1. Given data

• Right triangle sides: $5\text{ cm},\ 12\text{ cm},\ 13\text{ cm}$
• Current in loop: $I = 2\text{ A}$
• Magnetic field: $B = 0.75\text{ T}$
• Field direction is parallel to the current in the $13\text{ cm}$ side.

We need the magnetic force on the $5\text{ cm}$ side.

2. Formula for force on a straight current-carrying wire

For a wire segment of length $L$ carrying current $I$ in magnetic field $B$:

$F = I L B \sin \theta$

where $\theta$ is the angle between the current direction in that segment and the magnetic field.

3. Find the angle between the 5 cm side and the magnetic field

Since the magnetic field is parallel to the $13\text{ cm}$ side, the required angle is the angle between the $5\text{ cm}$ side and the $13\text{ cm}$ side of the triangle.

In the $5$-$12$-$13$ right triangle:

• Hypotenuse $=13$
• One leg $=5$

Let $\theta$ be the angle between sides $5$ and $13$. Then,

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

4. Substitute values

Length of the $5\text{ cm}$ side:

$L = 5\text{ cm} = 0.05\text{ m}$

Now,

$F = I L B \sin\theta$

$F = 2 \times 0.05 \times 0.75 \times \frac{12}{13}$

First,

$2 \times 0.05 = 0.1$

$0.1 \times 0.75 = 0.075$

So,

$F = 0.075 \times \frac{12}{13} = \frac{0.9}{130} = \frac{9}{1300}$

Now write in the form given in the question:

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

So,

$\frac{x}{130} = \frac{9}{1300}$

This gives

$x = 0.9$

But since the intended simplification is usually: