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A proton accelerated by a potential dif...

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^(@)`

Text Solution

Verified by Experts

The correct Answer is:
2
`Bqv=(mv^(2))/(r),(1)/(2)mv^(2)=Vq` and `sin theta==(d)/(r)`
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