Two infinite length wires carry currents 8 A and 6 A respectively and are placed along X and Y axes respectively. Magnetic field at a point $\mathrm{P}(0,0, d)$ will be

• A

  $\frac{7 \mu_0}{\pi d}$
• B

  $\frac{10 \mu_0}{\pi d}$
• C

  $\frac{14 \mu_0}{\pi d}$
• Correct answer
  D

  $\frac{5 \mu_0}{\pi d}$

Solution

$\begin{aligned} & \mathrm{I}_1=8 \mathrm{~A}, \mathrm{I}_2=6 \mathrm{~A} \\ & \mathrm{r}=\mathrm{d} \\ & \therefore \overrightarrow{\mathrm{B}}_1=\frac{2 \mathrm{kI}_1}{\mathrm{r}} \hat{\mathrm{i}} \\ & \therefore \overrightarrow{\mathrm{B}}_2=-\frac{2 \mathrm{kI}_2}{\mathrm{r}} \hat{\mathrm{j}} \\ & \therefore \overrightarrow{\mathrm{B}}_{\mathrm{p}}=\overrightarrow{\mathrm{B}}_1+\overrightarrow{\mathrm{B}}_2 \\ & =\frac{2 \mathrm{k}}{\mathrm{r}}\left(\mathrm{I}_1 \hat{\mathrm{i}}-\mathrm{I}_2 \hat{\mathrm{j}}\right)=\frac{2 \times \mu_0}{4 \pi \mathrm{~d}}(8 \hat{\mathrm{i}}-6 \hat{\mathrm{j}}) \\ & \therefore \mathrm{B}_{\mathrm{p}}=\frac{\mu_0}{2 \pi \mathrm{~d}} \sqrt{(8)^2+(-6)^2}=\frac{5 \mu_0}{\pi \mathrm{~d}}\end{aligned}$

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