The resultant force on the current loop $P Q R S$ due to a long current carrying conductor will be

• A

  $10^{-4} \mathrm{~N}$
• B

  $3.6 \times 10^{-4} \mathrm{~N}$
• C

  $1.8 \times 10^{-4} \mathrm{~N}$
• Correct answer
  D

  $5 \times 10^{-4} \mathrm{~N}$

Solution

$$ \begin{aligned} I_{1} &=20 \mathrm{~A}, I_{2}=20 \mathrm{~A} \\ r_{1} &=2 \mathrm{~cm} \\ r_{2} &=(10+2) \mathrm{cm}=12 \mathrm{~cm} \end{aligned} $$

The given situation is shown in the figure. The forces on arm $S R$ and $P Q$ are equal in magnitude and opposite in direction and also having same line of action, hence they cancel to each other. $\therefore$ Resultant force on the current loop PQRS, $$ \begin{aligned} F &=F_{P S}-F_{Q R} \\ &=\frac{\mu_{0}}{2 \pi} \cdot \frac{I_{1} I_{2} \times l}{r_{1}}-\frac{\mu_{0}}{2 \pi} \cdot \frac{I_{1} I_{2} l}{r_{2}} \\ &=\frac{\mu_{0}}{2 \pi} \cdot I_{1} I_{2} l\left[\frac{1}{r_{1}}-\frac{1}{r_{2}}\right] \\ =2 \times 10^{-7} \times 20 \times 20 \times 15 \times 10^{-2}\left[\frac{1}{2 \times 10^{-2}}-\frac{1}{12 \times 10^{-2}}\right] \\ =5 \times 10^{-4} \mathrm{~N} \end{aligned} $$

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