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Magnetics question

2025 · 24 Jan · Shift 1 · Q72
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Magnetics question

2025 · 24 Jan · Shift 1 · Q72

JEE MainPhysicsMagneticsNumerical+4 / −1
A current of 5 A exists in a square loop of side 12 m\frac{1}{\sqrt{2}} \mathrm{~m}2​1​ m. Then the magnitude of the magnetic field BBB at the centre of the square loop will be p×10−6 Tp \times 10^{-6} \mathrm{~T}p×10−6 T. where, value of p is ‾[\underline{\hspace{2cm}}\left[\right.​[ Take μ0=4π×10−7 T mA−1\mu_0=4 \pi \times 10^{-7} \mathrm{~T} \mathrm{~mA}^{-1}μ0​=4π×10−7 T mA−1 ].
Numerical answer
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