Question 204 of 276

A single turn current loop in the shape of a right angle triangle with sides $5cm,12cm,13cm$ is carrying a current of $2A$. The loop is in a uniform magnetic field of magnitude $0.75T$ whose direction is parallel to the current in the $13cm$ side of the loop. The magnitude of the magnetic force on the $5cm$ side will be $\dfrac{x}{130}N$. The value of $x$ is _______.

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Answer: 9

Here, $B=0.75T$
The magnitude of magnetic force on the $5cm$ side is {F}_{AB}=i\times \left({l}_{AB}\right)B\sin \theta \left(\text{angle between}AC&AB\right)
$=2\times \left(\dfrac{5}{100}\right)\times 0.75\times \dfrac{12}{13}$
$=\dfrac{9}{130}N$
Hence, the value of $x=9$.