A particle having a charge of `5.0 mu C` and a mass of `5.0xx 10^(-12)kg` is projected with a speed of `1.0km s^(-1)` in a magnetic field of a magnetic field of magnitude 5.0 m T. The angle between the magnetic field and the velocity is `sin^(-1)(0.90)`. show that the path of the particle will be a helix. Find the diameter of the helix and its pitch.

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 .