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 helical path taken by the proton beam.

Explanation

The components of the proton's velocity along
and perpendicular to the magnetic filed are
`v_(||) = ((4xx10^5 ms^(-1) cos 60^@ = 2xx10^5 ms(-1).
and `v_|_ = (4X10^5ms^(-1)) sin60^@ = 2sqrt3 X 10^5 ms^(-1).`
As the force vecqvX vecB is perpendicular to the magnetic field, the component `v_11` will remain constant. IN the plane perpendicular to the field, the proton will equation `qv_|_ B = (mv_|_^2)/(r) or, r = (mv_|_ )/(qB) = ((1.67X 10 ^(-27) kg)X (2sqrt3 X10^5ms^(-1)) /(1.6X 10^(-19) C) X (0.3T) ~~ 0.012m = 1.2 cm`
. The time taken in one compete revolution in the plane perpendiuclar to B is T=(2pir)/(v_|_) = (2X3.14X0.12m )/ (2sqrt3 X 10^5 ms^(-1)
. The distance moved along the filed during this period, i.e., the pitch =(2X10^5ms^(-1) X2X3.14X 0.044m = 4.4cm.
The qualitative nature of the path of the protons is shown in figure.