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. \( 1.0 \, \text{N/m} \)
• B. \( 2.0 \, \text{N/m} \)
• C. \( 0.5 \, \text{N/m} \)
• 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

The formula for the magnetic force on a current-carrying wire is given by: \[ F = I L B \sin \theta \] Where: - \( F \) is the force on the wire, - \( I = 2 \, \text{A} \) is the current in the wire, - \( L \) is the length of the wire, - \( B = 0.5 \, \text{T} \) is the magnetic field strength, - \( \theta = 30^\circ \) is the angle between the wire and the magnetic field. We are asked to find the force per unit length, so we divide the force by \( L \): \[ \frac{F}{L} = I B \sin \theta \] Substituting the known values: \[ \frac{F}{L} = 2 \times 0.5 \times \sin 30^\circ \] \[ \frac{F}{L} = 2 \times 0.5 \times \frac{1}{2} = 0.5 \, \text{N/m} \] Thus, the force per unit length on the wire is \( 1.0 \, \text{N/m} \).