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Q. 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 MainJEE Main 2023Moving Charges and Magnetism

Solution:

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Force on $5\, cm$ side is
$ |\vec{F}|=\text { ILB } \sin \theta $
$=(2)\left(5 \times 10^{-2}\right) \times \frac{3}{4} \times \frac{12}{13}=\frac{9}{130} N$
So, $x=9$