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Question
A particle of charge ‘q’ and mass ‘m’ is moving with velocity .`vecV` It is subjected to a uniform magnetic field `vecB` directed perpendicular to its velocity. Show that it describes a circular path. Write the expression for its radius.
Solution
A charge ‘q’ projected perpendicular to the uniform magnetic field ‘B’ with velocity ‘v’. The perpendicular force F = qv × B acts like a centripetal force perpendicular to the magnetic field. Then the path followed by charge is circular as shown in the figure below.
The Lorentz magnetic force acts as centripetal force thus,
`qvB =(mv^2)/r`
`r =(mv)/(qB)`
Here, r = radius of the circular path followed by charge projected perpendicular to the uniform magnetic field.
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