In uniform magnetic field, if angle between `vec(v) and vec(B) is 0^(@) lt 0 lt 90^(@)`, the path of particle is helix. Let `v_(1)` be the component of `vec(v) along vec(B) and v_(2)` be the component perpendicular to `vec(B)`. Suppose p is the pitch. T is the time period and r is the radius of helix. Then
`T = (2pim)/(qB), r = (mv_(2))/(qB), P = (v_(1))T`
Assume a charged particle of charge q and mass m is released from the origin with velocity `vec(v) = v_(0) hat(i) - v_(0) hat(k)` in a uniform magnetic field `vec(B) = -B_(0) hat(k)`.
Angle between v and B is

Options

A. `45^(@)`
B. `30^(@)`
C. `60^(@)`
D. `120^(@)`

Answer

A

Explanation

`vec(v)*vec(B)=|v||B| cos theta`
or `cos theta =(vec(v)*vec(B))/(|v||B|)=(B_(0)v_(0))/(sqrt(2)v_(0)B_(0))=(1)/(sqrt(2))`
or `theta=45^(@)`
Hence choice (a) is correct.