Moving Charge and Magnetism Chapter-Wise Test 6

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A circular coil of radius \( 0.05 \, \text{m} \) with 30 turns carries \( 2.5 \, \text{A} \). What is the magnetic field at the center? (\( \mu_0 = 4 \pi \times 10^{-7} \, \text{T m/A} \))

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

\( B = \frac{4 \pi \times 10^{-7} \times 30 \times 2.5}{2 \times 0.05} = \frac{30 \pi \times 10^{-6}}{0.1} = 3 \pi \times 10^{-4} \approx 9.42 \times 10^{-4} \, \text{T} \).

4.71 × 10⁻⁴ T
1.88 × 10⁻³ T
7.06 × 10⁻⁴ T
9.42 × 10⁻⁴ T
4

A rectangular loop of area \( 0.05 \, \text{m}^2 \) with 16 turns carries \( 2.2 \, \text{A} \) in a field of \( 0.7 \, \text{T} \) at \( 45^\circ \) to the plane. What is the torque?

\( \tau = N I A B \sin \theta \), where \( \theta = 45^\circ \) to plane means \( \sin 45^\circ \) with normal.

\( \tau = 16 \times 2.2 \times 0.05 \times 0.7 \times \sin 45^\circ = 1.232 \times 0.707 = 0.871 \approx 0.87 \, \text{N m} \).

0.62 N m
1.74 N m
0.87 N m
1.23 N m
3

A proton moves at \( 3 \times 10^7 \, \text{m/s} \) perpendicular to a field of \( 0.1 \, \text{T} \). What is the radius of its path? (Mass = \( 1.67 \times 10^{-27} \, \text{kg} \), charge = \( 1.6 \times 10^{-19} \, \text{C} \))

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

\( r = \frac{1.67 \times 10^{-27} \times 3 \times 10^7}{1.6 \times 10^{-19} \times 0.1} = \frac{5.01 \times 10^{-20}}{1.6 \times 10^{-20}} = 3.13125 \, \text{m} \approx 3.13 \, \text{m} \).

2.5 m
4.0 m
3.13 m
1.8 m
3

A square loop of side \( 0.2 \, \text{m} \) with 30 turns carries \( 1.5 \, \text{A} \) in a magnetic field of \( 0.4 \, \text{T} \). The plane of the loop is at \( 60^\circ \) to the field. What is the torque?

Torque \( \tau = N I A B \sin \theta \), where \( A = 0.2 \times 0.2 = 0.04 \, \text{m}^2 \).

\( \tau = 30 \times 1.5 \times 0.04 \times 0.4 \times \sin 60^\circ = 1.8 \times 0.4 \times 0.866 = 0.6235 \approx 0.62 \, \text{N m} \).

0.31 N m
0.62 N m
1.24 N m
0.93 N m
2

Which rule is used to determine the direction of the magnetic field produced by a current-carrying straight wire?

The direction of the magnetic field around a current-carrying straight wire is determined by the right-hand rule: grasp the wire with your right hand, thumb pointing in the direction of the current, and your fingers curl in the direction of the magnetic field.

Left-hand rule
Right-hand rule
Fleming’s left-hand rule
Ampere’s swimming rule
1

A wire of length \( 2 \, \text{m} \) carries a current of \( 5 \, \text{A} \) and is placed perpendicular to a magnetic field of \( 0.4 \, \text{T} \). What is the magnitude of the force on the wire?

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

\( F = 5 \times 2 \times 0.4 = 4 \, \text{N} \).

4 N
2 N
6 N
8 N
1

A long wire carries \( 20 \, \text{A} \). At what distance from the wire is the magnetic field \( 4 \times 10^{-6} \, \text{T} \)? (\( \mu_0 = 4 \pi \times 10^{-7} \, \text{T m/A} \))

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

\( r = \frac{4 \pi \times 10^{-7} \times 20}{2 \pi \times 4 \times 10^{-6}} = \frac{8 \times 10^{-6}}{8 \times 10^{-6}} = 1 \, \text{m} \).

0.5 m
2 m
1 m
1.5 m
3

A proton moves at \( 6 \times 10^7 \, \text{m/s} \) perpendicular to a field of \( 0.1 \, \text{T} \). What is the magnetic force? (Charge = \( 1.6 \times 10^{-19} \, \text{C} \))

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

\( F = 1.6 \times 10^{-19} \times 6 \times 10^7 \times 0.1 = 9.6 \times 10^{-13} \, \text{N} \).

4.8 × 10⁻¹³ N
1.92 × 10⁻¹² N
9.6 × 10⁻¹³ N
1.44 × 10⁻¹² N
3

A long straight wire carries a current of \( 28 \, \text{A} \). What is the magnetic field at a distance of \( 0.35 \, \text{m} \) from it? (\( \mu_0 = 4 \pi \times 10^{-7} \, \text{T m/A} \))

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

\( B = \frac{4 \pi \times 10^{-7} \times 28}{2 \pi \times 0.35} = \frac{112 \times 10^{-7}}{0.7} = 1.6 \times 10^{-5} \, \text{T} \).

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

A straight wire of length \( 0.7 \, \text{m} \) carries a current of \( 5 \, \text{A} \) perpendicular to a uniform magnetic field of \( 0.6 \, \text{T} \). What is the force on the wire?

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

\( F = 5 \times 0.7 \times 0.6 = 2.1 \, \text{N} \).

2.1 N
4.2 N
1.05 N
3.15 N
1

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