Question

If a proton enters perpendicularly a magnetic field with velocity V, then time period of revolution is T. If proton enters with velocity 2v. Then time period will be.

• A. T
• B. 2 T
• C. 3 T
• D. 4 T

Answer 1

Answer: (A)

T

Given : If a proton enters perpendicularly a magnetic field with velocity V, then time period of revolution is T. If proton enters with velocity 2v.

Solution:

We have formula for radius of circular path R=$\frac{mv}{qB}$

and Time Period $\frac{2\pi R}{v}$ ,

where v is velocity, m is mass , B is magnetic field and q is charge.

substituting value of R in time period formula we get T=$\frac{2\pi m}{qB}$

From here we can say that time period doesnot depends on velocity of particle .

The correct Opt=A