A square loop of side a hangs from an insulating hanger of spring balance. The magnetic field of strength `B` occurs only at the lower edge. It carries a current `I` . Find the change in the reading of the spring balance if the direction of current is reversed.

Options

A. IaB
B. 2 IaB
C. `(laB)/(2)`
D. `(3)/(2)` laB

Answer

B

Explanation

Intially `F_(1) = mg+ I alpha B` (downwards)
When current is reversed
`F_(2) = mg - I alpha B` (downwards)
`therefore Delta F = 2 I alpha B`