A beam of protons enters a uniform magnetic field of 0.3 T,with a velocity of `4xx10^(5)ms^(-5),` in a direction making an angle of`60^(@)` with the direction of magnetic field.Determine `
` (i) the path of motion of the particles `
` (ii) the radius of the path of the particles `
` (iii) the pitch of the helix.

Explanation

(i) When the charged particle enters a magnetic field at angle other than `90^(@)` , it follow a helical path.
(ii) Force on proton =Bev
centripetal force `=(mv^(2))/("r")` , `
:. Bev=(mv^(2))/(.r")` or
`r=(mv)/(Be)=(1.67xx10^(-27)xx4xx10^(5))/(0.3xx(1.6xx10^(-19))`
Radius of the path, r=0.014m
(iii) Pitch of the path,It is the liner distance covered by the charged particles , in the one time period in the direction of magnetic field .
Pitch `=(2pimvcos theta)/(eB)=(2picos theta)xxr`
`=2xx3.14xxcos60^(@)xx0.014=0.04m`