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JEE Mains · Physics · STD 12 - Moving charges and magnetism
  • JEE Mains 2019
  • Medium
  • Numerical

The magnitude of the magnetic field at the centre of an equilateral triangular loop of side \(1\,m\) which is carrying a current of \(10\,A\) is:......\(\mu T\) [Take \(\mu _0 = 4\pi  \times 10^{-7}\,NA^{-2}\)]

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Answer & Solution

Correct Answer

(C) \(3\)

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Detailed explanation

\(i=10\, \mathrm{A}\)

\(\ell=1\, \mathrm{m}\)

\(\mu_{0}=4 \pi \times 10^{-7} \,\frac{\mathrm{N}}{\mathrm{A}^{2}}\)

\(\mathrm{B}=\frac{\mu_{0} \mathrm{i}}{\frac{4 \pi \sqrt{3} \ell}{2}} \times 3\)

\(=\frac{\mu_{0} \mathrm{i} \sqrt{3}}{2 \pi \ell}=\frac{4 \pi \times 10^{-7} \times 10 \times \sqrt{3}}{2 \pi \times 1}=20 \sqrt{3} \times 10^{-7}\)

\(=3 \,\mu \mathrm{T}\)

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