A horizontal semi-circular wire of radius r i | vedclass

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(C) The magnetic force $F_B$ acting on a current-carrying wire in a magnetic field is given by $F_B = I \vec{L}_{eff} 	imes \vec{B}$.For a semi-circu

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$BIr$

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(C) The magnetic force $F_B$ acting on a current-carrying wire in a magnetic field is given by $F_B = I \vec{L}_{eff} 	imes \vec{B}$.For a semi-circular wire of radius $r$, the effective length $\vec{L}_{eff}$ is the straight-line distance between the two ends, which is the diameter $2r$.Therefore, the magnitude of the magnetic force is $F_B = I(2r)B = 2IrB$.This total upward magnetic force is shared equally by the two identical springs $X$ and $Y$ attached to the ends of the wire.Let $F$ be the force acting on each spring. Then, $2F = F_B$.Substituting the value of $F_B$, we get $2F = 2IrB$.Thus, the force on each spring is $F = I r B$.

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