Magnetic Effects & Magnetism

JEE Main level, numerical answer, medium

30 of 276

Question 30 of 276

A current of 5A exists in a square loop of side 12 m\frac{1}{\sqrt{2}} \mathrm{~m}. Then the magnitude of the magnetic field BB at the centre of the square loop will be p×10−6 Tp \times 10^{-6} \mathrm{~T}. where, value of p is _____ .
[\left[\right. Take μ0=4π×10−7TmA−1]\left.\mu_0=4 \pi \times 10^{-7} \mathrm{TmA}^{-1}\right].

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Answer: 8



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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