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(C) The magnetic force on a current-carrying wire in a uniform magnetic field is given by $\vec{F} = i(\vec{L}_{eff} 	imes \vec{B})$,where $\vec{L}_{

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$2iBL$

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(C) The magnetic force on a current-carrying wire in a uniform magnetic field is given by $\vec{F} = i(\vec{L}_{eff} 	imes \vec{B})$,where $\vec{L}_{eff}$ is the effective length vector from the starting point to the ending point of the wire.For the given curve $y = a \sin \left( \frac{\pi x}{L} ight)$,the wire starts at $x = 0$ and ends at $x = 2L$.The starting point is $(0, 0)$ and the ending point is $(2L, 0)$.Thus,the effective length vector is $\vec{L}_{eff} = (2L - 0)\hat{i} = 2L\hat{i}$.The magnetic field is uniform and perpendicular to the plane of the wire,so $\vec{B} = B(-\hat{k})$ (assuming the field is into the page).The force is $\vec{F} = i(2L\hat{i} 	imes B(-\hat{k})) = 2iLB(\hat{i} 	imes \hat{k}) = 2iLB(-\hat{j})$.The magnitude of the force is $F = 2iLB$.

$A$ wire carrying a current $i$ is placed in a uniform magnetic field in the form of the curve $y = a \sin \left( \frac{\pi x}{L} \right)$ for $0 \le x \le 2L$. The force acting on the wire is

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