Magnetism and Matter Chapter-Wise Test 1

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A material’s ability to retain strong magnetism after an external field is removed is a property of:

Ferromagnetic materials, especially hard ones, retain strong magnetism after the external field is removed due to the persistence of aligned domains, making them suitable for permanent magnets. This distinguishes them from soft ferromagnetic or other materials.

Diamagnetic materials
Hard ferromagnetic materials
Paramagnetic materials
Superconductors
2

A solenoid produces \( B = 1.2 \, \text{T} \) with a core of \( \mu_r = 400 \) 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: \( B = 1.2 \, \text{T} \), \( \mu_r = 400 \), \( n = 2000 \, \text{m}^{-1} \), \( \mu_0 = 4\pi \times 10^{-7} \).

\( I = \frac{1.2}{4\pi \times 10^{-7} \times 400 \times 2000} = \frac{1.2}{1.005 \times 10^{0}} \approx 1.19 \, \text{A} \approx 1.2 \, \text{A} \).

1.0 A
1.1 A
1.2 A
1.3 A
3

A dipole with \( m = 0.9 \, \text{A m}^2 \) in \( B = 0.5 \, \text{T} \) at \( 90^\circ \) has torque:

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

Given: \( m = 0.9 \, \text{A m}^2 \), \( B = 0.5 \, \text{T} \), \( \theta = 90^\circ \), \( \sin 90^\circ = 1 \).

\( \tau = 0.9 \times 0.5 \times 1 = 0.45 \, \text{N m} \).

0.40 N m
0.42 N m
0.43 N m
0.45 N m
4

A bar magnet produces a field of \( 3 \times 10^{-6} \, \text{T} \) at \( 0.3 \, \text{m} \) on its equatorial line. What is its magnetic moment? (Take \( \mu_0 = 4\pi \times 10^{-7} \, \text{T m A}^{-1} \)).

\( B = \frac{\mu_0}{4\pi} \frac{m}{r^3} \), so \( m = \frac{B r^3}{\frac{\mu_0}{4\pi}} \).

Given: \( B = 3 \times 10^{-6} \, \text{T} \), \( r = 0.3 \, \text{m} \), \( \frac{\mu_0}{4\pi} = 10^{-7} \).

\( m = \frac{3 \times 10^{-6} \times (0.3)^3}{10^{-7}} = \frac{3 \times 10^{-6} \times 0.027}{10^{-7}} = 0.81 \, \text{A m}^2 \).

0.75 A m²
0.81 A m²
0.9 A m²
1.0 A m²
2

A bar magnet with \( m = 1.0 \, \text{A m}^2 \) produces a field at \( 0.25 \, \text{m} \) on its equatorial line. What is \( B \)? (Take \( \mu_0 = 4\pi \times 10^{-7} \, \text{T m A}^{-1} \)).

\( B = \frac{\mu_0}{4\pi} \frac{m}{r^3} \).

Given: \( m = 1.0 \, \text{A m}^2 \), \( r = 0.25 \, \text{m} \), \( \frac{\mu_0}{4\pi} = 10^{-7} \).

\( B = 10^{-7} \times \frac{1.0}{(0.25)^3} = 10^{-7} \times \frac{1.0}{0.015625} = 6.4 \times 10^{-6} \, \text{T} \).

6.0 × 10⁻⁶ T
6.2 × 10⁻⁶ T
6.4 × 10⁻⁶ T
6.6 × 10⁻⁶ T
3

A bar magnet with \( m = 0.6 \, \text{A m}^2 \) produces a field at \( 0.3 \, \text{m} \) on its equatorial line. What is \( B \)? (Take \( \mu_0 = 4\pi \times 10^{-7} \, \text{T m A}^{-1} \)).

\( B = \frac{\mu_0}{4\pi} \frac{m}{r^3} \).

Given: \( m = 0.6 \, \text{A m}^2 \), \( r = 0.3 \, \text{m} \), \( \frac{\mu_0}{4\pi} = 10^{-7} \).

\( B = 10^{-7} \times \frac{0.6}{(0.3)^3} = 10^{-7} \times \frac{0.6}{0.027} \approx 2.222 \times 10^{-6} \, \text{T} \approx 2.22 × 10^{-6} \, \text{T} \).

2.0 × 10⁻⁶ T
2.1 × 10⁻⁶ T
2.22 × 10⁻⁶ T
2.3 × 10⁻⁶ T
3

Why do magnetic field lines never intersect, unlike some electric field lines in complex charge distributions?

Magnetic field lines represent the direction of the magnetic field at each point. If they intersected, it would imply two different directions for the field at the same point, which is physically impossible. This uniqueness stems from the vector nature of the magnetic field and the absence of magnetic monopoles.

