JEE Physics 2023 Question with Solution

Given: current $I = 2 \, \text{A}$, side length $L = 5 \, \text{cm} = 0.05 \, \text{m}$, magnetic field $B = 0.75 \, \text{T}$.

Find: the required numerical value for the magnetic force on the $5 \, \text{cm}$ side.

The force on a straight current-carrying conductor is

$$F = ILB \sin \theta$$

Since the magnetic field is parallel to the current in the $13 \, \text{cm}$ side, and the triangle is a $5\text{-}12\text{-}13$ right triangle, the angle between the $5 \, \text{cm}$ side and the $13 \, \text{cm}$ side satisfies

$$\sin \theta = \frac{12}{13}$$

So,

$$F = (2)(0.05)(0.75)\left(\frac{12}{13}\right) = \frac{9}{130} \, \text{N}$$

Hence the required numerical value is $9$.

Therefore, the answer is $9$.

Given: the loop is a right angle triangle with sides $5 \, \text{cm}$, $12 \, \text{cm}$, $13 \, \text{cm}$ and carries current $2 \, \text{A}$. The magnetic field is parallel to the $13 \, \text{cm}$ side.

Find: magnetic force on the $5 \, \text{cm}$ side.

For a wire segment in a magnetic field,

$$F = ILB \sin \theta$$

Here $\theta$ is the angle between the current direction in the $5 \, \text{cm}$ side and the magnetic field. Because $B$ is parallel to the hypotenuse, this angle is the acute angle between the base and hypotenuse of the $5\text{-}12\text{-}13$ triangle.

From the triangle,

$$\sin \theta = \frac{\text{perpendicular}}{\text{hypotenuse}} = \frac{12}{13}$$

Now substitute:

$$F = 2 \times 5 \times 10^{-2} \times 0.75 \times \frac{12}{13}$$ $$F = \frac{9}{130} \, \text{N}$$

If the force is written as $\frac{x}{130} \, \text{N}$, then $x = 9$.

Therefore, the correct numerical answer is $9$.