A particle of mass m and charge q moves with a constant velocity V along the positive x-direction . It enters a regio containing a uniform magnetic field B directed along the negative z-direction, extending from x = a to x = b . The minimum value of V required, so that the particle can just enter the region of `x gt b ` is

Options

A. `(qbB)/(m)`
B. `(qaB)/(m)`
C. `(q(b-a)B)/(m)`
D. `(q(b+a)B)/(2m)`

Answer

C

Explanation

Given : Mass of particle = m, charge on particle = q , velocity of particle = v and uniform magnetic field = B . We know that when a charged particle enters in a uniform magnetic field , directed vertically upwards, then the particale will move in a circular path . And radius of this circular path
`(r) = b - a = (mv)/(qB)`
`:.` Particle cannot enter the region `x gt` b. And if the particles is to enter the region `x gt ` b , then ` r gt (b - a)` .