Physics · Moving Charges and Magnetic Field

JEE Main 2025 — 24 January, Morning Shift — Question 70

A current of 5 A 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

$\mathrm{p} \times 10^{-6} \mathrm{~T}$ . where, value of p is $\qquad$ .[0pt]

[Take $\mu_{0}=4 \pi \times 10^{-7} \mathrm{~T} \mathrm{~mA}^{-1}$ ].

Answer: 8

Numerical answer — enter this value.

Let B be the magnetic field due to single side

then

$B=\frac{\mu_{0} \mathrm{i}}{4 \pi \mathrm{~d}}\left(\sin \theta_{1}+\sin \theta_{2}\right)$

$=\frac{10^{-7} \times 5 \times 2}{\frac{1}{2 \sqrt{2}}} \times \frac{1}{\sqrt{2}}=2 \times 10^{-6}$

$\therefore \mathrm{B}_{\text {net }}$ at centre $\mathrm{O}=4 \mathrm{~B}$

$=8 \times 10^{-6}$

$\therefore \mathrm{P}=8$

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