Question

A direct current flowing through the winding of a long cylindrical solenoid of radius R produces in it a uniform magnetic field of induction B. An electron flies into the solenoid along the radius between its turns (at right angles to the solenoid axis) at a velocity v (Fig. 107).

After a certain time, the electron deflected by the magnetic field leaves the solenoid.

Determine the time t during which the electron moves in the solenoid.

Answer 1

The magnetic induction of the solenoid is directed along its axis. Therefore, the Lorentz force acting on the electron at any instant of time will lie in the plane perpendicular to the solenoid axis. Since the electron velocity at the initial moment is directed at right angles to the solenoid axis, the electron trajectory will lie in the plane perpendicular to the solenoid axis. The Lorentz force can be found from the formula F = evB. The trajectory of the electron in the solenoid is an arc of the circle whose radius can be determined from the relation evB = mv²/r, whence

The trajectory of the electron is shown in Fig. 223 (top view), where 01 is the centre of the arc AC described by the electron, v' is the velocity at which the electron leaves the solenoid. The segments OA and OC are tangents to the electron

trajectory at points A and C. The angle between v and v' is obviously φ = ∠AO₁C since ∠OAO₁ = ∠OCO₁.

In order to find φ, let us consider the right triangle OAO₁ : side OA = R and side AO₁ = r. Therefore, tan (φ/2) = R/r = eBR/(mv). Therefore,

Obviously, the magnitude of the velocity remains unchanged over the entire trajectory since the Lorentz force is perpendicular to the velocity at any instant. Therefore, the transit time of electron in the solenoid can be determined from the relation