Magnetism and Matter Chapter-Wise Test 2

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A material has \( B = 0.54 \, \text{T} \) and \( M = 4.0 \times 10^5 \, \text{A m}^{-1} \). 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: \( B = 0.54 \, \text{T} \), \( M = 4.0 \times 10^5 \, \text{A m}^{-1} \), \( \mu_0 = 4\pi \times 10^{-7} \).

\( \frac{B}{\mu_0} = \frac{0.54}{4\pi \times 10^{-7}} \approx 4.297 \times 10^5 \, \text{A m}^{-1} \).

\( H = 4.297 \times 10^5 - 4.0 \times 10^5 = 2.97 \times 10^4 \, \text{A m}^{-1} \approx 3.0 × 10^4 \, \text{A m}^{-1} \).

2.8 × 10⁴ A m⁻¹
3.0 × 10⁴ A m⁻¹
3.2 × 10⁴ A m⁻¹
3.4 × 10⁴ A m⁻¹
2

A material that conducts electricity perfectly while expelling magnetic fields relies on:

Superconductors achieve perfect conductivity and expel magnetic fields (Meissner effect) below their critical temperature, where induced surface currents counteract the external field, a phenomenon tied to zero resistance and perfect diamagnetism.

Weak repulsion
Domain alignment
Atomic moment alignment
Induced surface currents
4

A bar magnet with magnetic moment \( 2.0 \, \text{A m}^2 \) is placed at a distance of \( 0.2 \, \text{m} \) 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: \( m = 2.0 \, \text{A m}^2 \), \( r = 0.2 \, \text{m} \), \( \frac{\mu_0}{4\pi} = 10^{-7} \).

Substitute: \( B = 10^{-7} \times \frac{2 \times 2.0}{(0.2)^3} = 10^{-7} \times \frac{4.0}{0.008} = 5.0 \times 10^{-5} \, \text{T} \).

5.0 × 10⁻⁵ T
5.5 × 10⁻⁵ T
6.0 × 10⁻⁵ T
6.5 × 10⁻⁵ T
1

A material that conducts electricity perfectly and repels magnetic fields is likely operating as:

Superconductors, when below their critical temperature, exhibit zero electrical resistance (perfect conductivity) and perfect diamagnetism (repelling magnetic fields via the Meissner effect), a unique combination not seen in other materials.

A diamagnetic material
A paramagnetic material
A ferromagnetic material
A superconductor
4

A bar magnet’s field resembles that of a solenoid because:

A bar magnet’s magnetic field mimics that of a solenoid due to the analogy that its magnetization can be modeled as arising from circulating currents within the material, similar to the current loops in a solenoid, producing comparable field patterns with north and south poles.

It can be modeled as circulating currents
It has a uniform field throughout
Its field lines are straight and parallel
It lacks magnetic poles
1

The magnetic field contribution \( B_m \) due to a material with \( M = 3 \times 10^5 \, \text{A m}^{-1} \) is: (Take \( \mu_0 = 4\pi \times 10^{-7} \, \text{T m A}^{-1} \)).

\( B_m = \mu_0 M \).

Given: \( M = 3 \times 10^5 \, \text{A m}^{-1} \), \( \mu_0 = 4\pi \times 10^{-7} \).

\( B_m = 4\pi \times 10^{-7} \times 3 \times 10^5 = 0.3768 \, \text{T} \approx 0.38 \, \text{T} \).

0.35 T
0.36 T
0.38 T
0.40 T
3

A material that exhibits strong magnetism due to the growth and alignment of microscopic domains when placed in an external field is:

Ferromagnetic materials have microscopic domains that align and grow in the presence of an external magnetic field, leading to strong magnetization. This cooperative behavior amplifies the field inside the material significantly.

Diamagnetic
Paramagnetic
Superconductor
Ferromagnetic
4

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 magnet strength
Zero
4

A material with \( \mu_r = 600 \) and \( H = 300 \, \text{A m}^{-1} \) has \( B \): (Take \( \mu_0 = 4\pi \times 10^{-7} \, \text{T m A}^{-1} \)).

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

Given: \( \mu_r = 600 \), \( H = 300 \, \text{A m}^{-1} \), \( \mu_0 = 4\pi \times 10^{-7} \).

\( B = 4\pi \times 10^{-7} \times 600 \times 300 = 0.22608 \, \text{T} \approx 0.23 \, \text{T} \).

0.20 T
0.22 T
0.23 T
0.25 T
3

The magnetic potential energy of a dipole with \( m = 0.6 \, \text{A m}^2 \) in a field \( B = 0.3 \, \text{T} \) at \( 60^\circ \) is:

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

Given: \( m = 0.6 \, \text{A m}^2 \), \( B = 0.3 \, \text{T} \), \( \theta = 60^\circ \), \( \cos 60^\circ = 0.5 \).

Substitute: \( U_m = -0.6 \times 0.3 \times 0.5 = -0.09 \, \text{J} \).

-0.09 J
-0.12 J
-0.15 J
-0.18 J
1

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