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)`.
Pitch of helical path described by particle is

Options

A. `(2mv_(0))/(qB_(0))`
B. `(2pimv_(0))/(qB_(0))`
C. `(sqrt(2)pimv_(0))/(qB_(0))`
D. `(sqrt(3)pimv_(0))/(qB_(0))`

Answer

B

Explanation

Pitch of helix `P=V_(||) xx T`
`P=v_()((2pim)/(qB_(0)))=(2 pi mv_(0))/(qB_(0)`
Hence choice (b) is correct.