Question

A proton is accelerated through a potential difference V, subjected to a uniform magnetic field acting normal to the velocity of the proton. If the potential difference is doubled, how will the radius of the circular path described by the proton in the magnetic field change?

Answer 1

Answer:

root 2 times increase in magnetic field

Explanation:

The radius of the Circular path by the proton in the magnetic field change can be described as

r= 1/B(2mV/q)^½

B = magnetic field

V= potential difference

So, r'/r = (2V/V)^½ = root 2