Two identical charged particles moving with same speed enter a region of uniform magnetic field. If one of these enters normal to the field direction and the other enters along a direction at `30^@` with the field, what would be the ratio of their angular frequencies?

Explanation

When a charged particle of charge q, mass m, moving with velocity v enters a uniform magnetic field B, acting perpendicular to the direction of motion of charged particle, then the magnetic force on the particle provides centripetal force. So
`qvB=mv^2//r` or `v//r=qB//m`
The particle will describe a circular path of radius r. Angular frequency of particle,
`omega=v/r=(qB)/(m)`
When the charged particle is moving, making an angle `30^@` with the direction of magnetic field, then due to component velocity `v cos 30^@` (along the direction of field) no force acts on the particle. Due to component velocity ` v sin 30^@`, (acting normally to the direction of field ) of force is acting on the particle which provides the required centripetal force for circular motion of particle. Thus,
`q(v sin 30^@)B=(m(v sin 30^@)^2)/(r)`
or `(vsin 30^@)/(r)=(qB)/(r)`
Angular frequency `omega=(v sin 30^@)/(r)=(qB)/(r)`
Thus, the angular frequency of both the particles is same. So ratio of their angular frequency
`=1:1`