A rectangular loop of sides 25 cm and 10 cm carrying a current of 10 A is placed with its longer side…

A rectangular loop of sides 25 cm and 10 cm carrying a current of 10 A is placed with its longer side parallel to a long straight conductor 10 cm apart carrying current 25 A. The net force on the loop is
  1. $6.25 \times 10^{-5} \mathrm{~N}$
  2. $5.5 \times 10^{-5} \mathrm{~N}$
  3. $3.75 \times 10^{-5} \mathrm{~N}$
  4. $8.75 \times 10^{-11} \mathrm{~N}$

Solution


Due to symmetry, $\begin{aligned} & \mathrm{F}_3=\mathrm{F}_4 \\ & \mathrm{~F}_1=\frac{\mu_0 \mathrm{I}_1 \mathrm{I}_2 1}{2 \pi \mathrm{r}_1} \\ & \mathrm{~F}_2=\frac{\mu_0 \mathrm{I}_1 \mathrm{I}_2 1}{2 \pi \mathrm{r}_2} \end{aligned}$ $\therefore \quad$ Net force on the loop, $\begin{aligned} & \mathrm{F}=\left(\mathrm{F}_1-\mathrm{F}_2\right)+\left(\mathrm{F}_3-\mathrm{F}_4\right) \\ & =\frac{\mu_0 \mathrm{I}_1 \mathrm{I}_2 1}{2 \pi}\left(\frac{1}{\mathrm{r}_1}-\frac{1}{\mathrm{r}_2}\right) \\ & =\frac{4 \pi \times 10^{-7} \times 25 \times 10 \times 25 \times 10^{-2}}{2 \pi}\left(\frac{1}{10}-\frac{1}{20}\right) \times 10^2 \\ & =6.25 \times 10^{-5} \mathrm{~N} \end{aligned}$

Asked in: AP EAMCET 2024 (21 May Shift 1)

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