Magnetism and Matter Chapter-Wise Test 11

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A bar magnet with \( m = 3.0 \, \text{A m}^2 \) is at \( 0.5 \, \text{m} \) along its axis. What is \( B \)? (Take \( \mu_0 = 4\pi \times 10^{-7} \, \text{T m A}^{-1} \)).

\( B = \frac{\mu_0}{4\pi} \frac{2m}{r^3} \).

Given: \( m = 3.0 \, \text{A m}^2 \), \( r = 0.5 \, \text{m} \), \( \frac{\mu_0}{4\pi} = 10^{-7} \).

\( B = 10^{-7} \times \frac{2 \times 3.0}{(0.5)^3} = 10^{-7} \times \frac{6.0}{0.125} = 4.8 \times 10^{-6} \, \text{T} \).

4.2 × 10⁻⁶ T
4.5 × 10⁻⁶ T
4.7 × 10⁻⁶ T
4.8 × 10⁻⁶ T
4

A solenoid produces \( B = 0.9 \, \text{T} \) with a core of \( \mu_r = 300 \) and \( n = 1500 \, \text{m}^{-1} \). What is the current \( I \)? (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: \( B = 0.9 \, \text{T} \), \( \mu_r = 300 \), \( n = 1500 \, \text{m}^{-1} \), \( \mu_0 = 4\pi \times 10^{-7} \).

\( I = \frac{0.9}{4\pi \times 10^{-7} \times 300 \times 1500} = \frac{0.9}{5.654 \times 10^{-1}} \approx 1.59 \, \text{A} \approx 1.6 \, \text{A} \).

1.4 A
1.5 A
1.6 A
1.7 A
3

A material that enhances the magnetic field inside it significantly when placed in an external field is likely:

Ferromagnetic materials greatly enhance the magnetic field inside them due to the strong alignment and growth of magnetic domains, resulting in a high susceptibility and permeability, far exceeding that of paramagnetic or diamagnetic materials.

Diamagnetic
Ferromagnetic
Paramagnetic
Non-magnetic
2

A material that repels a magnet and is repelled by it in return is likely:

Superconductors, exhibiting perfect diamagnetism (Meissner effect), expel magnetic fields completely, causing mutual repulsion with a magnet. This is distinct from weak diamagnetic repulsion and stronger than other material responses.

Paramagnetic
Ferromagnetic
Superconductor
Non-magnetic
3

Which material has a small positive \( \chi \) and aligns weakly with an external magnetic field?

Paramagnetic materials have a small positive \( \chi \) and align weakly with an external magnetic field due to the alignment of atomic dipoles.

Diamagnetic
Ferromagnetic
Paramagnetic
Superconductor
3

A material with \( B = 0.45 \, \text{T} \) and \( H = 3000 \, \text{A m}^{-1} \) 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: \( B = 0.45 \, \text{T} \), \( H = 3000 \, \text{A m}^{-1} \), \( \mu_0 = 4\pi \times 10^{-7} \).

\( \frac{B}{\mu_0} = \frac{0.45}{4\pi \times 10^{-7}} \approx 3.581 \times 10^5 \, \text{A m}^{-1} \).

\( M = 3.581 \times 10^5 - 3000 \approx 3.551 \times 10^5 \, \text{A m}^{-1} \).

3.5 × 10⁵ A m⁻¹
3.52 × 10⁵ A m⁻¹
3.54 × 10⁵ A m⁻¹
3.551 × 10⁵ A m⁻¹
4

A magnetic dipole of moment \( 0.25 \, \text{A m}^2 \) is in a uniform field of \( 0.5 \, \text{T} \) at \( 30^\circ \). What is the torque on it?

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

Given: \( m = 0.25 \, \text{A m}^2 \), \( B = 0.5 \, \text{T} \), \( \theta = 30^\circ \), \( \sin 30^\circ = 0.5 \).

Substitute: \( \tau = 0.25 \times 0.5 \times 0.5 = 0.0625 \, \text{N m} \).

0.0625 N m
0.075 N m
0.085 N m
0.1 N m
1

A solenoid with 300 turns per meter carries a current of \( 6 \, \text{A} \). What is the magnetic intensity \( H \) inside?

Magnetic intensity \( H = n I \).

Given: \( n = 300 \, \text{m}^{-1} \), \( I = 6 \, \text{A} \).

Substitute: \( H = 300 \times 6 = 1800 \, \text{A m}^{-1} \).

1800 A m⁻¹
2000 A m⁻¹
2200 A m⁻¹
2400 A m⁻¹
1

A dipole with \( m = 0.35 \, \text{A m}^2 \) in \( B = 0.9 \, \text{T} \) at \( 30^\circ \) has torque:

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

Given: \( m = 0.35 \, \text{A m}^2 \), \( B = 0.9 \, \text{T} \), \( \theta = 30^\circ \), \( \sin 30^\circ = 0.5 \).

\( \tau = 0.35 \times 0.9 \times 0.5 = 0.1575 \, \text{N m} \).

0.14 N m
0.15 N m
0.155 N m
0.1575 N m
4

A bar magnet with \( m = 2.5 \, \text{A m}^2 \) is at \( 0.5 \, \text{m} \) along its axis. What is \( B \)? (Take \( \mu_0 = 4\pi \times 10^{-7} \, \text{T m A}^{-1} \)).

\( B = \frac{\mu_0}{4\pi} \frac{2m}{r^3} \).

Given: \( m = 2.5 \, \text{A m}^2 \), \( r = 0.5 \, \text{m} \), \( \frac{\mu_0}{4\pi} = 10^{-7} \).

\( B = 10^{-7} \times \frac{2 \times 2.5}{(0.5)^3} = 10^{-7} \times \frac{5.0}{0.125} = 4.0 \times 10^{-6} \, \text{T} \).

3.6 × 10⁻⁶ T
3.8 × 10⁻⁶ T
3.9 × 10⁻⁶ T
4.0 × 10⁻⁶ T
4

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