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(C) The magnetic field is $\vec{B} = B\hat{j}$. The force on a current-carrying wire is given by $\vec{F} = i(\vec{L} 	imes \vec{B})$.For segments $P

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$iBL$

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(C) The magnetic field is $\vec{B} = B\hat{j}$. The force on a current-carrying wire is given by $\vec{F} = i(\vec{L} 	imes \vec{B})$.For segments $PQ$ and $TU$,the length vector is parallel to the magnetic field (along $Y$-axis),so $\vec{F}_{PQ} = 0$ and $\vec{F}_{TU} = 0$.For segment $QR$,the length vector is along the $Z$-axis $(\vec{L} = L\hat{k})$,so $\vec{F}_{QR} = i(L\hat{k} 	imes B\hat{j}) = -iLB\hat{i}$.For segment $RS$,the length vector is along the $-X$-axis $(\vec{L} = -L\hat{i})$,so $\vec{F}_{RS} = i(-L\hat{i} 	imes B\hat{j}) = -iLB\hat{k}$.For segment $ST$,the length vector is along the $-Z$-axis $(\vec{L} = -L\hat{k})$,so $\vec{F}_{ST} = i(-L\hat{k} 	imes B\hat{j}) = iLB\hat{i}$.The net force is $\vec{F}_{net} = \vec{F}_{PQ} + \vec{F}_{QR} + \vec{F}_{RS} + \vec{F}_{ST} + \vec{F}_{TU} = 0 + (-iLB\hat{i}) + (-iLB\hat{k}) + (iLB\hat{i}) + 0 = -iLB\hat{k}$.The magnitude of the net force is $iLB$.

$A$ magnetic field is in the $+Y$ direction. $A$ current $i$ flows through the wire $PQRSTU$. Each side has a length $L$. What is the net force on the wire?

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