Moving Charge and Magnetism Chapter-Wise Test 9

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A circular coil of 80 turns and radius \( 4 \, \text{cm} \) carries a current of \( 0.5 \, \text{A} \). What is the magnetic field at its center? (\( \mu_0 = 4 \pi \times 10^{-7} \, \text{T m/A} \))

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

\( B = \frac{4 \pi \times 10^{-7} \times 80 \times 0.5}{2 \times 0.04} = \frac{16 \pi \times 10^{-6}}{0.08} = 2 \pi \times 10^{-4} \approx 6.28 \times 10^{-4} \, \text{T} \).

6.28 × 10⁻⁴ T
3.14 × 10⁻⁴ T
1.57 × 10⁻⁴ T
9.42 × 10⁻⁴ T
1

A square loop of side \( 0.16 \, \text{m} \) with 25 turns carries \( 3 \, \text{A} \) in a magnetic field of \( 0.7 \, \text{T} \). 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 \( A = 0.16 \times 0.16 = 0.0256 \, \text{m}^2 \).

\( \tau = 25 \times 3 \times 0.0256 \times 0.7 \times \sin 45^\circ = 1.344 \times 0.707 = 0.9502 \approx 0.95 \, \text{N m} \).

0.48 N m
0.95 N m
1.90 N m
1.43 N m
2

A long wire carries \( 18 \, \text{A} \). At what distance is the magnetic field \( 6 \times 10^{-6} \, \text{T} \)? (\( \mu_0 = 4 \pi \times 10^{-7} \, \text{T m/A} \))

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

\( r = \frac{4 \pi \times 10^{-7} \times 18}{2 \pi \times 6 \times 10^{-6}} = \frac{72 \times 10^{-7}}{12 \times 10^{-6}} = 0.6 \, \text{m} \).

0.3 m
1.2 m
0.6 m
0.9 m
3

What is the nature of the force between two parallel wires carrying currents in opposite directions?

Two parallel wires with currents in opposite directions experience a repulsive force because the magnetic field produced by one wire interacts with the current in the other, causing them to push apart.

Repulsive
Attractive
No force
Oscillatory
1

A solenoid has 2000 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 \( B = \mu_0 n I \).

\( B = 4 \pi \times 10^{-7} \times 2000 \times 3 = 12 \pi \times 10^{-4} \approx 7.54 \times 10^{-3} \, \text{T} \).

7.54 × 10⁻³ T
3.77 × 10⁻³ T
1.26 × 10⁻³ T
5.02 × 10⁻³ T
1

What is the direction of the magnetic force on a positive charge moving parallel to a uniform magnetic field?

The magnetic force is given by \( \mathbf{F} = q (\mathbf{v} \times \mathbf{B}) \). If the velocity \( \mathbf{v} \) is parallel to the magnetic field \( \mathbf{B} \), the angle between them is \( 0^\circ \), so \( \sin 0^\circ = 0 \), and the force magnitude is zero. Thus, there is no force.

No force acts
Perpendicular to both velocity and field
Opposite to the velocity
Along the magnetic field
1

A proton moves with a speed of \( 1.8 \times 10^6 \, \text{m/s} \) perpendicular to a magnetic field of \( 0.5 \, \text{T} \). What is the radius of its path? (Mass = \( 1.67 \times 10^{-27} \, \text{kg} \), charge = \( 1.6 \times 10^{-19} \, \text{C} \))

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

\( r = \frac{1.67 \times 10^{-27} \times 1.8 \times 10^6}{1.6 \times 10^{-19} \times 0.5} = \frac{3.006 \times 10^{-21}}{8 \times 10^{-20}} = 3.7575 \times 10^{-2} \, \text{m} = 3.76 \, \text{cm} \).

3.76 cm
7.52 cm
1.88 cm
5.64 cm
1

A solenoid has 800 turns per meter and carries a current of \( 2.5 \, \text{A} \). What is the magnetic field inside it? (\( \mu_0 = 4 \pi \times 10^{-7} \, \text{T m/A} \))

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

\( B = 4 \pi \times 10^{-7} \times 800 \times 2.5 = 8 \pi \times 10^{-4} \approx 2.51 \times 10^{-3} \, \text{T} \).

1.26 × 10⁻³ T
2.51 × 10⁻³ T
5.02 × 10⁻³ T
3.77 × 10⁻³ T
2

What is the role of the magnetic field in a moving coil galvanometer?

The magnetic field interacts with the current in the coil to produce a torque, causing the coil to rotate, which is then balanced by the spring to indicate the current magnitude.

To increase resistance
To reduce friction
To align the coil
To produce torque
4

A long straight wire carries a current of \( 25 \, \text{A} \). What is the magnetic field at a distance of \( 0.25 \, \text{m} \) from it? (\( \mu_0 = 4 \pi \times 10^{-7} \, \text{T m/A} \))

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

\( B = \frac{4 \pi \times 10^{-7} \times 25}{2 \pi \times 0.25} = \frac{100 \times 10^{-7}}{0.5} = 2 \times 10^{-5} \, \text{T} \).

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

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