The path of a charged particle in a uniform magnetic field depends on the angle `theta` between velocity vector and magnetic field, When `theta is 0^(@) or 180^(@), F_(m) = 0` hence path of a charged particle will be linear.
When `theta = 90^(@)`, the magnetic force is perpendicular to velocity at every instant. Hence path is a circle of radius `r = (mv)/(qB)`.
The time period for circular path will be `T = (2pim)/(qB)`
When `theta` is other than `0^(@), 180^(@) and 90^(@)`, velocity can be resolved into two components, one along `vec(B)` and perpendicular to B.
`v_(|/|)=cos theta`
`v_(^)= v sin theta`
The `v_(_|_)` component gives circular path and `v_(|/|)` givestraingt line path. The resultant path is a helical path. The radius of helical path
`r=(mv sin theta)/(qB)`
ich of helix is defined as `P=v_(|/|)T`
`P=(2 i mv cos theta)`
`p=(2 pi mv cos theta)/(qB)`
A charged particle moves in a uniform magnetic field. The velocity of particle at some instant makes acute angle with magnetic field. The path of the particle will be

Options

A. A stralight line
B. A circle
C. A helix with uniform pitch
D. A helix with non-uniform pitch

Answer

C

Explanation

In the given question, the angel between `vec(V) and vec(B)` are acute, hence path will be helical. In the direction of magnetic field, particle moves with constant speed and hence the pitch is uniform. So choice (c) is correct.