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(C) The magnetic field is given by $\vec{B} = B_0 \left(1 + \frac{x}{l}\right) \hat{k}$.For a square loop of side length $l$ placed in the $x-y$ plane

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$i B_0 l$

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(C) The magnetic field is given by $\vec{B} = B_0 \left(1 + \frac{x}{l}ight) \hat{k}$.For a square loop of side length $l$ placed in the $x-y$ plane with one corner at the origin,the vertical segments are at $x = 0$ and $x = l$.The force on a current-carrying wire is given by $\vec{F} = i(\vec{L} 	imes \vec{B})$.At $x = 0$,the magnetic field is $\vec{B}_1 = B_0 \hat{k}$. The force on the vertical segment of length $l$ (carrying current in the $+y$ direction) is $\vec{F}_1 = i(l \hat{j} 	imes B_0 \hat{k}) = i l B_0 \hat{i}$.At $x = l$,the magnetic field is $\vec{B}_2 = B_0 \left(1 + \frac{l}{l}ight) \hat{k} = 2 B_0 \hat{k}$. The force on the vertical segment of length $l$ (carrying current in the $-y$ direction) is $\vec{F}_2 = i(-l \hat{j} 	imes 2 B_0 \hat{k}) = -2 i l B_0 \hat{i}$.The horizontal segments experience forces in the $y$-direction which cancel each other out due to symmetry.The net force is $\vec{F}_{net} = \vec{F}_1 + \vec{F}_2 = i l B_0 \hat{i} - 2 i l B_0 \hat{i} = -i l B_0 \hat{i}$.The magnitude of the net force is $|\vec{F}_{net}| = i B_0 l$.

The magnetic field existing in a region is given by $\vec{B} = B_0 \left(1 + \frac{x}{l}\right) \hat{k}$. $A$ square loop of edge length $l$ and carrying a current $i$ is placed with its edges parallel to the $x$ and $y$ axes. Find the magnitude of the net magnetic force experienced by the loop.

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