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 _____.

Magnetic force on a current carrying conductor is given by:

F=BILsin⁡θ

where
$$B=0.75T$$
$$I=2A$$
L= length of the conductor
θ=angle between current direction and magnetic field.

The magnetic field is parallel to the current in the 13 cm side.
Hence, the magnetic field is along the hypotenuse of the right triangle.

We need the force on the 5 cm side.

In the $$5\text{-}12\text{-}13$$ right triangle, the angle between the 5 cm side and the 13 cm side is obtained using:

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

Length of the 5 cm side:

$$L=5cm=\frac{5}{100}m$$

Now,

$$F=0.75\times2\times\frac{5}{100}\times\frac{12}{13}$$

$$F=\frac{90}{1300}$$

$$F=\frac{9}{130}N$$

Given,

$$F=\frac{x}{130}N$$

Therefore,

$$x=9$$