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(B) 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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(B) 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 vector displacement from the starting point to the end point of the wire.The wire starts at $x = 0$ and ends at $x = 2L$.At $x = 0$,$y = a \sin(0) = 0$.At $x = 2L$,$y = a \sin\left(\frac{\pi(2L)}{L}ight) = a \sin(2\pi) = 0$.Thus,the starting point is $(0, 0)$ and the ending point is $(2L, 0)$.The effective length vector $\vec{L}_{eff}$ is the displacement vector from $(0, 0)$ to $(2L, 0)$,which is $\vec{L}_{eff} = 2L \hat{i}$.The magnetic field is uniform and directed into the plane,so $\vec{B} = -B \hat{k}$ (assuming the plane is the $xy$-plane).The magnetic force is $\vec{F} = i(2L \hat{i} 	imes -B \hat{k}) = -2iLB(\hat{i} 	imes \hat{k}) = -2iLB(-\hat{j}) = 2iLB \hat{j}$.The magnitude of the force is $2iLB$.

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

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