A particle having a mass of 0.5g carries a charge of `2.5xx10^-8C`. The particle is given an initial horizontal velocity of `6xx10^4ms^-1`. To keep the particle moving in a horizontal direction

Options

A. The magnetic field may be perpendicular to the direction of the velocity
B. the magnetic field should be along the direction of the velocity
C. magnetic field should have a minimum value of 3.27T
D. no magnetic field is required

Answer

(a,c)

Explanation

In the absence of a magnetic field, the particle will experience gravitational force mg. As a result, the particle will not continue moving in the horizontal direction but will describe a parabolic path. So, a magnetic field must be present and its direction must be perpendicular to the direction of the velocity. The magnetic force experience by the particle is give by `vecF=q(vecvxxvecB)`.
The magnitude of the force is `F=qvB sin theta`. If the particle is to move in the horizontal direction, this force must balance the force of gravity, i.e., `mg=qvB sin theta`
The minimum value of B corresponds to `sin theta=1` or `theta=90^@`
Thus, mg=qvB
or `B=(mg)/(qv)=(0.5xx10^-3xx9.8)/(2.5xx10^-8xx6xx10^4)=3.27T`
Hence, the correct options are (a) and (c).