An electron of charge e moves in a circular orbit of radius r around a nucleus. The magnetic field due to orbital motion of the electron at the site of the nucleus is B. The angular velocity ω of the electron is
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a
4πrBμ0e
b
2πrBμ0e
c
μ0eBπr
d
μ0eB4πr
answer is A.
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Detailed Solution
The electron moving in a circular orbit is equivalent to a current carrying loop. If v be its frequency, theni=ve=e/T (T= time period )We know that T=(2πr/v)∴ i=ev2πr=e(rω)2πr=eω2π (where ω = angular frequency)The magnetic field at the centre of the loopB=μ0i2ror B=μ02reω2π=μ0eω4πr∴ ω=4πrBμ0e