Force on Current-Carrying Conductors question from JEE Main 2024 (31 Jan, Shift 1)

Answer: (b)

1. In a uniform field, force on a bent wire = force on the straight line joining its ends inside the field.
2. Inside the field the wire goes up the left straight part, round the semicircle, down the right part; forces on the two straight parts cancel.
3. Effective length is $2R$ along $+\hat{i}$: $\vec{F} = i(2R\hat{i})\times B_0\hat{k} = 2iRB_0(\hat{i}\times\hat{k})$
4. $\hat{i}\times\hat{k} = -\hat{j} \Rightarrow \vec{F} = -2iBR\hat{j}$
5. So the answer is (b) $-2iBR\hat{j}$.