Question

A beam of protons with a velocity of `4 X 10 ^5 ms^(-1)` enters a uniform magnetic field of 0.3 T. The velocity makes an angle of `60^@` with the magnetic field. Find the radius of the helicla path taken by the proton beam and the pitch of the helix.

Answer 1

Correct Answer - A::B::D
`v_(1)` is responsible for horizontal motion of proton
`v_(2)` is responsible for circular motion of proton
:. `(mv_(2)^(2))/(qB) = ( 1.76xx10^(-27)xx4xx10^(5)xxsqrt(3))/(1.6xx10^(-19)xx0.3xx2) = 0.012 m`
Pitch of helix `= v_(1)xxT`
where ` T = (2pir)/(v_(2)) = (2 pir)/(v sin theta)`
rArr Pitch of helix `= v cos thetaxx(2pir)/(v sin theta)`
`= 2pir cot theta = 2xx3.14xx0.012xxcot60^(@) = 0.044 m`