Magnetism and Matter Chapter-Wise Test 6

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A paramagnetic material with \( \chi = 1.5 \times 10^{-3} \) in \( H = 600 \, \text{A m}^{-1} \) has magnetization \( M \):

\( M = \chi H \).

Given: \( \chi = 1.5 \times 10^{-3} \), \( H = 600 \, \text{A m}^{-1} \).

\( M = 1.5 \times 10^{-3} \times 600 = 0.9 \, \text{A m}^{-1} \).

0.7 A m⁻¹
0.9 A m⁻¹
1.0 A m⁻¹
1.2 A m⁻¹
2

A material’s weak attraction to a magnetic field is due to:

Paramagnetic materials exhibit weak attraction to a magnetic field because their atomic or molecular magnetic moments, which are randomly oriented in the absence of a field, partially align with an external field, producing a small positive magnetization.

Partial alignment of atomic magnetic moments
Complete expulsion of field lines
Spontaneous domain formation
Induced opposing currents
1

A paramagnetic material with \( \chi = 7 \times 10^{-4} \) in \( H = 3000 \, \text{A m}^{-1} \) has magnetization \( M \):

\( M = \chi H \).

Given: \( \chi = 7 \times 10^{-4} \), \( H = 3000 \, \text{A m}^{-1} \).

\( M = 7 \times 10^{-4} \times 3000 = 2.1 \, \text{A m}^{-1} \).

1.8 A m⁻¹
2.1 A m⁻¹
2.4 A m⁻¹
2.7 A m⁻¹
2

The magnetic field inside a long solenoid is uniform because:

In a long solenoid, the magnetic field is uniform inside due to the symmetrical arrangement of current-carrying loops, which produce overlapping field lines that are parallel and evenly spaced along the solenoid’s axis, minimizing edge effects in the central region.

The field lines are parallel and evenly spaced
The field lines diverge at the ends
The current varies along its length
The field cancels outside the solenoid
1

The magnetic field inside a material increases significantly when placed in an external field due to:

In ferromagnetic materials, the magnetic field inside increases significantly because the external field aligns and expands pre-existing magnetic domains, enhancing the overall magnetization and field strength far beyond the applied field.

Alignment and growth of magnetic domains
Expulsion of field lines
Weak opposing induced currents
Random thermal motion of atoms
1

The reason a bar magnet cannot exert a torque on itself is:

A bar magnet cannot exert a torque on itself because torque requires an external field acting on the dipole. The field produced by the magnet itself does not generate a net rotational effect on its own structure, as internal forces cancel out.

Its poles are fixed
Its field is uniform
Its field strength is zero
It lacks an external field to act on it
4

A material has \( B = 0.65 \, \text{T} \) and \( M = 4.5 \times 10^5 \, \text{A m}^{-1} \). 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: \( B = 0.65 \, \text{T} \), \( M = 4.5 \times 10^5 \, \text{A m}^{-1} \), \( \mu_0 = 4\pi \times 10^{-7} \).

\( \frac{B}{\mu_0} = \frac{0.65}{4\pi \times 10^{-7}} \approx 5.167 \times 10^5 \, \text{A m}^{-1} \).

\( H = 5.167 \times 10^5 - 4.5 \times 10^5 = 6.67 \times 10^4 \, \text{A m}^{-1} \).

6.5 × 10⁴ A m⁻¹
6.67 × 10⁴ A m⁻¹
6.8 × 10⁴ A m⁻¹
7.0 × 10⁴ A m⁻¹
2

A magnetic dipole of moment \( 0.2 \, \text{A m}^2 \) is placed in a uniform field of \( 0.3 \, \text{T} \) at \( 45^\circ \). What is the torque on the dipole?

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

Given: \( m = 0.2 \, \text{A m}^2 \), \( B = 0.3 \, \text{T} \), \( \theta = 45^\circ \), \( \sin 45^\circ = \frac{1}{\sqrt{2}} \approx 0.707 \).

Substitute: \( \tau = 0.2 \times 0.3 \times 0.707 = 0.04242 \, \text{N m} \approx 0.042 \, \text{N m} \).

0.042 N m
0.05 N m
0.06 N m
0.08 N m
1

What is the magnetic field at a point on the equatorial line of a bar magnet with magnetic moment \( 2 \, \text{A m}^2 \) at a distance of \( 10 \, \text{cm} \) from its center? (Take \( \mu_0 = 4\pi \times 10^{-7} \, \text{T m A}^{-1} \)).

The magnetic field on the equatorial line is \( B = \frac{\mu_0}{4\pi} \frac{m}{r^3} \).

Given: \( m = 2 \, \text{A m}^2 \), \( r = 0.1 \, \text{m} \), \( \frac{\mu_0}{4\pi} = 10^{-7} \, \text{T m A}^{-1} \).

Substitute: \( B = 10^{-7} \times \frac{2}{(0.1)^3} = 10^{-7} \times \frac{2}{0.001} = 2 \times 10^{-4} \, \text{T} \).

1 × 10⁻⁴ T
2 × 10⁻⁴ T
3 × 10⁻⁴ T
4 × 10⁻⁴ T
2

A material with susceptibility \( \chi = 4 \times 10^{-3} \) has a relative permeability \( \mu_r \) of:

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

Given: \( \chi = 4 \times 10^{-3} \).

Substitute: \( \mu_r = 1 + 4 \times 10^{-3} = 1.004 \).

1.004
1.002
1.006
1.008
1

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