Magnetism And Matter Chapter-Wise Test 1

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A material with susceptibility \( \chi = 5 \times 10^{-3} \) has a relative permeability \( \mu_r \) of:

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

Given: \( \chi = 5 \times 10^{-3} \).

Substitute: \( \mu_r = 1 + 5 \times 10^{-3} = 1.005 \).

1.005
1.003
1.002
1.001
1

A dipole with \( m = 0.2 \, \text{A m}^2 \) in \( B = 0.8 \, \text{T} \) at \( 45^\circ \) has torque:

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

Given: \( m = 0.2 \, \text{A m}^2 \), \( B = 0.8 \, \text{T} \), \( \theta = 45^\circ \), \( \sin 45^\circ = \frac{1}{\sqrt{2}} \approx 0.707 \).

\( \tau = 0.2 \times 0.8 \times 0.707 \approx 0.11312 \, \text{N m} \approx 0.11 \, \text{N m} \).

0.09 N m
0.10 N m
0.105 N m
0.11 N m
4

A material’s susceptibility becomes negative and large when:

A superconductor has a susceptibility of \( \chi = -1 \) (large and negative) when it transitions to the superconducting state below its critical temperature, expelling all magnetic fields via the Meissner effect, a unique property among materials.

It is heated above a critical point
It aligns weakly with the field
It forms magnetic domains
It becomes superconducting
4

A paramagnetic material with \( \chi = 3 \times 10^{-3} \) in \( H = 400 \, \text{A m}^{-1} \) has magnetization \( M \):

\( M = \chi H \).

Given: \( \chi = 3 \times 10^{-3} \), \( H = 400 \, \text{A m}^{-1} \).

\( M = 3 \times 10^{-3} \times 400 = 1.2 \, \text{A m}^{-1} \).

1.0 A m⁻¹
1.2 A m⁻¹
1.4 A m⁻¹
1.6 A m⁻¹
2

Which material retains magnetization after the external field is removed and is used in permanent magnets?

Ferromagnetic materials, particularly hard ferromagnets, retain magnetization after the external field is removed and are used in permanent magnets.

Diamagnetic
Paramagnetic
Ferromagnetic
Superconductor
3

A bar magnet with magnetic moment \( 1.0 \, \text{A m}^2 \) is placed at a distance of \( 0.5 \, \text{m} \) 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: \( m = 1.0 \, \text{A m}^2 \), \( r = 0.5 \, \text{m} \), \( \frac{\mu_0}{4\pi} = 10^{-7} \).

Substitute: \( B = 10^{-7} \times \frac{2 \times 1.0}{(0.5)^3} = 10^{-7} \times \frac{2.0}{0.125} = 1.6 \times 10^{-6} \, \text{T} \).

1.6 × 10⁻⁶ T
1.8 × 10⁻⁶ T
2.0 × 10⁻⁶ T
2.2 × 10⁻⁶ T
1

A material with \( B = 0.2 \, \text{T} \) and \( H = 1000 \, \text{A m}^{-1} \) has magnetization \( M \). What is \( M \)? (Take \( \mu_0 = 4\pi \times 10^{-7} \, \text{T m A}^{-1} \)).

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

Given: \( B = 0.2 \, \text{T} \), \( H = 1000 \, \text{A m}^{-1} \), \( \mu_0 = 4\pi \times 10^{-7} \).

\( \frac{B}{\mu_0} = \frac{0.2}{4\pi \times 10^{-7}} \approx 1.59 \times 10^5 \, \text{A m}^{-1} \).

\( M = 1.59 \times 10^5 - 1000 = 1.58 \times 10^5 \, \text{A m}^{-1} \).

1.5 × 10⁵ A m⁻¹
1.58 × 10⁵ A m⁻¹
1.6 × 10⁵ A m⁻¹
1.7 × 10⁵ A m⁻¹
2

A magnetic needle with magnetic moment \( 0.1 \, \text{A m}^2 \) is placed in a uniform magnetic field of \( 0.2 \, \text{T} \) at an angle of \( 60^\circ \) with the field. What is the torque acting on it?

Torque on a magnetic dipole is \( \tau = m B \sin\theta \).

Given: \( m = 0.1 \, \text{A m}^2 \), \( B = 0.2 \, \text{T} \), \( \theta = 60^\circ \), \( \sin 60^\circ = \frac{\sqrt{3}}{2} \approx 0.866 \).

Substitute: \( \tau = 0.1 \times 0.2 \times 0.866 = 0.01732 \, \text{N m} \approx 0.017 \, \text{N m} \).

0.01 N m
0.017 N m
0.02 N m
0.03 N m
2

A material with \( B = 0.55 \, \text{T} \) and \( H = 3500 \, \text{A m}^{-1} \) has \( M \): (Take \( \mu_0 = 4\pi \times 10^{-7} \, \text{T m A}^{-1} \)).

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

Given: \( B = 0.55 \, \text{T} \), \( H = 3500 \, \text{A m}^{-1} \), \( \mu_0 = 4\pi \times 10^{-7} \).

\( \frac{B}{\mu_0} = \frac{0.55}{4\pi \times 10^{-7}} \approx 4.375 \times 10^5 \, \text{A m}^{-1} \).

\( M = 4.375 \times 10^5 - 3500 \approx 4.34 \times 10^5 \, \text{A m}^{-1} \).

4.3 × 10⁵ A m⁻¹
4.32 × 10⁵ A m⁻¹
4.33 × 10⁵ A m⁻¹
4.34 × 10⁵ A m⁻¹
4

A solenoid produces \( B = 1.5 \, \text{T} \) with a core of \( \mu_r = 500 \) and \( n = 2000 \, \text{m}^{-1} \). What is the current \( I \)? (Take \( \mu_0 = 4\pi \times 10^{-7} \, \text{T m A}^{-1} \)).

\( B = \mu_0 \mu_r n I \), so \( I = \frac{B}{\mu_0 \mu_r n} \).

Given: \( B = 1.5 \, \text{T} \), \( \mu_r = 500 \), \( n = 2000 \, \text{m}^{-1} \), \( \mu_0 = 4\pi \times 10^{-7} \).

\( I = \frac{1.5}{4\pi \times 10^{-7} \times 500 \times 2000} = \frac{1.5}{1.256 \times 10^{0}} \approx 1.194 \, \text{A} \approx 1.2 \, \text{A} \).

1.0 A
1.1 A
1.2 A
1.3 A
3

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