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# Q.  An electron revolves with frequency 6.6 * 1015 r.p.s. around nucleas in circular orbit of radius 0.53A0 of hydrogen atom, then magnetic field produced at the centre of orbit is-                                            1. 0.125T              2.   1.25T               3.  12.5T              4.    125T​

deepak
84 Points
4 years ago
see the current due to an electron revolving with the frequency if 6.6x1015 r.p.s is same as that when 6.6x1015 electrons move around the orbit in a second.

hence the total current in the orbit is 6.6 x 1015 x 1.6 x10-19 = 10.56 x 10-4 A

now consedering the orbit to be a circular one,

$\overrightarrow{B}$= $\nu$I / 2R

$\overrightarrow{B}$= 12.5 T

hence option 3 is correct
one year ago
Dear student,

The total current in the orbit, I = q/T = f x q = 6.6 x 1015 x 1.6 x10-19 = 10.56 x 10-4 A
now consedering the orbit to be a circular one,
Magnetic field at the center of orbit, B = μ0I/2R
Hence, B = 12.5 T
Hence option 3 is correct

Hope it helps,
Thanks and regards,
Kushagra