
8 × 10⁻⁶ T
2.4 × 10⁻⁵ T
1

A solenoid has 600 turns per meter and carries a current of \( 3 \, \text{A} \). What is the magnetic field inside it? (\( \mu_0 = 4 \pi \times 10^{-7} \, \text{T m/A} \))

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

\( B = 4 \pi \times 10^{-7} \times 600 \times 3 = 7.2 \pi \times 10^{-4} \approx 2.26 \times 10^{-3} \, \text{T} \).

1.13 × 10⁻³ T
2.26 × 10⁻³ T
4.52 × 10⁻³ T
3.39 × 10⁻³ T
2

A solenoid has 1250 turns per meter and 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 \( B = \mu_0 n I \).

\( B = 4 \pi \times 10^{-7} \times 1250 \times 1.5 = 7.5 \pi \times 10^{-4} \approx 2.36 × 10^{-3} \, \text{T} \).

1.18 × 10⁻³ T
2.36 × 10⁻³ T
4.72 × 10⁻³ T
3.54 × 10⁻³ T
2

A solenoid with 1100 turns per meter carries \( 1.8 \, \text{A} \). What is the magnetic field inside? (\( \mu_0 = 4 \pi \times 10^{-7} \, \text{T m/A} \))

\( B = \mu_0 n I \).

\( B = 4 \pi \times 10^{-7} \times 1100 \times 1.8 = 7.92 \pi \times 10^{-4} \approx 2.49 \times 10^{-3} \, \text{T} \).

1.24 × 10⁻³ T
4.98 × 10⁻³ T
2.09 × 10⁻³ T
2.49 × 10⁻³ T
4

Two parallel wires \( 0.12 \, \text{m} \) apart carry currents of \( 4 \, \text{A} \) and \( 7 \, \text{A} \) 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 \( f = \frac{\mu_0 I_1 I_2}{2 \pi d} \).

\( f = \frac{4 \pi \times 10^{-7} \times 4 \times 7}{2 \pi \times 0.12} = \frac{112 \times 10^{-7}}{0.24} = 4.67 \times 10^{-6} \, \text{N/m} \).

2.33 × 10⁻⁶ N/m
4.67 × 10⁻⁶ N/m
9.33 × 10⁻⁶ N/m
7.00 × 10⁻⁶ N/m
2

What is the shape of the magnetic field lines produced by a straight current-carrying wire?

The magnetic field lines around a straight current-carrying wire form concentric circles centered on the wire, as derived from the Biot-Savart law or Ampere’s law.

Straight lines
Helical paths
Concentric circles
Parabolic curves
3

An electron moves at \( 2.5 \times 10^6 \, \text{m/s} \) perpendicular to a field of \( 0.5 \, \text{T} \). What is the radius of its path? (Mass = \( 9.1 \times 10^{-31} \, \text{kg} \), charge = \( 1.6 \times 10^{-19} \, \text{C} \))

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

\( r = \frac{9.1 \times 10^{-31} \times 2.5 \times 10^6}{1.6 \times 10^{-19} \times 0.5} = \frac{2.275 \times 10^{-24}}{8 \times 10^{-20}} = 2.84375 \times 10^{-5} \, \text{m} \approx 2.84 \times 10^{-3} \, \text{cm} \).

1.42 × 10⁻³ cm
5.68 × 10⁻³ cm
2.84 × 10⁻³ cm
4.26 × 10⁻³ cm
3

An electron moves at \( 6.5 \times 10^6 \, \text{m/s} \) perpendicular to a field of \( 0.2 \, \text{T} \). What is the radius of its path? (Mass = \( 9.1 \times 10^{-31} \, \text{kg} \), charge = \( 1.6 \times 10^{-19} \, \text{C} \))

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

\( r = \frac{9.1 \times 10^{-31} \times 6.5 \times 10^6}{1.6 \times 10^{-19} \times 0.2} = \frac{5.915 \times 10^{-24}}{3.2 \times 10^{-20}} = 1.848 \times 10^{-4} \, \text{m} \approx 0.0185 \, \text{cm} \).

0.0092 cm
0.0370 cm
0.0185 cm
0.0278 cm
3

A rectangular loop of area \( 0.06 \, \text{m}^2 \) with 15 turns carries \( 2.5 \, \text{A} \) in a field of \( 0.8 \, \text{T} \) at \( 60^\circ \) to the plane. What is the torque?

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

\( \tau = 15 \times 2.5 \times 0.06 \times 0.8 \times \sin 60^\circ = 1.8 \times 0.866 = 1.5588 \approx 1.56 \, \text{N m} \).

0.9 N m
3.12 N m
1.56 N m
2.34 N m
3

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