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In the bohrmodel of hydrogen atom, the electron circles around the nucleus with a speed 5×10^5 m/s In an orbit of radius 5×10^-11 m. What is the magnetic field at the centre of the orbit. If possible can somebody please solve it using the formula B=mv/qr Thank you

In the bohrmodel of hydrogen atom,  the electron circles around the nucleus with a speed 5×10^5 m/s In an orbit of radius 5×10^-11 m. What is the magnetic field at the centre of the orbit. 
If possible can somebody please solve it using the formula B=mv/qr
Thank you

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2 Answers

Khimraj
3007 Points
5 years ago
 
AS B = mv/qr
m = 9.1*10-31Kg
q = 1.6*10-19 C
v = 5*105 m/s
r = 5*10-11 m
so B = 5.7*104 T
Hope it clears.
Rohit
103 Points
5 years ago
See as the electron revolves around the nucleus in circular path, velocity will be perpendicular to magnetic field. In this case CPetal force is balanced by magnetic Lorentz force.
So we get, B= mv/qr
Now Mass of electron= 9.1×10^-31 kg.
e = 1.6 × 10^-19 C( take only magnitude of e).
By substitution we get,
B = 5.69 × 10^4 T.
 
 
Hope it helps
Regards 
Rohit.

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