JEE Mains · Physics · STD 12 - Moving charges and magnetism
- JEE Mains 2025
- Medium
- Numerical
A current of 5A exists in a square loop of side \(\frac{1}{\sqrt{2}} \mathrm{~m}\). Then the magnitude of the magnetic field \(B\) at the centre of the square loop will be \(p \times 10^{-6} \mathrm{~T}\). where, value of p is _____ .
\(\left[\right.\) Take \(\left.\mu_0=4 \pi \times 10^{-7} \mathrm{TmA}^{-1}\right]\).
Pick an option to check your answer.
Answer & Solution
Correct Answer
(B) 8
Step-by-step Solution
Detailed explanation

Let B be the magnetic field due to single side
\(\begin{aligned}
& \text { then } B=\frac{\mu_0 \mathrm{i}}{4 \pi \mathrm{~d}}\left(\sin \theta_1+\sin \theta_2\right) \\
& =\frac{10^{-7} \times 5 \times 2}{\frac{1}{2 \sqrt{2}}} \times \frac{1}{\sqrt{2}}=2 \times 10^{-6} \\
& \therefore B_{\text {net }} \text { at centre } O=4 B \\
& =8 \times 10^{-6}
\end{aligned}\)
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