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(A) The force on a current-carrying wire in a magnetic field is given by $\vec{F} = \int I (d\vec{l} 	imes \vec{B})$.
Given $B = B_0 (1 + \frac{xy}{

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$\frac{\sqrt{26}}{2} I B_0 L$

Vedclass · Answer

(A) The force on a current-carrying wire in a magnetic field is given by $\vec{F} = \int I (d\vec{l} 	imes \vec{B})$.
Given $B = B_0 (1 + \frac{xy}{L^2}) \hat{k}$.
For segment $AB$ $(x=0, y: 0 	o L)$: $\vec{B} = B_0(1+0)\hat{k} = B_0\hat{k}$. $d\vec{l} = dy\hat{j}$. $\vec{F}_{AB} = \int_0^L I(dy\hat{j} 	imes B_0\hat{k}) = I B_0 L \hat{i}$.
For segment $BC$ $(y=L, x: 0 	o L)$: $\vec{B} = B_0(1+\frac{xL}{L^2})\hat{k} = B_0(1+\frac{x}{L})\hat{k}$. $d\vec{l} = dx\hat{i}$. $\vec{F}_{BC} = \int_0^L I(dx\hat{i} 	imes B_0(1+\frac{x}{L})\hat{k}) = -I B_0 \int_0^L (1+\frac{x}{L}) dx \hat{j} = -I B_0 [x + \frac{x^2}{2L}]_0^L \hat{j} = -\frac{3}{2} I B_0 L \hat{j}$.
For segment $CD$ $(x=L, y: L 	o 0)$: $\vec{B} = B_0(1+\frac{Ly}{L^2})\hat{k} = B_0(1+\frac{y}{L})\hat{k}$. $d\vec{l} = dy(-\hat{j})$. $\vec{F}_{CD} = \int_L^0 I(dy(-\hat{j}) 	imes B_0(1+\frac{y}{L})\hat{k}) = -I B_0 \int_L^0 (1+\frac{y}{L}) dy \hat{i} = I B_0 [y + \frac{y^2}{2L}]_0^L \hat{i} = \frac{3}{2} I B_0 L \hat{i}$.
For segment $DA$ $(y=0, x: L 	o 0)$: $\vec{B} = B_0(1+0)\hat{k} = B_0\hat{k}$. $d\vec{l} = dx(-\hat{i})$. $\vec{F}_{DA} = \int_L^0 I(dx(-\hat{i}) 	imes B_0\hat{k}) = -I B_0 \int_L^0 dx \hat{j} = I B_0 L \hat{j}$.
Net force $\vec{F}_{net} = \vec{F}_{AB} + \vec{F}_{BC} + \vec{F}_{CD} + \vec{F}_{DA} = (I B_0 L + \frac{3}{2} I B_0 L)\hat{i} + (I B_0 L - \frac{3}{2} I B_0 L)\hat{j} = \frac{5}{2} I B_0 L \hat{i} - \frac{1}{2} I B_0 L \hat{j}$.
The magnitude is $|\vec{F}| = \sqrt{(\frac{5}{2} I B_0 L)^2 + (-\frac{1}{2} I B_0 L)^2} = \frac{I B_0 L}{2} \sqrt{25+1} = \frac{\sqrt{26}}{2} I B_0 L$.

$A$ square loop of length $L$ is placed with its edges parallel to the $XY$-axes. The loop carries a current $I$. If the magnetic field in the region varies as $B = B_0 \left(1 + \frac{xy}{L^2}\right) \hat{k}$, then the magnitude of the net force on the loop will be:

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