A proton and an `alpha` -particle are simultaneously projected in opposite direction into a region of uniform magnetic field of 2 mT perpendicular to the direction of the field. After some time it is found that the velocity of proton has changed in direction by `90^(@)`. Then at this time, the angle between the velocity vectors of proton and  `alpha` - particle is

Options

A. `60^(@)`
B. `90^(@)`
C. `45^(@)`
D. `180^(@)`

Answer

C

Explanation

Key Idea In a circular motion a body changes its direction by 90° in one fourth of its time period.
Given, magnetic field, `B = 2mT`
Let `T_(p)`, be the time-period of revolution of proton in the magnetic field.
`T_(p) = (2pim_(p))/(eB)" "...(1)`
`m_(alpha) cong 4 mp and q alpha = 2q ^(p)`
and `T_(alpha)` , be the time-period of revolution of `alpha`-particle in the magnetic field,
`T_(alpha) = (2 pi m _(p))/(qB) = (2 pi(4m_(p)))/((2e)B)" " ...(ii)`
Now, the ratio of time period of proton and `alpha` -particle, by dividing Eq. (i) and (ii)
`(T_(p))/T_(alpha) = 1/2`
`rArr T_(alpha ) = 2 T_(p)`
Hence, the time-period of `alpha`-particle is double of the proton, i.e. if proton covers `90^(@)` of angle from its starting, then`alpha` -particle will cover `45^(@)` of the angle.
`therefore` Hence, the correct option is ( c).