JEE MAIN - Physics (2024 - 9th April Morning Shift - No. 21)
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 $$B_o=5 T$$. The net magnetic force experienced by the loop is _________ $$\mathrm{N}$$.
Answer
160
Explanation
Due to constant component of magnetic field $$F = 0$$
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Due to variable component
$$\begin{aligned} & F_1=0 \\ & \text { and, } F_2+F_3=0 \\ & \begin{aligned} \text { and, } F_4 & =\left(\mathrm{B}_0 4_x\right) i \mathrm{~L} \\ & =5 \times 4 \times 2 \times 2 \times 2 \\ & =160 \mathrm{~N} \end{aligned} \end{aligned}$$
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