A single turn current loop in the shape of a right angle triangle with side \(5 \mathrm{~cm}, 12 \mathrm{~cm}, 13 \mathrm{~cm}\) is carrying a current of \(2 \mathrm{~A}\). The loop is in a uniform magnetic field of magnitude \(0.75 \mathrm{~T}\) whose direction is parallel to the current in the \(13 \mathrm{~cm}\) side of the loop. The magnitude of the magnetic force on the \(5 \mathrm{~cm}\) side will be \(\frac{\mathrm{x}}{130} \mathrm{~N}\). The value of ' \(x\) ' is

The net magnetic field is acting in the direction of GF as shown in figures:
Resolving \(\overrightarrow{\mathrm{B}}\) into its components, amongst the components, only \(\overrightarrow{\mathrm{B}} \sin \theta\) exerts force on side EF of current carrying loop.
\(
\therefore \quad \mathrm{F}_{\mathrm{EF}}=\mathrm{I} \times \mathrm{d}(\mathrm{EF}) \times \mathrm{B} \sin \theta
\)
From figure \((\mathrm{a}), \sin \theta=\frac{12}{13}\)
\(
\begin{array}{ll}
\therefore & \mathrm{F}_{\mathrm{EF}}=2 \times 0.05 \times 0.75 \times \frac{12}{13}, \\
\therefore & \mathrm{F}_{\mathrm{EF}}=\frac{9}{130} \mathrm{~N} \\
\therefore & \mathrm{x}=9
\end{array}
\)