Solution:
The time period of revolution in a magnetic field is \[T = \frac{2\pi m}{qB}\]. It is independent of the velocity \(v\) and inversely proportional to \(B\). Thus, doubling \(B\) makes the time period half of its initial value.
A charged particle is moving on circular path with velocity \(v\) in a uniform magnetic field \(B\). If velocity of the particle and strength of magnetic field is doubled, then time taken to complete one revolution becomes
Leave a Reply