Two long straight parallel conductors A and B carrying currents 4.5 A and 8 A respectively are separated by…

Two long straight parallel conductors A and B carrying currents 4.5 A and 8 A respectively are separated by 25 cm in air. The resultant magnetic field at a point which is at a distance of 15 cm from conductor A and 20 cm from conductor B is
  1. $2 \times 10^{-5} \mathrm{~N}$
  2. $2 \times 10^{-4} \mathrm{~N}$
  3. $10^{-5} \mathrm{~N}$
  4. $10^{-4} \mathrm{~N}$

Solution


$\begin{aligned} & \mathrm{I}_1=4.5 \mathrm{~A}, \mathrm{I}_2=8 \mathrm{~A} \\ & \mathrm{r}_1=15 \mathrm{~cm}, \mathrm{r}_2=10 \mathrm{~cm} \end{aligned}$ $\therefore \quad$ Magnetic field at P is $\begin{aligned} & \mathrm{B}=\mathrm{B}_2-\mathrm{B}_1 \\ & =\frac{\mu_0}{2 \pi}\left(\frac{\mathrm{I}_2}{\mathrm{r}_2}-\frac{\mathrm{I}_1}{\mathrm{r}_1}\right) \\ & =2 \times 10^{-7}\left(\frac{8}{10}-\frac{4.5}{15}\right) \times 10^2=10^{-5} \mathrm{~T} \end{aligned}$

Asked in: AP EAMCET 2024 (20 May Shift 2)

Practice more Magnetic Fields due to Electric Current questions on Aicharya