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 \(..........\)

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Answer & Solution

Correct Answer

(C) \(9\)

Detailed explanation

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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