Magnetism and Matter Chapter-Wise Test 7

Correct answer Carries: 4.

Wrong Answer Carries: -1.

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

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

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

Substitute: \( \mu_r = 1 - 3 \times 10^{-5} = 0.99997 \).

0.99997
1.00003
0.99995
1.00005
1

A solenoid with 700 turns per meter carries a current of \( 2 \, \text{A} \). What is the magnetic intensity \( H \) inside?

Magnetic intensity \( H = n I \).

Given: \( n = 700 \, \text{m}^{-1} \), \( I = 2 \, \text{A} \).

Substitute: \( H = 700 \times 2 = 1400 \, \text{A m}^{-1} \).

1400 A m⁻¹
1600 A m⁻¹
1800 A m⁻¹
2000 A m⁻¹
1

A material with susceptibility \( \chi = -8 \times 10^{-5} \) has a relative permeability \( \mu_r \) of:

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

Given: \( \chi = -8 \times 10^{-5} \).

Substitute: \( \mu_r = 1 - 8 \times 10^{-5} = 0.99992 \).

0.99992
0.99994
0.99996
1.00008
1

A dipole with \( m = 0.8 \, \text{A m}^2 \) in a field \( B = 0.5 \, \text{T} \) at \( 45^\circ \) has potential energy:

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

Given: \( m = 0.8 \, \text{A m}^2 \), \( B = 0.5 \, \text{T} \), \( \theta = 45^\circ \), \( \cos 45^\circ = \frac{1}{\sqrt{2}} \approx 0.707 \).

\( U_m = -0.8 \times 0.5 \times 0.707 \approx -0.2828 \, \text{J} \approx -0.28 \, \text{J} \).

-0.25 J
-0.28 J
-0.30 J
-0.35 J
2

The net magnetic flux through a closed surface surrounding a toroid with current 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 current
Zero
4

A bar magnet with original \( m = 1.4 \, \text{A m}^2 \) is cut along its length into two equal parts. What is \( m \) of each part?

When cut along its length, each part typically has half the original magnetic moment in standard problems.

Given: \( m = 1.4 \, \text{A m}^2 \).

Each part: \( m' = \frac{1.4}{2} = 0.7 \, \text{A m}^2 \).

0.5 A m²
0.6 A m²
0.65 A m²
0.7 A m²
4

The magnetic field contribution \( B_m \) due to a material with \( M = 2.8 \times 10^5 \, \text{A m}^{-1} \) is: (Take \( \mu_0 = 4\pi \times 10^{-7} \, \text{T m A}^{-1} \)).

\( B_m = \mu_0 M \).

Given: \( M = 2.8 \times 10^5 \, \text{A m}^{-1} \), \( \mu_0 = 4\pi \times 10^{-7} \).

\( B_m = 4\pi \times 10^{-7} \times 2.8 \times 10^5 = 0.35168 \, \text{T} \approx 0.35 \, \text{T} \).

0.32 T
0.34 T
0.35 T
0.37 T
3

A bar magnet of magnetic moment \( 0.5 \, \text{A m}^2 \) is placed at a distance of \( 20 \, \text{cm} \) from its center along its axis. Calculate 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 of a bar magnet is given by \( B = \frac{\mu_0}{4\pi} \frac{2m}{r^3} \).

Given: \( m = 0.5 \, \text{A m}^2 \), \( r = 20 \, \text{cm} = 0.2 \, \text{m} \), \( \frac{\mu_0}{4\pi} = 10^{-7} \, \text{T m A}^{-1} \).

Substitute: \( B = 10^{-7} \times \frac{2 \times 0.5}{(0.2)^3} = 10^{-7} \times \frac{1}{0.008} = 10^{-7} \times 125 = 1.25 \times 10^{-5} \, \text{T} \).

1.0 × 10⁻⁵ T
1.25 × 10⁻⁵ T
1.5 × 10⁻⁵ T
2.0 × 10⁻⁵ T
2

A bar magnet with \( m = 1.8 \, \text{A m}^2 \) is at \( 0.6 \, \text{m} \) 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: \( m = 1.8 \, \text{A m}^2 \), \( r = 0.6 \, \text{m} \), \( \frac{\mu_0}{4\pi} = 10^{-7} \).

\( B = 10^{-7} \times \frac{2 \times 1.8}{(0.6)^3} = 10^{-7} \times \frac{3.6}{0.216} = 1.67 \times 10^{-6} \, \text{T} \approx 1.7 \times 10^{-6} \, \text{T} \).

1.4 × 10⁻⁶ T
1.5 × 10⁻⁶ T
1.6 × 10⁻⁶ T
1.7 × 10⁻⁶ T
4

A bar magnet produces a field of \( 8 \times 10^{-6} \, \text{T} \) 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: \( B = 8 \times 10^{-6} \, \text{T} \), \( r = 0.4 \, \text{m} \), \( \frac{\mu_0}{4\pi} = 10^{-7} \).

\( m = \frac{8 \times 10^{-6} \times (0.4)^3}{10^{-7}} = \frac{8 \times 10^{-6} \times 0.064}{10^{-7}} = 5.12 \, \text{A m}^2 \).

4.5 A m²
5.12 A m²
5.5 A m²
6.0 A m²
2

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