Magnetism and Matter Chapter-Wise Test 16

Correct answer Carries: 4.

Wrong Answer Carries: -1.

The magnetic field contribution \( B_m \) due to a material with \( M = 1.8 \times 10^5 \, \text{A m}^{-1} \) is: (Take \( \mu_0 = 4\pi \times 10^{-7} \, \text{T m A}^{-1} \)).

\( B_m = \mu_0 M \).

Given: \( M = 1.8 \times 10^5 \, \text{A m}^{-1} \), \( \mu_0 = 4\pi \times 10^{-7} \).

\( B_m = 4\pi \times 10^{-7} \times 1.8 \times 10^5 = 0.22608 \, \text{T} \approx 0.23 \, \text{T} \).

0.20 T
0.22 T
0.23 T
0.25 T
3

A material with \( \mu_r = 0 \) completely expels magnetic fields and is classified as:

A material with \( \mu_r = 0 \) (and \( \chi = -1 \)) exhibits perfect diamagnetism, characteristic of a superconductor due to the Meissner effect.

Diamagnetic
Paramagnetic
Ferromagnetic
Superconductor
4

A material has \( B = 0.42 \, \text{T} \) and \( M = 3.0 \times 10^5 \, \text{A m}^{-1} \). What is \( H \)? (Take \( \mu_0 = 4\pi \times 10^{-7} \, \text{T m A}^{-1} \)).

\( B = \mu_0 (H + M) \), so \( H = \frac{B}{\mu_0} - M \).

Given: \( B = 0.42 \, \text{T} \), \( M = 3.0 \times 10^5 \, \text{A m}^{-1} \), \( \mu_0 = 4\pi \times 10^{-7} \).

\( \frac{B}{\mu_0} = \frac{0.42}{4\pi \times 10^{-7}} \approx 3.338 \times 10^5 \, \text{A m}^{-1} \).

\( H = 3.338 \times 10^5 - 3.0 \times 10^5 = 3.38 \times 10^4 \, \text{A m}^{-1} \).

3.2 × 10⁴ A m⁻¹
3.38 × 10⁴ A m⁻¹
3.5 × 10⁴ A m⁻¹
3.6 × 10⁴ A m⁻¹
2

The potential energy of a magnetic dipole in a uniform field is highest when:

The potential energy \( U = -m B \cos\theta \) is highest when \( \cos\theta = -1 \), i.e., \( \theta = 180^\circ \), when the dipole is anti-parallel to the field. This is the least stable position, as energy is maximized.

It is perpendicular to the field
It is anti-parallel to the field
It is parallel to the field
The field is zero
2

The absence of a net force on a magnetic dipole in a uniform field implies:

In a uniform magnetic field, the forces on the dipole’s poles are equal in magnitude and opposite in direction, resulting in no net translational force. This occurs because the field strength does not vary, unlike in a non-uniform field where a gradient would produce a net force.

The field varies across the dipole
The field strength is constant
The dipole is inherently balanced
The dipole cancels the field
2

A material’s magnetic susceptibility becomes zero when:

For a superconductor below its critical temperature, the susceptibility \( \chi = -1 \) and relative permeability \( \mu_r = 0 \), but in a broader context, susceptibility can approach zero for non-magnetic materials or when magnetization is negligible compared to the applied field. However, the question implies a specific state, and superconductors achieve this uniquely with perfect diamagnetism.

It is heated beyond its melting point
It is placed in a vacuum
It becomes ferromagnetic
It exhibits perfect diamagnetism
4

A material with susceptibility \( \chi = -5 \times 10^{-5} \) has a relative permeability \( \mu_r \) of:

\( \mu_r = 1 + \chi \).

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

Substitute: \( \mu_r = 1 - 5 \times 10^{-5} = 0.99995 \).

0.99995
1.00005
0.99990
1.00010
1

The net magnetic flux through a closed surface around a toroid is:

Gauss’s law for magnetism states that the net magnetic flux through any closed surface is zero, even for a toroid, as magnetic field lines form closed loops.

Positive
Negative
Depends on current
Zero
4

A solenoid with 800 turns per meter carries a current of \( 2.5 \, \text{A} \). What is the magnetic intensity \( H \) inside?

Magnetic intensity \( H = n I \).

Given: \( n = 800 \, \text{m}^{-1} \), \( I = 2.5 \, \text{A} \).

Substitute: \( H = 800 \times 2.5 = 2000 \, \text{A m}^{-1} \).

2000 A m⁻¹
2100 A m⁻¹
2200 A m⁻¹
2300 A m⁻¹
1

A bar magnet with \( m = 0.9 \, \text{A m}^2 \) is at \( 0.2 \, \text{m} \) 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: \( m = 0.9 \, \text{A m}^2 \), \( r = 0.2 \, \text{m} \), \( \frac{\mu_0}{4\pi} = 10^{-7} \).

\( B = 10^{-7} \times \frac{2 \times 0.9}{(0.2)^3} = 10^{-7} \times \frac{1.8}{0.008} = 2.25 \times 10^{-5} \, \text{T} \).

1.5 × 10⁻⁵ T
1.8 × 10⁻⁵ T
2.0 × 10⁻⁵ T
2.25 × 10⁻⁵ T
4

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