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

Hint: 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.