JEE MAIN - Physics (2023 - 24th January Evening Shift - No. 26)
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 ____________.
Answer
9
Explanation
$$ \begin{aligned} & =2 \times \frac{5}{100} \times 0.75 \times \frac{12}{13} \\\\ & =\frac{9}{130} \\\\ & \therefore \quad x=9 \end{aligned} $$
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