You have been tasked to use the magnetic field of a solenoid

You have been tasked to use the magnetic field of a solenoid to bend the path of a beam of electrons to form a 90o turn to the right. The radius of the solenoid is 2.0 cm. The electrons are traveling with a speed of 3.00 x 105 m/s. The solenoid has small holes in the windings so that the electron beams can enter and exit through the wall of the solenoid. The solenoid has 10 windings per centimeter. What is the current flowing through the solenoid and in what direction does the current travel around the solenoid?

Solution

magneic force,

                     F= Bvq

             mv^2/r = Bvq

              mv/r = Bq

hence, the magnetic field is

               9.11x10^-31 * 3.00 x 10⁵ / 0.02 = B*1.6x10^-19

then,

       B = 8.54e-5 T

Thus, the current in the solenoid is,

   B = u0nI

8.54e-5 = 4*pi*10^-7 *[10/10^-2]*I

Current is, I = 0.06796 A = 67.96 mA