A square loop of side 'a' carrying a current | vedclass

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(B) The magnetic force on a current-carrying wire is given by $\vec{F} = I(\vec{L} 	imes \vec{B})$.For the bottom side of the loop of length '$a$',th

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$2IaB$

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(B) The magnetic force on a current-carrying wire is given by $\vec{F} = I(\vec{L} 	imes \vec{B})$.For the bottom side of the loop of length '$a$',the magnetic force is $F = IaB$.According to Fleming's Left-Hand Rule,if the current flows from left to right,the force is directed upwards. If the current flows from right to left,the force is directed downwards.Let the initial reading of the spring balance be $W_1 = mg - F$ (assuming upward force).When the current direction is reversed,the force becomes $F' = -F$ (downward force).The new reading of the spring balance becomes $W_2 = mg + F$.The change in the reading of the spring balance is $\Delta W = |W_2 - W_1| = |(mg + F) - (mg - F)| = 2F = 2IaB$.

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