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In the Hydrogen atom, the electron is making
$6.6 \times {10^{15}}\,r.p.s.$
If the radius of the orbit is
$0.53 \times {10^{ - 10}}\,metre,$
then magnetic field produced at the centre of the orbit is

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answer is B.

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Detailed Solution

B = \frac{{{\mu _0}i}}{{2r}} = \frac{{{\mu _0}q\nu }}{{2r}} = \frac{{4\pi \times {{10}^{ - 7}} \times 1.6 \times {{10}^{ - 19}} \times 6.6 \times {{10}^{15}}}}{{2 \times 0.53 \times {{10}^{ - 10}}}}

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