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  4. Magnetism and Matter Chapter-Wise Test 3

Magnetism and Matter Chapter-Wise Test 3

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A bar magnet with \( m = 0.75 \, \text{A m}^2 \) produces a field at \( 0.15 \, \text{m} \) on its equatorial line. What is \( B \)? (Take \( \mu_0 = 4\pi \times 10^{-7} \, \text{T m A}^{-1} \)).

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

Given: \( m = 0.75 \, \text{A m}^2 \), \( r = 0.15 \, \text{m} \), \( \frac{\mu_0}{4\pi} = 10^{-7} \).

\( B = 10^{-7} \times \frac{0.75}{(0.15)^3} = 10^{-7} \times \frac{0.75}{0.003375} \approx 2.222 \times 10^{-5} \, \text{T} \approx 2.22 \times 10^{-5} \, \text{T} \).

2.0 × 10⁻⁵ T
2.1 × 10⁻⁵ T
2.22 × 10⁻⁵ T
2.3 × 10⁻⁵ T
3

A bar magnet with magnetic moment \( 0.9 \, \text{A m}^2 \) is placed at a distance of \( 0.3 \, \text{m} \) 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: \( m = 0.9 \, \text{A m}^2 \), \( r = 0.3 \, \text{m} \), \( \frac{\mu_0}{4\pi} = 10^{-7} \).

Substitute: \( B = 10^{-7} \times \frac{2 \times 0.9}{(0.3)^3} = 10^{-7} \times \frac{1.8}{0.027} \approx 6.67 \times 10^{-6} \, \text{T} \).

6.67 × 10⁻⁶ T
7.0 × 10⁻⁶ T
7.5 × 10⁻⁶ T
8.0 × 10⁻⁶ T
1

A dipole with \( m = 0.8 \, \text{A m}^2 \) in a field \( B = 0.3 \, \text{T} \) at \( 90^\circ \) has potential energy:

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

Given: \( m = 0.8 \, \text{A m}^2 \), \( B = 0.3 \, \text{T} \), \( \theta = 90^\circ \), \( \cos 90^\circ = 0 \).

\( U_m = -0.8 \times 0.3 \times 0 = 0 \, \text{J} \).

-0.24 J
0 J
0.24 J
-0.12 J
2

The magnetic field contribution \( B_m \) due to a material with \( M = 1.6 \times 10^5 \, \text{A m}^{-1} \) is: (Take \( \mu_0 = 4\pi \times 10^{-7} \, \text{T m A}^{-1} \)).

\( B_m = \mu_0 M \).

Given: \( M = 1.6 \times 10^5 \, \text{A m}^{-1} \), \( \mu_0 = 4\pi \times 10^{-7} \).

\( B_m = 4\pi \times 10^{-7} \times 1.6 \times 10^5 = 0.20096 \, \text{T} \approx 0.20 \, \text{T} \).

0.18 T
0.19 T
0.20 T
0.22 T
3

The magnetic potential energy of a dipole with moment \( 0.4 \, \text{A m}^2 \) in a uniform magnetic field of \( 0.5 \, \text{T} \) when aligned at \( 30^\circ \) to the field is:

Magnetic potential energy is \( U_m = -m B \cos\theta \).

Given: \( m = 0.4 \, \text{A m}^2 \), \( B = 0.5 \, \text{T} \), \( \theta = 30^\circ \), \( \cos 30^\circ = \frac{\sqrt{3}}{2} \approx 0.866 \).

Substitute: \( U_m = -0.4 \times 0.5 \times 0.866 = -0.1732 \, \text{J} \approx -0.17 \, \text{J} \).

-0.15 J
-0.17 J
-0.20 J
-0.25 J
2

A bar magnet with magnetic moment \( 0.8 \, \text{A m}^2 \) is placed at a distance of \( 0.4 \, \text{m} \) 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: \( m = 0.8 \, \text{A m}^2 \), \( r = 0.4 \, \text{m} \), \( \frac{\mu_0}{4\pi} = 10^{-7} \).

Substitute: \( B = 10^{-7} \times \frac{2 \times 0.8}{(0.4)^3} = 10^{-7} \times \frac{1.6}{0.064} = 2.5 \times 10^{-6} \, \text{T} \).

2.5 × 10⁻⁶ T
3.0 × 10⁻⁶ T
3.5 × 10⁻⁶ T
4.0 × 10⁻⁶ T
1

A paramagnetic material with \( \chi = 8 \times 10^{-4} \) in \( H = 2500 \, \text{A m}^{-1} \) has magnetization \( M \):

\( M = \chi H \).

Given: \( \chi = 8 \times 10^{-4} \), \( H = 2500 \, \text{A m}^{-1} \).

\( M = 8 \times 10^{-4} \times 2500 = 2 \, \text{A m}^{-1} \).

1.5 A m⁻¹
2.0 A m⁻¹
2.5 A m⁻¹
3.0 A m⁻¹
2

A bar magnet produces a field of \( 5 \times 10^{-6} \, \text{T} \) at \( 0.5 \, \text{m} \) on its equatorial line. What is its magnetic moment? (Take \( \mu_0 = 4\pi \times 10^{-7} \, \text{T m A}^{-1} \)).

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

Given: \( B = 5 \times 10^{-6} \, \text{T} \), \( r = 0.5 \, \text{m} \), \( \frac{\mu_0}{4\pi} = 10^{-7} \).

\( m = \frac{5 \times 10^{-6} \times (0.5)^3}{10^{-7}} = \frac{5 \times 10^{-6} \times 0.125}{10^{-7}} = 6.25 \, \text{A m}^2 \).

5.0 A m²
6.25 A m²
7.5 A m²
8.0 A m²
2

A solenoid with 1100 turns per meter and current \( 2 \, \text{A} \) has a core with \( \mu_r = 250 \). What is \( B \) inside? (Take \( \mu_0 = 4\pi \times 10^{-7} \, \text{T m A}^{-1} \)).

\( B = \mu_0 \mu_r n I \).

Given: \( n = 1100 \, \text{m}^{-1} \), \( I = 2 \, \text{A} \), \( \mu_r = 250 \), \( \mu_0 = 4\pi \times 10^{-7} \).

\( B = 4\pi \times 10^{-7} \times 250 \times 1100 \times 2 = 0.6908 \, \text{T} \approx 0.69 \, \text{T} \).

0.65 T
0.69 T
0.72 T
0.75 T
2

A dipole with \( m = 0.4 \, \text{A m}^2 \) in \( B = 0.8 \, \text{T} \) at \( 60^\circ \) has torque:

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

Given: \( m = 0.4 \, \text{A m}^2 \), \( B = 0.8 \, \text{T} \), \( \theta = 60^\circ \), \( \sin 60^\circ = \frac{\sqrt{3}}{2} \approx 0.866 \).

\( \tau = 0.4 \times 0.8 \times 0.866 \approx 0.277 \, \text{N m} \approx 0.28 \, \text{N m} \).

0.24 N m
0.26 N m
0.27 N m
0.28 N m
4

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