JEE Mains · Physics · STD 12 - 4. Moving charges and magnetism
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 \(......T\).
- A \(40\)
- B \(50\)
- C \(30\)
- D \(20\)
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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