The magnetic field existing in a region is given by $\overrightarrow{\mathrm{B}}={\mathrm{B}}_0\left[1+\frac{\mathrm{X}}{\mathrm{l}}\right]\widehat{\mathrm{k}}$. A square loop of edge l and carrying current i, is placed with its edges parallel to the x and y-axis. Find the magnitude of the net magnetic force experienced by the loop

Let current in the loop is CW direction force on bc and ad are equal in magnitude but opposite in direction.

$\therefore \text{ i.e. }\mathrm{dF}=\mathrm{Bidx}$
$\mathrm{dF}={\mathrm{B}}_0\left[1+\frac{\mathrm{X}}{\mathrm{l}}\right]\mathrm{i}$

${\mathrm{F}}_{\mathrm{bc}}={\mathrm{B}}_0\mathrm{i}{\int }_0^{\mathrm{l}} \left[1+\frac{\mathrm{x}}{\mathrm{L}}\right]\mathrm{dx}(-\widehat{\mathrm{j}})$

${\mathrm{F}}_{\mathrm{da}}={\mathrm{B}}_0\mathrm{i}{\int }_0^{\mathrm{l}} \left[1+\frac{\mathrm{x}}{\mathrm{L}}\right]\mathrm{dx}(+\widehat{\mathrm{j}})$

$\therefore {\mathrm{F}}_{\mathrm{net}}=0$

$\text{ force on }\mathrm{ab}\text{ is }{\mathrm{F}}_{\mathrm{ab}}={\mathrm{B}}_0\left[1+\frac{0}{\mathrm{L}}\right]\mathrm{il}\left(\widehat{\mathrm{i}}\right)$

$={\mathrm{B}}_0\mathrm{il}(\text{ along }+\mathrm{ve} x\text{ - axis })$

force on dc is ${\mathrm{F}}_{\mathrm{dc}}={\mathrm{B}}_0\left[1+\frac{\mathrm{l}}{\mathrm{l}}\right]\mathrm{il}(-\widehat{\mathrm{i}})$

$=2{\mathrm{B}}_0\mathrm{i}l$ (along –ve x -axis)

$\therefore {\mathrm{F}}_{\mathrm{net}}={\mathrm{B}}_0\mathrm{i}l$         (along –ve x-axis)