Question

A conductor of length l carrying current $i$ is placed perpendicular to the magnetic field of induction $B$ . The force experienced by it is:
A) $ilB$
B) $iB/l$
C) $il/B$
D) $lB/i$

Answer 1

To answer this question, we have to consider the force acting on a current-carrying wire in a magnetic field. When the current-carrying cable is placed perpendicular to the magnetic field, the force acting on it will be maximum.

Complete step by step solution:
We’ve been given that a current-carrying conductor cable of length $l$ carrying current $i$ is placed perpendicular to the magnetic field of induction $B$ .
We know that the force acting on a current-carrying cable placed in a magnetic field is given as:
$F = I(l \times B)$ .
Now in our case, the wire is placed perpendicular to the magnetic field. This implies that the angle between the current-carrying cable and the magnetic field is $90^\circ$ .
So, we can simplify the cross product in our calculation as:
$l \times B = lb\sin 90^\circ$
Since $\sin 90^\circ = 1$ , we can calculate the force as:
$F = IlB$
Hence the force in a current-carrying cable of length l that is carrying a current $i$ when placed perpendicular to a magnetic field of induction $B$ will experience a force $F = IlB$ which corresponds to option (A).

Note:
The force we calculated in the above case will be maximum when the wire is placed perpendicular to the magnetic field that is $\theta = 90^\circ$ . When the current-carrying cable is placed in the direction of the magnetic field, that is $\theta = 0^\circ$ , the force acting on the current-carrying cable will be zero. Even if the current-carrying cable is kept anti-parallel to the direction of the magnetic field that is $\theta = 180^\circ$ , the force acting on it will be zero.