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

• A

  $4$
• Correct answer
  B

  $9$
• C

  $12$
• D

  $15$

Solution

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 $(\mathrm{a}), \sin \theta=\frac{12}{13}$ $$ \begin{array}{ll} \therefore & \mathrm{F}_{\mathrm{EF}}=2 \times 0.05 \times 0.75 \times \frac{12}{13}, \\ \therefore & \mathrm{F}_{\mathrm{EF}}=\frac{9}{130} \mathrm{~N} \\ \therefore & \mathrm{x}=9 \end{array} $$

Hence, the correct answer is option: 'B'.