A proton with a velocity of `4xx10^5m//s` enters a uniform magnetic field region of induction 0.3 T. The angle between the velocity vector and magnetic field is `30^@`. The pitch of helix in this case, will be

Explanation

`q = 5 muC = 5 xx 10^(-6) C`
`m = 5 xx 10^(-12) kg`
`u = 1 Km/s`
`Q = sin ^(-1) (0.9)`,
`B = 5 xx 10^(-2) T`,
We have `(mu^(2) /r = qvB sin sin theta`
`r = (mu)/(qB sin sin theta) = 0.18`
Hence distance `= 36 cm`.
The velocity has on `x` component along wich there is no force so the particle move with uniform velocity .
The velocity has a `y` component with which is accelarates accaleration a eith vertical `x` component it move in a circlar cross section . Thus it move in a helix .