Discuss the motion of charged particle in uniform magnetic field, when its moves at an angle `theta` with the direction of magnetic field. Prove that its path is helical. Calculate the pitch of helical path. What will be the nature of path if `theta=90^@`

Explanation

Let at any instant particle is moving with velocity v inside the uniform magnetic field `vecB` making an angle `theta` with the direction of magnetic field.
Resolve `vecv` into two rectangular components
(i) `v_(B) = v cos theta =` the component of `vecv` along the direction of magnetic field.
(ii) `v_(n) = v cos theta =` the component of velocity `vecv` normal to the direction of magnetic field.
As the magnetic field act perpendicular to the velocity component `v_(n)`, the charged particle moves along a circular path.


As centripetal force is provided by magnetic force
`therefore Bqv_(n) = (mv_(n)^(2))/(r)` [`v_(n)` is responsible for their motion in circular path]
`r = (mv_(n))/(Bq) = (mv sin theta)/(Bq)" "...(i)`
The period of the circular path is given by
`T = (2pir)/(v_(n)) = (2pi)/(v sin theta) xx ( mv sin theta)/(Bq)`
or `T = (2pim)/(Bq) " "...(ii)`
As the charged particle moves along circular path in XY-plane due to the velocity component `v_(n)`, it also advances linearly along OZ due to the velocity component `v_(B)`. As a result, the charged particle will move along the path as shown in Fig. Such a path is called helical path.
Pitch : The distance travelled by the charged particle. along the direction of magnetic field in a time it completes one revolution. is called pitch of the helical path. Therefore
Pitch of the helical path ` = v_(B) xx T = v cos theta xx (2pim)/(Bq)`
Pitch ` = (2pimvcos theta)/(Bq)`
Pitch of the helical path is equal to the spacing between its two consecutive circular paths
(i) if `theta = 90^(@)` i.e., when charged particle moves in a direction perpendicular to the direction of magnetic field `(vecB)` then
`r = (mv sin 90)/(Bq)`
`r = (mv)/(Bq)`
`v_(B) = v cos 90^(@) = 0`
It means that charged particle moves in a circular path.
(ii) If `theta = 180^(@)` then F `F = Bqv sin 180^(@) =0`. Magnetic force acting on the charged particle ix zero and hence it will continue to move along the same path antiparallel to `vecB` with same velocity `vecv` .