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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​

Naina Singh , 8 Years ago
Grade 12th pass
anser 2 Answers
deepak

Last Activity: 8 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}= \nuI / 2R
 
\overrightarrow{B}= 12.5 T
 
hence option 3 is correct
Kushagra Madhukar

Last Activity: 5 Years ago

Dear student,
Please find the attached solution to your problem below.
 
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
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