Q. 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
A) $IaB$
B) $2\ IaB$
C) $\frac{IaB}{2}$
D) $\frac{3}{2}IaB$
Explanation
(b) Initially $F_{1} = mg + IaB\ $(downwards)
When the direction of current is reversed
$F_{2} = mg - IaB$ (downward) $\Rightarrow \mathrm{\Delta}F = 2IaB$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.