Question

A proton accelerated by a potential difference `500 KV` moves though a transverse field of `0.51 T` as shown in figure. The angle `theta` through which the proton deviates from the intial direction of its motion is

A. `15^(@)`
B. `30^(@)`
C. `45^(@)`
D. `60^(@)`

Answer 1

Correct Answer - 2
`Bqv=(mv^(2))/(r),(1)/(2)mv^(2)=Vq` and `sin theta==(d)/(r)`