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  4. Magnetism and Matter Chapter-Wise Test 2

Magnetism and Matter Chapter-Wise Test 2

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A bar magnet with \( m = 2.8 \, \text{A m}^2 \) is at \( 0.4 \, \text{m} \) 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: \( m = 2.8 \, \text{A m}^2 \), \( r = 0.4 \, \text{m} \), \( \frac{\mu_0}{4\pi} = 10^{-7} \).

\( B = 10^{-7} \times \frac{2 \times 2.8}{(0.4)^3} = 10^{-7} \times \frac{5.6}{0.064} = 8.75 \times 10^{-6} \, \text{T} \).

8.0 × 10⁻⁶ T
8.5 × 10⁻⁶ T
8.7 × 10⁻⁶ T
8.75 × 10⁻⁶ T
4

The magnetic field lines inside a solenoid are nearly parallel because:

Inside a solenoid, the magnetic field lines are nearly parallel due to the uniform distribution of current loops along its length, creating a consistent field direction and magnitude, especially in a long solenoid where end effects are minimal.

The field cancels outside
The solenoid has no poles
The current loops produce a uniform field
The field lines diverge internally
3

The primary reason a diamagnetic material develops a weak opposing magnetic moment in an external field is:

In diamagnetic materials, an external magnetic field induces orbital currents in atoms that oppose the applied field, per Lenz’s law. This results in a weak, negative magnetization, as all electrons contribute to this effect in materials with no net magnetic moment.

Alignment of intrinsic magnetic moments
Spontaneous domain formation
Thermal agitation of electrons
Induced currents opposing the field
4

A solenoid with 1300 turns per meter and current \( 2 \, \text{A} \) has a core with \( \mu_r = 100 \). 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: \( n = 1300 \, \text{m}^{-1} \), \( I = 2 \, \text{A} \), \( \mu_r = 100 \), \( \mu_0 = 4\pi \times 10^{-7} \).

\( B = 4\pi \times 10^{-7} \times 100 \times 1300 \times 2 = 0.32656 \, \text{T} \approx 0.33 \, \text{T} \).

0.30 T
0.33 T
0.35 T
0.38 T
2

A dipole with \( m = 0.7 \, \text{A m}^2 \) in \( B = 0.6 \, \text{T} \) at \( 30^\circ \) has torque:

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

Given: \( m = 0.7 \, \text{A m}^2 \), \( B = 0.6 \, \text{T} \), \( \theta = 30^\circ \), \( \sin 30^\circ = 0.5 \).

\( \tau = 0.7 \times 0.6 \times 0.5 = 0.21 \, \text{N m} \).

0.18 N m
0.19 N m
0.20 N m
0.21 N m
4

A dipole with \( m = 0.5 \, \text{A m}^2 \) in a field \( B = 0.6 \, \text{T} \) at \( 180^\circ \) has potential energy:

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

Given: \( m = 0.5 \, \text{A m}^2 \), \( B = 0.6 \, \text{T} \), \( \theta = 180^\circ \), \( \cos 180^\circ = -1 \).

\( U_m = -0.5 \times 0.6 \times (-1) = 0.3 \, \text{J} \).

0.25 J
0.3 J
0.35 J
0.4 J
2

Which material has \( \chi = -1 \) and completely expels magnetic field lines?

Superconductors have \( \chi = -1 \) and exhibit perfect diamagnetism, expelling all magnetic field lines (Meissner effect).

Diamagnetic
Paramagnetic
Superconductor
Ferromagnetic
3

A paramagnetic material with \( \chi = 4 \times 10^{-4} \) in \( H = 2000 \, \text{A m}^{-1} \) has magnetization \( M \):

\( M = \chi H \).

Given: \( \chi = 4 \times 10^{-4} \), \( H = 2000 \, \text{A m}^{-1} \).

\( M = 4 \times 10^{-4} \times 2000 = 0.8 \, \text{A m}^{-1} \).

0.6 A m⁻¹
0.8 A m⁻¹
1.0 A m⁻¹
1.2 A m⁻¹
2

A magnetic dipole of moment \( 0.2 \, \text{A m}^2 \) is in a uniform field of \( 0.9 \, \text{T} \) at \( 30^\circ \). What is the torque on it?

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

Given: \( m = 0.2 \, \text{A m}^2 \), \( B = 0.9 \, \text{T} \), \( \theta = 30^\circ \), \( \sin 30^\circ = 0.5 \).

Substitute: \( \tau = 0.2 \times 0.9 \times 0.5 = 0.09 \, \text{N m} \).

0.09 N m
0.10 N m
0.12 N m
0.15 N m
1

The magnetic potential energy of a dipole with \( m = 0.8 \, \text{A m}^2 \) in a field \( B = 0.25 \, \text{T} \) at \( 180^\circ \) is:

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

Given: \( m = 0.8 \, \text{A m}^2 \), \( B = 0.25 \, \text{T} \), \( \theta = 180^\circ \), \( \cos 180^\circ = -1 \).

Substitute: \( U_m = -0.8 \times 0.25 \times (-1) = 0.2 \, \text{J} \).

0.2 J
0.15 J
0.1 J
0.25 J
1

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