Question

In the electron gun of a TV picture tube the electrons (charge ?e, mass m) are accelerated by a voltage V. After leaving the electron gun, the electron beam travels a distance D to the screen; in this region there is a transverse magnetic field of magnitude B and no electric field.

Find the approximate deflection of the beam due to this magnetic field. (Hint: Place the origin at the center of the electron beam

Answer 1

As it is accelerated through V volts, by energy conservation eV= kinetic energy = mv^2/2 From here find the value of v. When an electron of velocity v is injected into a transverse magnetic field of strength B, it follows a circular path of radius ,r = mv/eB. We have to find the deflection which means the deviation from its path which in turn means the angle between initial velocity vector and the final velocity vector. Check the file attached. Since it is a circular path, the velocity vector which is tangent to trajectory, is therefore perpendicular to radius vectors at two points( drawn in red).So angle between velocity vectors is equal to angle between radius vectors (shown as theta). Calculating the value of r as above and given the distance between gun and screen as D , we can find the value of theta from simple trigonometry in the rt triangle ABC .