Magnetism and Matter Chapter-Wise Test 15

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A material’s weak repulsion from a magnetic field is due to:

Diamagnetic materials exhibit weak repulsion from a magnetic field because an external field induces small currents in their atoms that generate an opposing magnetic moment, per Lenz’s law, resulting in a slight reduction of the field inside the material.

Strong domain alignment
Partial moment alignment
Complete field expulsion
Induced opposing moments
4

A magnetic dipole of moment \( 0.15 \, \text{A m}^2 \) is in a uniform field of \( 0.8 \, \text{T} \) at \( 60^\circ \). What is the torque on it?

Torque is \( \tau = m B \sin\theta \).

Given: \( m = 0.15 \, \text{A m}^2 \), \( B = 0.8 \, \text{T} \), \( \theta = 60^\circ \), \( \sin 60^\circ = \frac{\sqrt{3}}{2} \approx 0.866 \).

Substitute: \( \tau = 0.15 \times 0.8 \times 0.866 \approx 0.1039 \, \text{N m} \approx 0.104 \, \text{N m} \).

0.104 N m
0.12 N m
0.15 N m
0.18 N m
1

A solenoid with magnetic field \( B = 0.8 \, \text{T} \) inside has a core with \( \mu_r = 400 \) and \( n = 800 \, \text{m}^{-1} \). What is the current \( I \) in the solenoid? (Take \( \mu_0 = 4\pi \times 10^{-7} \, \text{T m A}^{-1} \)).

\( B = \mu_0 \mu_r n I \), so \( I = \frac{B}{\mu_0 \mu_r n} \).

Given: \( B = 0.8 \, \text{T} \), \( \mu_r = 400 \), \( n = 800 \, \text{m}^{-1} \), \( \mu_0 = 4\pi \times 10^{-7} \).

Substitute: \( I = \frac{0.8}{4\pi \times 10^{-7} \times 400 \times 800} = \frac{0.8}{4\pi \times 3.2 \times 10^{-2}} \approx 1.989 \, \text{A} \approx 2 \, \text{A} \).

1.5 A
2 A
2.5 A
3 A
2

A bar magnet with original \( m = 2.0 \, \text{A m}^2 \) is cut transversely into two equal parts. What is \( m \) of each part?

When cut transversely, each part has half the original magnetic moment.

Given: \( m = 2.0 \, \text{A m}^2 \).

Each part: \( m' = \frac{2.0}{2} = 1.0 \, \text{A m}^2 \).

0.8 A m²
0.9 A m²
0.95 A m²
1.0 A m²
4

A material’s weak enhancement of the magnetic field inside it is due to:

Paramagnetic materials weakly enhance the magnetic field inside them because their atomic magnetic moments partially align with the external field, producing a small positive magnetization that adds to the applied field.

Complete domain alignment
Weak alignment of atomic moments
Opposing induced currents
Expulsion of field lines
2

A magnetic dipole placed in a uniform magnetic field experiences no net force but a torque. This is because:

In a uniform magnetic field, the forces on the north and south poles of a dipole are equal and opposite, canceling out to produce no net force. However, these forces act at different points, creating a torque that tends to align the dipole with the field.

Forces on the poles cancel out but produce a couple
The field exerts a net force in one direction
The dipole is inherently unstable in uniform fields
The field strength varies across the dipole
1

A solenoid with 600 turns per meter carries a current of \( 3.5 \, \text{A} \). What is the magnetic intensity \( H \) inside?

Magnetic intensity \( H = n I \).

Given: \( n = 600 \, \text{m}^{-1} \), \( I = 3.5 \, \text{A} \).

Substitute: \( H = 600 \times 3.5 = 2100 \, \text{A m}^{-1} \).

2100 A m⁻¹
2200 A m⁻¹
2300 A m⁻¹
2400 A m⁻¹
1

A material with \( B = 0.4 \, \text{T} \) and \( H = 2000 \, \text{A m}^{-1} \) has \( M \): (Take \( \mu_0 = 4\pi \times 10^{-7} \, \text{T m A}^{-1} \)).

\( B = \mu_0 (H + M) \), so \( M = \frac{B}{\mu_0} - H \).

Given: \( B = 0.4 \, \text{T} \), \( H = 2000 \, \text{A m}^{-1} \), \( \mu_0 = 4\pi \times 10^{-7} \).

\( \frac{B}{\mu_0} = \frac{0.4}{4\pi \times 10^{-7}} \approx 3.183 \times 10^5 \, \text{A m}^{-1} \).

\( M = 3.183 \times 10^5 - 2000 \approx 3.163 \times 10^5 \, \text{A m}^{-1} \).

3.0 × 10⁵ A m⁻¹
3.1 × 10⁵ A m⁻¹
3.15 × 10⁵ A m⁻¹
3.163 × 10⁵ A m⁻¹
4

What is the magnetic field contribution \( B_m \) due to a material with magnetization \( M = 4 \times 10^5 \, \text{A m}^{-1} \)? (Take \( \mu_0 = 4\pi \times 10^{-7} \, \text{T m A}^{-1} \)).

\( B_m = \mu_0 M \).

Given: \( M = 4 \times 10^5 \, \text{A m}^{-1} \), \( \mu_0 = 4\pi \times 10^{-7} \).

Substitute: \( B_m = 4\pi \times 10^{-7} \times 4 \times 10^5 = 1.6\pi \times 10^{-1} \approx 0.5 \, \text{T} \).

0.3 T
0.4 T
0.5 T
0.6 T
3

A material with \( B = 0.35 \, \text{T} \) and \( H = 1800 \, \text{A m}^{-1} \) has \( M \): (Take \( \mu_0 = 4\pi \times 10^{-7} \, \text{T m A}^{-1} \)).

\( B = \mu_0 (H + M) \), so \( M = \frac{B}{\mu_0} - H \).

Given: \( B = 0.35 \, \text{T} \), \( H = 1800 \, \text{A m}^{-1} \), \( \mu_0 = 4\pi \times 10^{-7} \).

\( \frac{B}{\mu_0} = \frac{0.35}{4\pi \times 10^{-7}} \approx 2.785 \times 10^5 \, \text{A m}^{-1} \).

\( M = 2.785 \times 10^5 - 1800 \approx 2.767 \times 10^5 \, \text{A m}^{-1} \).

2.7 × 10⁵ A m⁻¹
2.75 × 10⁵ A m⁻¹
2.76 × 10⁵ A m⁻¹
2.767 × 10⁵ A m⁻¹
4

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