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An (alpha)-particle and a proton are bot...

An `(alpha)`-particle and a proton are both simultaneously projected in opposite direction into a region of constant magnetic field perpendicular to the direction of the field. After some time it is found that the velocity of the `(alpha)`-particle has changed in a direction by `45^(@)`. Then at this time, the angle between velocity vectors of `(alpha)`-particle and proton is

A

`90^(@)`

B

`45^(@)`

C

`135^(@)`

D

none

Text Solution

Verified by Experts

The correct Answer is:
C
`omega=qB//m`
`omega_(p)=eB//m, omega_(aplha)=(2eB)/(4m)=2 omega_(p)`
so proton will cover doubel angle is same time as that of `(alpha)`-particle. Clearly from the above figure angle between `v_(alpha)` and `v_(p) is 135^(@)`
Path of electron is shown in fig. Deviation of electron will be `theta=90^(@)`
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