Question

A long straight current-carrying wire is placed in a uniform magnetic field of strength \( B = 0.5 \, \text{T} \). If the current in the wire is \( I = 2 \, \text{A} \) and the wire makes an angle of \( 30^\circ \) with the magnetic field, find the force per unit length on the wire.

• A. \( 0.5 \, \text{N/m} \)
• B. \( 2.0 \, \text{N/m} \)
• C. \( 1 \, \text{N/m} \)
• D. \( 3.0 \, \text{N/m} \)

Hint:

When calculating the force on a current-carrying wire in a magnetic field, remember that the force depends on the angle between the wire and the magnetic field. If the wire is parallel to the field, the force will be zero.

Answer 1

The Correct Option is A

Input:

• Magnetic field strength: \( B = 0.5 \, \text{T} \)
• Current: \( I = 2 \, \text{A} \)
• Angle: \( \theta = 30^\circ \)

Calculation:

The force per unit length (\( \frac{F}{L} \)) on a current-carrying wire in a magnetic field is given by: \[ \frac{F}{L} = B I \sin \theta \] Substituting the provided values: \[ \frac{F}{L} = (0.5 \, \text{T}) \times (2 \, \text{A}) \times \sin(30^\circ) \] As \( \sin(30^\circ) = 0.5 \): \[ \frac{F}{L} = 0.5 \times 2 \times 0.5 = 0.5 \, \text{N/m} \]

✅ Result:

The force per unit length on the wire is \( \boxed{0.5 \, \text{N/m}} \).