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

Velocity component along the field `V_(||) = 4 xx 10^(5) xx cos 60^(@)`
`= 2 xx 10^(5) m//s`
and `v_(bot) = (4 xx 10^(5)) sin 60^(@) = 2 sqrt3 xx 10^(5) m//s`
Proton will describe a circle in plane perpendicular to magnetic field with radius
`r = (mv_(bot))/(qB) = ((1.67 xx 10^(-27) kg) xx (2 sqrt3 xx 10^(5) m//s))/((1.6 xx 10^(-19) C) xx (0.3 T))`
`= 1.2 cm`
Time taken to complete one revolution is
`T = (2pi r)/(v_(bot)) = (2 xx 3.14 xx 0.012)/(2 sqrt3 xx 10^(5) m//s)`
Because of `v_(||)` protons will also move in the direction of magnetic field.
Pitch of helix `= v_(||) xx T`
`= (2 xx 10^(5) xx 2 xx 3.14 xx 0.012)/(2 sqrt3 xx 10^(5) m//s)`
`= 0.044 = 4.4 cm`