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JEE Mains · Physics · STD 12 - Moving charges and magnetism
  • JEE Mains 2023
  • Medium
  • Numerical

An electron in a hydrogen atom revolves around its nucleus with a speed of \(6.76 \times 10^6\,ms ^{-1}\) in an orbit of radius \(0.52\,\mathring A\). The magnetic field produced at the nucleus of the hydrogen atom is \(......T\).

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Answer & Solution

Correct Answer

(A) \(40\)

Step-by-step Solution

Detailed explanation

Magnetic field due to moving charge

\(B =\frac{\mu_0}{4 \pi} \frac{ q \sin \theta}{ r ^2}\)

\(B =\frac{\mu_0}{4 \pi} \frac{ ev \sin (\pi / 2)}{ r ^2}\)

\(B =\frac{10^{-7} \times 1.6 \times 10^{-19} \times 6.76 \times 10^6}{0.52 \times 0.52 \times 10^{-20}}\)

\(B =40\,T\)

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