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A particle with specific charge `q//m` is located inside a round solenoid at a distance `r` from its axis. With the current swichted into the winding, the magnetic induction of the field generated by the solenoid amounts to `B`. Find the velocity of the particle and the curvature radius of its trajectory,n assuming that during the increase of current flowing in the solenoid the particle shifts by a negligible distance.

Answer» When the magnetic field is being set up in the solenoid, and electric field will be induced in it, this will accelerate the charged paritcle. If `B` is the rate, at which the matgnetic field is increasing, then.
`pi r^(2) dotB = 2pi r E` or `E = (1)/(2) r dotB`
Thus, `m (dv)/(dt) = (1)/(2) r dotB q`, or `v = (qBr)/(2m)`,
After the field is set up, the parcticle will execute a circular motion of radius `rho`, where
`mv = B q rho`, or `rho = (1)/(2) r`


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