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 resolve this issue, we must ascertain the force exerted on a 0.5 m wire. This wire carries a current of 3 A and is positioned within a 2 T magnetic field, oriented perpendicularly to the wire. The relevant formula for the magnetic force ($F$) on a current-carrying wire of length ($L$) with current ($I$) in a magnetic field ($B$) is $F = I L B \sin \theta$, where $\theta$ denotes the angle between the wire and the magnetic field. The given parameters are $L = 0.5\, m$, $I = 3\, A$, and $B = 2\, T$. Since the magnetic field is perpendicular to the wire, $\theta = 90^\circ$. As $\sin 90^\circ = 1$, the calculation for the force is $F = 3 \times 0.5 \times 2 \times 1 = 3\, N$. Consequently, the force acting on the wire is $ {3\, N} $.