Magnetism and Matter Chapter-Wise Test 18

Correct answer Carries: 4.

Wrong Answer Carries: -1.

The reason a paramagnetic material’s magnetization increases with a stronger external field is:

In paramagnetic materials, magnetization increases with a stronger external field because more atomic magnetic moments align with the field, overcoming random thermal motion, though the alignment remains partial compared to ferromagnetic materials.

Domain growth
Field expulsion
Complete moment alignment
Increased moment alignment
4

Unlike a bar magnet, an iron nail near a magnet experiences both a net force and a torque. This is primarily due to:

An iron nail, being a ferromagnetic material, becomes magnetized in the non-uniform field of a bar magnet, developing an induced magnetic moment. The field's gradient causes a net force toward the stronger region (typically the pole), while the misalignment of the induced moment with the field produces a torque.

The uniform nature of the magnet’s field
The non-uniform field inducing a magnetic moment
The nail’s inherent magnetic poles
The cancellation of forces on the nail
2

A bar magnet with \( m = 1.6 \, \text{A m}^2 \) is at \( 0.8 \, \text{m} \) 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: \( m = 1.6 \, \text{A m}^2 \), \( r = 0.8 \, \text{m} \), \( \frac{\mu_0}{4\pi} = 10^{-7} \).

\( B = 10^{-7} \times \frac{2 \times 1.6}{(0.8)^3} = 10^{-7} \times \frac{3.2}{0.512} \approx 6.25 \times 10^{-7} \, \text{T} \).

5.8 × 10⁻⁷ T
6.0 × 10⁻⁷ T
6.2 × 10⁻⁷ T
6.25 × 10⁻⁷ T
4

The magnetic field contribution \( B_m \) due to a material with \( M = 3.2 \times 10^5 \, \text{A m}^{-1} \) is: (Take \( \mu_0 = 4\pi \times 10^{-7} \, \text{T m A}^{-1} \)).

\( B_m = \mu_0 M \).

Given: \( M = 3.2 \times 10^5 \, \text{A m}^{-1} \), \( \mu_0 = 4\pi \times 10^{-7} \).

\( B_m = 4\pi \times 10^{-7} \times 3.2 \times 10^5 = 0.40192 \, \text{T} \approx 0.40 \, \text{T} \).

0.38 T
0.39 T
0.40 T
0.42 T
3

A diamagnetic material has a susceptibility \( \chi = -2 \times 10^{-5} \). What is its magnetic permeability \( \mu \) in terms of \( \mu_0 \)?

Magnetic permeability \( \mu = \mu_0 (1 + \chi) \).

Given: \( \chi = -2 \times 10^{-5} \).

Substitute: \( \mu = \mu_0 (1 - 2 \times 10^{-5}) = \mu_0 \times 0.99998 \).

\( \mu_0 \times 1.00002 \)
\( \mu_0 \times 0.99998 \)
\( \mu_0 \times 1.00005 \)
\( \mu_0 \times 0.99995 \)
2

A material with \( \mu_r = 350 \) and \( H = 400 \, \text{A m}^{-1} \) has \( B \): (Take \( \mu_0 = 4\pi \times 10^{-7} \, \text{T m A}^{-1} \)).

\( B = \mu_0 \mu_r H \).

Given: \( \mu_r = 350 \), \( H = 400 \, \text{A m}^{-1} \), \( \mu_0 = 4\pi \times 10^{-7} \).

\( B = 4\pi \times 10^{-7} \times 350 \times 400 = 0.17584 \, \text{T} \approx 0.18 \, \text{T} \).

0.16 T
0.17 T
0.18 T
0.19 T
3

A bar magnet with magnetic moment \( 1.5 \, \text{A m}^2 \) is placed at a distance of \( 0.6 \, \text{m} \) along its axis. What is the magnetic field \( B \) at that point? (Take \( \mu_0 = 4\pi \times 10^{-7} \, \text{T m A}^{-1} \)).

The magnetic field along the axis is \( B = \frac{\mu_0}{4\pi} \frac{2m}{r^3} \).

Given: \( m = 1.5 \, \text{A m}^2 \), \( r = 0.6 \, \text{m} \), \( \frac{\mu_0}{4\pi} = 10^{-7} \).

Substitute: \( B = 10^{-7} \times \frac{2 \times 1.5}{(0.6)^3} = 10^{-7} \times \frac{3.0}{0.216} \approx 1.389 \times 10^{-6} \, \text{T} \approx 1.39 × 10^{-6} \, \text{T} \).

1.39 × 10⁻⁶ T
1.5 × 10⁻⁶ T
1.6 × 10⁻⁶ T
1.7 × 10⁻⁶ T
1

A dipole with \( m = 0.4 \, \text{A m}^2 \) in a field \( B = 0.8 \, \text{T} \) at \( 0^\circ \) has potential energy:

\( U_m = -m B \cos\theta \).

Given: \( m = 0.4 \, \text{A m}^2 \), \( B = 0.8 \, \text{T} \), \( \theta = 0^\circ \), \( \cos 0^\circ = 1 \).

\( U_m = -0.4 \times 0.8 \times 1 = -0.32 \, \text{J} \).

-0.35 J
-0.32 J
-0.30 J
-0.28 J
2

The magnetic field contribution \( B_m \) due to a material with \( M = 2 \times 10^5 \, \text{A m}^{-1} \) is: (Take \( \mu_0 = 4\pi \times 10^{-7} \, \text{T m A}^{-1} \)).

\( B_m = \mu_0 M \).

Given: \( M = 2 \times 10^5 \, \text{A m}^{-1} \), \( \mu_0 = 4\pi \times 10^{-7} \).

\( B_m = 4\pi \times 10^{-7} \times 2 \times 10^5 = 0.2512 \, \text{T} \approx 0.25 \, \text{T} \).

0.20 T
0.22 T
0.25 T
0.30 T
3

A dipole with \( m = 0.5 \, \text{A m}^2 \) in \( B = 0.1 \, \text{T} \) at \( 60^\circ \) has torque:

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

Given: \( m = 0.5 \, \text{A m}^2 \), \( B = 0.1 \, \text{T} \), \( \theta = 60^\circ \), \( \sin 60^\circ = \frac{\sqrt{3}}{2} \approx 0.866 \).

\( \tau = 0.5 \times 0.1 \times 0.866 = 0.0433 \, \text{N m} \approx 0.043 \, \text{N m} \).

0.03 N m
0.035 N m
0.04 N m
0.043 N m
4

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