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.

I a B

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2 I a B

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\frac{I a B}{2}

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\frac{3}{2} Ia B

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Solution

The correct option is D $2 I a B$
Initially $F_{1} = m g + I a B$ (downward)
When the direction is reversed $F_{2} = m g - I a B$ (downward)
So, $Δ F = F_{1} - F_{2} = 2 I a B$

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