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(C) The magnetic field is $\vec B = B_0 \left( 5 + \frac{x}{l} \right) \hat k$. Let the vertices of the square loop be $P(0,0)$,$Q(l,0)$,$R(l,l)$,and

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$B_0 il$

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(C) The magnetic field is $\vec B = B_0 \left( 5 + \frac{x}{l} ight) \hat k$. Let the vertices of the square loop be $P(0,0)$,$Q(l,0)$,$R(l,l)$,and $S(0,l)$.The force on a current-carrying wire is given by $\vec F = i \int d\vec l 	imes \vec B$.$1$. For segment $PQ$ (along $x$-axis,$y=0$,$x$ from $0$ to $l$): $d\vec l = dx \hat i$,$\vec B = B_0 (5 + x/l) \hat k$. $\vec F_{PQ} = i \int_0^l (dx \hat i) 	imes B_0 (5 + x/l) \hat k = -i B_0 \int_0^l (5 + x/l) dx \hat j = -i B_0 [5x + x^2/(2l)]_0^l \hat j = -i B_0 (5l + l/2) \hat j = -5.5 B_0 il \hat j$.$2$. For segment $RS$ (parallel to $x$-axis,$y=l$,$x$ from $l$ to $0$): $d\vec l = dx \hat i$,$\vec B = B_0 (5 + x/l) \hat k$. $\vec F_{RS} = i \int_l^0 (dx \hat i) 	imes B_0 (5 + x/l) \hat k = -i B_0 \int_l^0 (5 + x/l) dx \hat j = i B_0 [5x + x^2/(2l)]_0^l \hat j = 5.5 B_0 il \hat j$.$3$. For segment $QR$ (parallel to $y$-axis,$x=l$,$y$ from $0$ to $l$): $d\vec l = dy \hat j$,$\vec B = B_0 (5 + l/l) \hat k = 6 B_0 \hat k$. $\vec F_{QR} = i \int_0^l (dy \hat j) 	imes 6 B_0 \hat k = 6 B_0 il \hat i$.$4$. For segment $SP$ (parallel to $y$-axis,$x=0$,$y$ from $l$ to $0$): $d\vec l = dy \hat j$,$\vec B = B_0 (5 + 0/l) \hat k = 5 B_0 \hat k$. $\vec F_{SP} = i \int_l^0 (dy \hat j) 	imes 5 B_0 \hat k = -5 B_0 il \hat i$.Net force $\vec F_{net} = \vec F_{PQ} + \vec F_{RS} + \vec F_{QR} + \vec F_{SP} = (-5.5 + 5.5) B_0 il \hat j + (6 - 5) B_0 il \hat i = B_0 il \hat i$.The magnitude is $B_0 il$.

The magnetic field existing in a region is given by $\vec B = B_0 \left( 5 + \frac{x}{l} \right) \hat k$. $A$ square loop of edge $l$ and carrying a current $i$ is placed with its edges parallel to $x-y$ axes. Find the magnitude of the net magnetic force experienced by the loop.

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