A square loop of edge length $2 \mathrm{~m}$ carrying current of $2 \mathrm{~A}$ is placed with its edges parallel to the $x$ $y$ axis. A magnetic field is passing through the $x-y$ plane and expressed as $\vec{B}=B_0(1+4 x) \hat{k}$, where $B_0=5 \mathrm{~T}$. The net magnetic force experienced by the loop is _______ $\mathrm{N}$.

Solution

$\begin{aligned} & \mathrm{B}(\mathrm{x}=0)=\mathrm{B}_0, \quad \mathrm{~B}(\mathrm{x}=2)=9 \mathrm{~B}_0 \\ & \text { Also, } \mathrm{F}=\mathrm{i} \ell \mathrm{B} \\ & \Rightarrow \mathrm{F}_1=\mathrm{i} \ell \mathrm{B}_0 \quad \& \quad \mathrm{~F}_2=9 \mathrm{i} \ell \mathrm{B}_0 \\ & \mathrm{~F}=\mathrm{F}_2-\mathrm{F}_1=8 \mathrm{i} \ell \mathrm{B}_0=8 \times 2 \times 2 \times 5 \\ & \mathrm{~F}=160 \mathrm{~N}\end{aligned}$

Hence, the correct answer is 160.