WBJEE · Physics · Magnetic Effects of Current
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 \(\frac{\sqrt{2} \mu_0 I}{\pi \mathrm{L}}\)
- D \(\frac{2 \sqrt{2} \mu_{\mathrm{o}} \mathrm{I}}{\pi \mathrm{L}}\)
Answer & Solution
Correct Answer
(D) \(\frac{2 \sqrt{2} \mu_{\mathrm{o}} \mathrm{I}}{\pi \mathrm{L}}\)
Step-by-step Solution
Detailed explanation
\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…
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