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$\frac{3}{2}IaB$
Medium
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B$2\, IaB$
b (b) Initially ${F_1} = mg + IaB$ $(down\, wards)$
when the direction of current is reversed
${F_2} = mg - IaB$ $(down\, wards)$ $==>$ $\Delta F = 2IaB$
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