A particle having a charge of `100 mu C` and a mass of 10mg is projected in a uniform magnetic field of 25m T with a speed of `10ms^(-1)`. If the velocity is perpendicular to the magnetic field , how long will it take for the particle to come back to its original position for the first time after being projected.

Explanation

The particle move along a circle and returns
to its original position after completing to circle, that
is after one time period. The time period is
`T= (2pim)/(qB)`
`= (2pi xx (10X10^(-6)kg))/(100xx10^(-6)C) xx(25xx10^(-3) T) = 25s.`
If the velocity of the charge is not perpendicular to
the magnetic field, we can break the velocity in two
components- `v_||,` parallel to the field and `v_|_,`
perpendicular to the field. the component `v_11` remains
unchanged as the force `vec quX vec B` is perpendicular to it . in
the perpendicular to the field, the particle traces
a circle of radius `r=(mv_|_)/(qB)` as given by equation . The
resultant path is a helix .