The magnetic field at the point of intersection of diagonals of a square wire loop of side \(\mathrm{L}\) carrying a current \(\mathrm{I}\) is

• A

  \(\frac{\mu_0 \mathrm{I}}{\pi \mathrm{L}}\)
• B

  \(\frac{2 \mu_0 \mathrm{I}}{\pi \mathrm{L}}\)
• C

  \(\frac{\sqrt{2} \mu_0 I}{\pi \mathrm{L}}\)
• Correct answer
  D

  \(\frac{2 \sqrt{2} \mu_{\mathrm{o}} \mathrm{I}}{\pi \mathrm{L}}\)

Solution

\(\begin{aligned} & \text {Hints: } B=4\left[\frac{\mu_0}{4 \pi} \frac{I}{\left(\frac{\mathrm{L}}{2}\right)}\left(\sin 45^{\circ}+\sin 45^{\circ}\right)\right] \\ & =\frac{\mu_{\mathrm{o}}}{\pi} \frac{2 \mathrm{I}}{\mathrm{L}} \cdot \frac{2}{\sqrt{2}} ; \quad B=\frac{\mu_{\mathrm{o}}}{\pi} \frac{2 \sqrt{2} \mathrm{I}}{\mathrm{L}}\end{aligned}\)

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