Magnetic Effects & Magnetism

JEE Main level, numerical answer, medium

127 of 276

Question 127 of 276

An electron in a hydrogen atom revolves around its nucleus with a speed of 6.76×106ms–16.76\times {10}^{6}m{s}^{–1} in an orbit of radius 0.52A˚0.52Å. The magnetic field produced at the nucleus of the hydrogen atom is _______ TT.

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Answer: 40

The magnetic field due to the electron is given by
B=μ0qvsin⁡90∘4πr2B=\dfrac{{\mu }_{0}qv\sin 90^{\circ}}{4\pi {r}^{2}}
The given data is
v=6.76×106ms−1r=0.52×10−10mv=6.76\times {10}^{6}m{s}^{-1} r=0.52\times {10}^{-10}m
Thus, the magnetic field is
B=10−7×(1.6××10−19)×(6.76×106)0.52×0.52×10−20=40TB=\dfrac{{10}^{-7}\times \left(1.6\times \times {10}^{-19}\right)\times \left(6.76\times {10}^{6}\right)}{0.52\times 0.52\times {10}^{-20}}=40T

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