Moving Charge and Magnetism Chapter-Wise Test 1

Correct answer Carries: 4.

Wrong Answer Carries: -1.

Two parallel wires \( 0.04 \, \text{m} \) apart carry \( 5 \, \text{A} \) and \( 7 \, \text{A} \) in opposite directions. What is the force per unit length? (\( \mu_0 = 4 \pi \times 10^{-7} \, \text{T m/A} \))

\( f = \frac{\mu_0 I_1 I_2}{2 \pi d} \).

\( f = \frac{4 \pi \times 10^{-7} \times 5 \times 7}{2 \pi \times 0.04} = \frac{140 \times 10^{-7}}{0.08} = 1.75 \times 10^{-5} \, \text{N/m} \).

8.75 × 10⁻⁶ N/m
3.5 × 10⁻⁵ N/m
1.75 × 10⁻⁵ N/m
2.63 × 10⁻⁵ N/m
3

A wire of length \( 1.9 \, \text{m} \) carrying \( 4.5 \, \text{A} \) is at \( 45^\circ \) to a magnetic field of \( 0.3 \, \text{T} \). What is the force on the wire?

Force \( F = I l B \sin \theta \).

\( F = 4.5 \times 1.9 \times 0.3 \times \sin 45^\circ = 8.55 \times 0.3 \times 0.707 = 1.8127 \approx 1.81 \, \text{N} \).

2.56 N
1.81 N
0.91 N
3.62 N
2

A circular loop of radius \( 0.09 \, \text{m} \) with 45 turns carries a current of \( 1.2 \, \text{A} \). What is the magnetic field at the center? (\( \mu_0 = 4 \pi \times 10^{-7} \, \text{T m/A} \))

Magnetic field \( B = \frac{\mu_0 N I}{2 R} \).

\( B = \frac{4 \pi \times 10^{-7} \times 45 \times 1.2}{2 \times 0.09} = \frac{21.6 \pi \times 10^{-6}}{0.18} = 1.2 \pi \times 10^{-4} \approx 3.77 \times 10^{-4} \, \text{T} \).

1.88 × 10⁻⁴ T
3.77 × 10⁻⁴ T
7.54 × 10⁻⁴ T
5.65 × 10⁻⁴ T
2

A solenoid with 1200 turns per meter carries \( 1 \, \text{A} \). What is the magnetic field inside? (\( \mu_0 = 4 \pi \times 10^{-7} \, \text{T m/A} \))

\( B = \mu_0 n I \).

\( B = 4 \pi \times 10^{-7} \times 1200 \times 1 = 4.8 \pi \times 10^{-4} \approx 1.51 \times 10^{-3} \, \text{T} \).

7.54 × 10⁻⁴ T
3.02 × 10⁻³ T
1.26 × 10⁻³ T
1.51 × 10⁻³ T
4

A solenoid has 950 turns per meter and carries a current of \( 1.4 \, \text{A} \). What is the magnetic field inside it? (\( \mu_0 = 4 \pi \times 10^{-7} \, \text{T m/A} \))

Magnetic field \( B = \mu_0 n I \).

\( B = 4 \pi \times 10^{-7} \times 950 \times 1.4 = 5.32 \pi \times 10^{-4} \approx 1.67 \times 10^{-3} \, \text{T} \).

8.35 × 10⁻⁴ T
1.67 × 10⁻³ T
3.34 × 10⁻³ T
2.51 × 10⁻³ T
2

In a current-carrying loop placed in a magnetic field, when is the torque maximum?

The torque on a current-carrying loop in a magnetic field is given by \( \tau = N I A B \sin \theta \). It is maximum when \( \sin \theta = 1 \), i.e., \( \theta = 90^\circ \), meaning the plane of the loop is perpendicular to the magnetic field.

When the plane of the loop is perpendicular to the field
When the plane of the loop is parallel to the field
When the current is zero
When the magnetic field is zero
1

Why is the magnetic field outside a long solenoid considered negligible?

In a long solenoid, the magnetic field lines are concentrated inside, and outside, the fields from opposite sides of the coils cancel each other, making the external field nearly zero.

Due to high resistance
Due to short length
Due to current reversal
Due to field cancellation
4

A square loop of side \( 0.18 \, \text{m} \) with 30 turns carries \( 2 \, \text{A} \) in a magnetic field of \( 0.4 \, \text{T} \). The plane of the loop is at \( 60^\circ \) to the field. What is the torque?

Torque \( \tau = N I A B \sin \theta \), where \( A = 0.18 \times 0.18 = 0.0324 \, \text{m}^2 \).

\( \tau = 30 \times 2 \times 0.0324 \times 0.4 \times \sin 60^\circ = 0.7776 \times 0.866 = 0.6734 \approx 0.67 \, \text{N m} \).

0.34 N m
0.67 N m
1.34 N m
1.01 N m
2

What is the effect on the magnetic force if a charged particle’s velocity is doubled while keeping the magnetic field constant?

The magnetic force \( F = q v B \sin \theta \) is directly proportional to velocity \( v \). Doubling the velocity doubles the force.

It remains unchanged
It halves
It doubles
It becomes four times
3

An electron moves at \( 4.5 \times 10^6 \, \text{m/s} \) perpendicular to a field of \( 0.35 \, \text{T} \). What is the magnetic force? (Charge = \( 1.6 \times 10^{-19} \, \text{C} \))

Force \( F = q v B \sin \theta \), \( \theta = 90^\circ \), so \( \sin \theta = 1 \).

\( F = 1.6 \times 10^{-19} \times 4.5 \times 10^6 \times 0.35 = 2.52 \times 10^{-13} \, \text{N} \).

1.26 × 10⁻¹³ N
5.04 × 10⁻¹³ N
2.52 × 10⁻¹³ N
3.78 × 10⁻¹³ N
3

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