They are always parallel
They originate from monopoles
The field direction would be ambiguous at the intersection
They are shielded by materials
3

A bar magnet produces a field of \( 6 \times 10^{-6} \, \text{T} \) at \( 0.3 \, \text{m} \) on its equatorial line. What is its magnetic moment? (Take \( \mu_0 = 4\pi \times 10^{-7} \, \text{T m A}^{-1} \)).

\( B = \frac{\mu_0}{4\pi} \frac{m}{r^3} \), so \( m = \frac{B r^3}{\frac{\mu_0}{4\pi}} \).

Given: \( B = 6 \times 10^{-6} \, \text{T} \), \( r = 0.3 \, \text{m} \), \( \frac{\mu_0}{4\pi} = 10^{-7} \).

\( m = \frac{6 \times 10^{-6} \times (0.3)^3}{10^{-7}} = \frac{6 \times 10^{-6} \times 0.027}{10^{-7}} = 1.62 \, \text{A m}^2 \).

1.5 A m²
1.62 A m²
1.8 A m²
2.0 A m²
2

The reason a soft ferromagnetic material loses its magnetism when the external field is removed is:

Soft ferromagnetic materials have low coercivity and do not retain significant magnetization after the external field is removed. Their domains, which align strongly in the presence of a field, return to a random orientation due to minimal internal resistance, unlike hard ferromagnets.

Its susceptibility becomes negative
Its permeability becomes unity
Its domains remain permanently aligned
Its domains randomize without the field
4

A superconductor has a susceptibility \( \chi \) of:

Superconductors exhibit perfect diamagnetism with \( \chi = -1 \), as the magnetic field is completely expelled (Meissner effect).

0
1
-0.5
-1
4

Which material has a negative \( \chi \) and is weakly repelled by a magnetic field?

Diamagnetic materials have a small negative \( \chi \) and are weakly repelled from stronger to weaker field regions.

Ferromagnetic
Paramagnetic
Diamagnetic
Superconductor
3

A bar magnet produces a field of \( 5 \times 10^{-6} \, \text{T} \) at \( 0.4 \, \text{m} \) on its equatorial line. What is its magnetic moment? (Take \( \mu_0 = 4\pi \times 10^{-7} \, \text{T m A}^{-1} \)).

\( B = \frac{\mu_0}{4\pi} \frac{m}{r^3} \), so \( m = \frac{B r^3}{\frac{\mu_0}{4\pi}} \).

Given: \( B = 5 \times 10^{-6} \, \text{T} \), \( r = 0.4 \, \text{m} \), \( \frac{\mu_0}{4\pi} = 10^{-7} \).

\( m = \frac{5 \times 10^{-6} \times (0.4)^3}{10^{-7}} = \frac{5 \times 10^{-6} \times 0.064}{10^{-7}} = 3.2 \, \text{A m}^2 \).

3.0 A m²
3.2 A m²
3.4 A m²
3.6 A m²
2

Which of the following materials has \( \chi \) negative and tends to move from stronger to weaker magnetic field regions?

Diamagnetic materials have negative \( \chi \) and are repelled from stronger to weaker field regions.

Ferromagnetic
Paramagnetic
Diamagnetic
None of these
3

A solenoid with 1100 turns per meter and current \( 1.5 \, \text{A} \) has a core with \( \mu_r = 200 \). 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: \( n = 1100 \, \text{m}^{-1} \), \( I = 1.5 \, \text{A} \), \( \mu_r = 200 \), \( \mu_0 = 4\pi \times 10^{-7} \).

\( B = 4\pi \times 10^{-7} \times 200 \times 1100 \times 1.5 = 0.41448 \, \text{T} \approx 0.41 \, \text{T} \).

0.38 T
0.41 T
0.45 T
0.48 T
2

A bar magnet produces a magnetic field of \( 2 \times 10^{-5} \, \text{T} \) at a point \( 0.1 \, \text{m} \) along its axis. What is its magnetic moment? (Take \( \mu_0 = 4\pi \times 10^{-7} \, \text{T m A}^{-1} \)).

\( B = \frac{\mu_0}{4\pi} \frac{2m}{r^3} \), so \( m = \frac{B r^3}{\frac{\mu_0}{4\pi} \times 2} \).

Given: \( B = 2 \times 10^{-5} \, \text{T} \), \( r = 0.1 \, \text{m} \), \( \frac{\mu_0}{4\pi} = 10^{-7} \).

\( m = \frac{2 \times 10^{-5} \times (0.1)^3}{10^{-7} \times 2} = \frac{2 \times 10^{-5} \times 0.001}{2 \times 10^{-7}} = 0.05 \, \text{A m}^2 \).

0.02 A m²
0.05 A m²
0.1 A m²
0.2 A m²
2

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