Magnetism and Matter Chapter-Wise Test 17

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A magnetic dipole of moment \( 0.5 \, \text{A m}^2 \) is in a uniform field of \( 0.3 \, \text{T} \) at \( 60^\circ \). What is the torque on it?

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

Given: \( m = 0.5 \, \text{A m}^2 \), \( B = 0.3 \, \text{T} \), \( \theta = 60^\circ \), \( \sin 60^\circ = \frac{\sqrt{3}}{2} \approx 0.866 \).

Substitute: \( \tau = 0.5 \times 0.3 \times 0.866 \approx 0.1299 \, \text{N m} \approx 0.13 \, \text{N m} \).

0.13 N m
0.15 N m
0.17 N m
0.20 N m
1

The net magnetic flux through a closed surface surrounding a solenoid is:

Gauss’s law for magnetism states that the net magnetic flux through any closed surface is zero, as magnetic field lines form closed loops with no monopoles.

Positive
Negative
Depends on solenoid strength
Zero
4

A bar magnet’s field strength decreases with distance because:

The magnetic field strength of a bar magnet decreases with distance as the field lines spread out, reducing their density and thus the field intensity, following an inverse cube relationship (\( B \propto 1/r^3 \)) for a dipole at large distances.

The magnet loses its poles
The field becomes uniform
The field lines spread out
The field lines contract
3

A dipole with \( m = 0.7 \, \text{A m}^2 \) in a field \( B = 0.4 \, \text{T} \) at \( 180^\circ \) has potential energy:

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

Given: \( m = 0.7 \, \text{A m}^2 \), \( B = 0.4 \, \text{T} \), \( \theta = 180^\circ \), \( \cos 180^\circ = -1 \).

\( U_m = -0.7 \times 0.4 \times (-1) = 0.28 \, \text{J} \).

0.25 J
0.28 J
0.30 J
0.35 J
2

The net magnetic flux through a closed surface surrounding a bar magnet is:

Gauss’s law for magnetism states that the net magnetic flux through any closed surface is zero, as magnetic field lines form closed loops with no monopoles.

Positive
Negative
Depends on the magnet’s strength
Zero
4

A solenoid with 900 turns per meter and current \( 3 \, \text{A} \) 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: \( n = 900 \, \text{m}^{-1} \), \( I = 3 \, \text{A} \), \( \mu_r = 250 \), \( \mu_0 = 4\pi \times 10^{-7} \).

\( B = 4\pi \times 10^{-7} \times 250 \times 900 \times 3 = 0.8478 \, \text{T} \approx 0.85 \, \text{T} \).

0.80 T
0.85 T
0.90 T
0.95 T
2

A paramagnetic material has \( \chi = 5 \times 10^{-4} \) and \( H = 2 \times 10^3 \, \text{A m}^{-1} \). What is its magnetization \( M \)?

\( M = \chi H \).

Given: \( \chi = 5 \times 10^{-4} \), \( H = 2 \times 10^3 \, \text{A m}^{-1} \).

Substitute: \( M = 5 \times 10^{-4} \times 2 \times 10^3 = 1 \, \text{A m}^{-1} \).

0.5 A m⁻¹
1 A m⁻¹
1.5 A m⁻¹
2 A m⁻¹
2

A bar magnet with \( m = 2.5 \, \text{A m}^2 \) is at \( 0.7 \, \text{m} \) 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: \( m = 2.5 \, \text{A m}^2 \), \( r = 0.7 \, \text{m} \), \( \frac{\mu_0}{4\pi} = 10^{-7} \).

\( B = 10^{-7} \times \frac{2 \times 2.5}{(0.7)^3} = 10^{-7} \times \frac{5.0}{0.343} \approx 1.46 \times 10^{-6} \, \text{T} \).

1.2 × 10⁻⁶ T
1.3 × 10⁻⁶ T
1.4 × 10⁻⁶ T
1.46 × 10⁻⁶ T
4

The magnetic potential energy of a dipole with \( m = 0.7 \, \text{A m}^2 \) in a field \( B = 0.2 \, \text{T} \) at \( 0^\circ \) is:

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

Given: \( m = 0.7 \, \text{A m}^2 \), \( B = 0.2 \, \text{T} \), \( \theta = 0^\circ \), \( \cos 0^\circ = 1 \).

Substitute: \( U_m = -0.7 \times 0.2 \times 1 = -0.14 \, \text{J} \).

-0.14 J
-0.16 J
-0.18 J
-0.20 J
1

A solenoid has 500 turns per meter and carries a current of \( 3 \, \text{A} \). What is the magnetic intensity \( H \) inside the solenoid?

Magnetic intensity \( H = n I \), where \( n \) is turns per unit length and \( I \) is current.

Given: \( n = 500 \, \text{m}^{-1} \), \( I = 3 \, \text{A} \).

Substitute: \( H = 500 \times 3 = 1500 \, \text{A m}^{-1} \).

1000 A m⁻¹
1200 A m⁻¹
1500 A m⁻¹
1800 A m⁻¹
3

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