JEE MAIN - Physics (2024 - 4th April Morning Shift - No. 26)
The magnetic field existing in a region is given by $$\vec{B}=0.2(1+2 x) \hat{k}$$. A square loop of edge $$50 \mathrm{~cm}$$ carrying 0.5 A current is placed in $$x$$-$$y$$ plane with its edges parallel to the $$x$$-$$y$$ axes, as shown in figure. The magnitude of the net magnetic force experienced by the loop is _________ $$\mathrm{mN}$$.
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Answer
50
Explanation
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$$\begin{aligned} & \vec{F}_{B C}+\vec{F}_{D A}=0 \\ & \vec{F}_{A B}=i l B=0.5 \times 0.5(5)=1.25 \mathrm{~N} \times 0.2=0.25 \mathrm{~N} \\ & \vec{F}_{C D}=0.5 \times 0.5(6)=1.5 \times 0.2=0.3 \mathrm{~N} \\ & F_{\text {net }}=0.05 \mathrm{~N} \\ & \quad=50 \mathrm{mN} \end{aligned}$$
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