A charged particle of mass m and charge q is projected into a uniform magnetic field of induciton `vecB` with speed v which is perpendicular to `vecB`. The width of the magnetic field is d. The impulse imparted to the particle by the field is `(dlt lt mv//qB)`

Options

A. `qBd`
B. `qBv//d`
C. `mv//(qBd)`
D. `mv^2//(qBd)`

Answer

A

Explanation

Force on the charged particle in magnetic field is `F=qvB`, acting in the direction as shown in figure. It provides the required centripetal force. Due to it, the charged particle describes a circular path AC of radius r with centre at O. Then

`qvB=(mv^2)/(r)` or `r=(mv)/(qB)` or `v/r=(qB)/(m)`
In figure, `d=r sin theta`
When `dltltr`, `sin theta~~theta=d//r`
Change in velocity of charged particle from A to C is, `Deltav=vsin theta=v theta` (for small `theta`), acting along CF.
Thus impulse imparted =change in momentum
`=Deltap=mDeltav=mvtheta=mv(d//r)=md(qB//m)`
`=qBd`