Question

The magnetic field existing in a region is given by
$\overrightarrow{B}=B_0⟮1+\frac{x}{l}⟯\hat{k}$.
A square loop of edge I and carrying a current i, is placed with its edges parallel to the X-Y axes, Find the magnitude of the net magnetic force experienced by the loop.

Answer 1

${\overrightarrow{B}}_{x_0}={\overrightarrow{B}}_0(1+\frac{x_0}{l})\hat{k}$

${\overrightarrow{B}}_{x_0+l}={\overrightarrow{B}}_0(1+\frac{x_0+l}{l})\hat{k}$

Now,

${\overrightarrow{F}}_{x_0}=i(\overrightarrow{l}\times \overrightarrow{B_{x_0}})$

${\overrightarrow{F}}_{x_0}=il{B}_{x_0}(\hat{j}\times \hat{k})$

${\overrightarrow{F}}_{x_0}=il{B}_{x_0}\hat{i}$

${\overrightarrow{F}}_{x_0}=il{\overrightarrow{B}}_0(1+\frac{x_0}{l})\hat{i}$

Similarly

${\overrightarrow{F}}_{x_0+l}=il{\overrightarrow{B}}_0(1+\frac{x_0+l}{l})\hat{i}$

Thus, the net force is

${\overrightarrow{F}}_{x_0}+{\overrightarrow{F}}_{x_0+l}=il{\overrightarrow{B}}_0(1+\frac{x_0}{l})\hat{i}-il{\overrightarrow{B}}_0(1+\frac{x_0+l}{l})\hat{i}$

${\overrightarrow{F}}_{x_0}+{\overrightarrow{F}}_{x_0+l}=il{\overrightarrow{B}}_0\hat{i}$

Thus, the magnitude of the net force is $il{B}_0$