Q 12-04-116JEE MainJEE Main 2023 (24 Jan, Shift 2)Medium
A single turn current loop in the shape of a right angle triangle with sides $5\ \text{cm}$, $12\ \text{cm}$, $13\ \text{cm}$ is carrying a current of $2\ \text{A}$. The loop is in a uniform magnetic field of magnitude $0.75\ \text{T}$ whose direction is parallel to the current in the $13\ \text{cm}$ side of the loop. The magnitude of the magnetic force on the $5\ \text{cm}$ side will be $\dfrac{x}{130}$ N. The value of $x$ is ______.
Numerical value type. Enter your answer.
Answer: 9
The angle $\theta$ between the $5\ \text{cm}$ side and the hypotenuse satisfies $\sin\theta=\dfrac{12}{13}$.
$$F=IlB\sin\theta=2\times0.05\times0.75\times\frac{12}{13}=\frac{0.9}{13}=\frac{9}{130}\ \text{N}$$
Solution by Sreeraj P, M.Sc Physics