Question

A wire of length 0.5 m carrying a current of 3 A is placed in a magnetic field of 2 T perpendicular to the wire. What is the force on the wire?

• A. 1 N
• B. 3 N
• C. 0.75 N
• D. 6 N

Hint:

Use $F = BIL \sin\theta$ to find magnetic force. For perpendicular orientation, $\sin\theta = 1$.

Answer 1

The Correct Option is B

To solve the problem, we need to find the force on a wire of length 0.5 m carrying a current of 3 A placed in a magnetic field of 2 T perpendicular to the wire. 

1. Formula for Force on a Current-Carrying Wire:
The magnetic force $F$ on a wire of length $L$ carrying current $I$ in a magnetic field $B$ is given by:
$ F = I L B \sin \theta $
where $\theta$ is the angle between the wire and the magnetic field.

2. Given Data:
$L = 0.5\, m$
$I = 3\, A$
$B = 2\, T$
$\theta = 90^\circ$ (since the field is perpendicular to the wire)

3. Calculating the Force:
Since $\sin 90^\circ = 1$,
$ F = 3 \times 0.5 \times 2 \times 1 = 3\, N $

Final Answer:
The force on the wire is $ {3\, N} $.