A square loop of side a hangs from an insulat | vedclass

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(B) Let the mass of the loop be $m$. The magnetic force acting on the lower edge of the loop is $F_m = IaB$.Initially,let the current flow such that t

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$2IaB$

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(B) Let the mass of the loop be $m$. The magnetic force acting on the lower edge of the loop is $F_m = IaB$.Initially,let the current flow such that the magnetic force acts downwards. The reading of the spring balance is $F_1 = mg + IaB$.When the direction of the current is reversed,the magnetic force acts upwards. The new reading of the spring balance is $F_2 = mg - IaB$.The change in the reading of the spring balance is $\Delta F = F_1 - F_2 = (mg + IaB) - (mg - IaB) = 2IaB$.

$A$ square loop of side $a$ hangs from an insulating hanger of a spring balance. The magnetic field of strength $B$ exists 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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