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  4. Moving Charge and Magnetism Chapter-Wise Test 3

Moving Charge and Magnetism Chapter-Wise Test 3

Correct answer Carries: 4.

Wrong Answer Carries: -1.

What happens to the torque on a rectangular current loop if the magnetic field direction is reversed?

Torque is given by \( \boldsymbol{\tau} = \mathbf{m} \times \mathbf{B} \). Reversing the magnetic field \( \mathbf{B} \) reverses the direction of the torque vector, but its magnitude remains the same.

Magnitude doubles
Direction reverses, magnitude unchanged
Becomes zero
Magnitude halves
2

A circular loop of radius \( 0.15 \, \text{m} \) with 15 turns carries a current of \( 2 \, \text{A} \). What is the magnetic field at the center? (\( \mu_0 = 4 \pi \times 10^{-7} \, \text{T m/A} \))

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

\( B = \frac{4 \pi \times 10^{-7} \times 15 \times 2}{2 \times 0.15} = \frac{12 \pi \times 10^{-6}}{0.3} = 4 \pi \times 10^{-5} \approx 1.26 \times 10^{-4} \, \text{T} \).

6.28 × 10⁻⁵ T
1.26 × 10⁻⁴ T
2.51 × 10⁻⁴ T
3.77 × 10⁻⁴ T
2

A proton moves with a speed of \( 2 \times 10^6 \, \text{m/s} \) perpendicular to a uniform magnetic field of \( 0.5 \, \text{T} \). What is the radius of its circular path? (Mass of proton = \( 1.67 \times 10^{-27} \, \text{kg} \), charge = \( 1.6 \times 10^{-19} \, \text{C} \))

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

Substitute: \( r = \frac{1.67 \times 10^{-27} \times 2 \times 10^6}{1.6 \times 10^{-19} \times 0.5} = \frac{3.34 \times 10^{-21}}{8 \times 10^{-20}} = 4.175 \times 10^{-2} \, \text{m} = 4.18 \, \text{cm} \).

4.18 cm
5.25 cm
3.12 cm
6.40 cm
1

A long wire carries \( 12 \, \text{A} \). At what distance is the magnetic field \( 2.4 \times 10^{-6} \, \text{T} \)? (\( \mu_0 = 4 \pi \times 10^{-7} \, \text{T m/A} \))

\( B = \frac{\mu_0 I}{2 \pi r} \), so \( r = \frac{\mu_0 I}{2 \pi B} \).

\( r = \frac{4 \pi \times 10^{-7} \times 12}{2 \pi \times 2.4 \times 10^{-6}} = \frac{48 \times 10^{-7}}{4.8 \times 10^{-6}} = 1 \, \text{m} \).

0.5 m
2 m
1 m
1.5 m
3

Which factor does not affect the magnetic field produced by a current element according to the Biot-Savart law?

The Biot-Savart law states \( d\mathbf{B} \propto \frac{I d\mathbf{l} \times \mathbf{r}}{r^3} \). The field depends on current, length of the element, distance, and angle, but not on the mass of the conductor.

Current in the element
Mass of the conductor
Distance from the element
Angle between current and position vector
2

A circular loop of radius \( 0.13 \, \text{m} \) with 20 turns carries a current of \( 3.5 \, \text{A} \). What is the magnetic field at the center? (\( \mu_0 = 4 \pi \times 10^{-7} \, \text{T m/A} \))

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

\( B = \frac{4 \pi \times 10^{-7} \times 20 \times 3.5}{2 \times 0.13} = \frac{28 \pi \times 10^{-6}}{0.26} = 1.0769 \pi \times 10^{-4} \approx 3.38 \times 10^{-4} \, \text{T} \).

1.69 × 10⁻⁴ T
3.38 × 10⁻⁴ T
6.76 × 10⁻⁴ T
5.07 × 10⁻⁴ T
2

A wire of length \( 1.5 \, \text{m} \) carrying \( 8 \, \text{A} \) is at \( 60^\circ \) to a magnetic field of \( 0.5 \, \text{T} \). What is the force on the wire?

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

\( F = 8 \times 1.5 \times 0.5 \times \sin 60^\circ = 12 \times 0.5 \times 0.866 = 5.196 \approx 5.2 \, \text{N} \).

3 N
6 N
4.5 N
5.2 N
4

A circular coil of 70 turns and radius \( 7 \, \text{cm} \) carries a current of \( 0.8 \, \text{A} \). What is the magnetic field at its center? (\( \mu_0 = 4 \pi \times 10^{-7} \, \text{T m/A} \))

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

\( B = \frac{4 \pi \times 10^{-7} \times 70 \times 0.8}{2 \times 0.07} = \frac{22.4 \pi \times 10^{-6}}{0.14} = 1.6 \pi \times 10^{-4} \approx 5.03 \times 10^{-4} \, \text{T} \).

5.03 × 10⁻⁴ T
2.51 × 10⁻⁴ T
1.26 × 10⁻⁴ T
7.54 × 10⁻⁴ T
1

An electron moves with a speed of \( 2 \times 10^6 \, \text{m/s} \) perpendicular to a magnetic field of \( 0.7 \, \text{T} \). What is the radius of its path? (Mass = \( 9.1 \times 10^{-31} \, \text{kg} \), charge = \( 1.6 \times 10^{-19} \, \text{C} \))

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

\( r = \frac{9.1 \times 10^{-31} \times 2 \times 10^6}{1.6 \times 10^{-19} \times 0.7} = \frac{1.82 \times 10^{-24}}{1.12 \times 10^{-19}} = 1.625 \times 10^{-5} \, \text{m} = 1.625 \times 10^{-3} \, \text{cm} \).

1.625 × 10⁻³ cm
3.25 × 10⁻³ cm
8.13 × 10⁻⁴ cm
2.44 × 10⁻³ cm
1

A circular coil of radius \( 0.12 \, \text{m} \) with 35 turns carries \( 1.5 \, \text{A} \). What is the magnetic field at the center? (\( \mu_0 = 4 \pi \times 10^{-7} \, \text{T m/A} \))

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

\( B = \frac{4 \pi \times 10^{-7} \times 35 \times 1.5}{2 \times 0.12} = \frac{21 \pi \times 10^{-6}}{0.24} = 8.75 \pi \times 10^{-5} \approx 2.75 \times 10^{-4} \, \text{T} \).

1.38 × 10⁻⁴ T
5.50 × 10⁻⁴ T
2.06 × 10⁻⁴ T
2.75 × 10⁻⁴ T
4

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