A single turn current loop in the shape of a right angle triangle with sides $5\,cm , 12\,cm , 13\,cm$ is carrying a current of $2\,A$. The loop is in a uniform magnetic field of magnitude $0.75\,T$ whose direction is parallel to the current in the $13\,cm$ side of the loop. The magnitude of the magnetic force on the $5\,cm$ side will be $\frac{ x }{130}\,N$. The value of $x$ is $..........$
JEE MAIN 2023, Diffcult
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Force on $5 cm$ side is
$|\overrightarrow{ F }|= ILB \sin \theta$
$=(2)\left(5 \times 10^{-2}\right) \times \frac{3}{4} \times \frac{12}{13}=\frac{9}{130}\,N$
So, $x=9$
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