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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.

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$IaB$

$2 IaB$

$\frac{IaB}{2}$

$\frac{3}{2} IaB$

answer is B.

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Detailed Solution

Initially $F_{1} = mg + IaB$ (downwards)
When the direction of current is
reversed $F_{2} = mg - IaB$ (downwards)

$\Rightarrow ΔF = 2 IaB$ .

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