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......$Tesla$
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(b) $i = q\nu $
$B = \frac{{{\mu _0}i}}{{2r}} = \frac{{{\mu _0}q\nu }}{{2r}} = \frac{{4\pi \times {{10}^{ - 7}} \times 1.6 \times {{10}^{ - 19}} \times 6.6 \times {{10}^{15}}}}{{2 \times 0.53 \times {{10}^{ - 10}}}}$
$ = \frac{{2\pi \times 1.6 \times 6.6}}{{5.3}} = 12.518\, Tesla$
$B = \frac{{{\mu _0}i}}{{2r}} = \frac{{{\mu _0}q\nu }}{{2r}} = \frac{{4\pi \times {{10}^{ - 7}} \times 1.6 \times {{10}^{ - 19}} \times 6.6 \times {{10}^{15}}}}{{2 \times 0.53 \times {{10}^{ - 10}}}}$
$ = \frac{{2\pi \times 1.6 \times 6.6}}{{5.3}} = 12.518\, Tesla$
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