AP Physics C E and M

Question

An electron is accelerated by a potential difference of 3000V and enters a region of a uniform magnetic field. As a result the electron bends along a path with a radius of curvature of 26.0 cm.

1. Find the speed of the electron in [Mm/s]as it enters the magnetic field.
2. Find the magnitude of the magnetic field in [µT].
3. Find the speed of the electron as it leaves the magnetic field in [Mm/s].

Answer 1

1) Find the kinetic energy of the accelerated electron by multiplying the potential difference by the electdron charge in coulombs. Next, set this equal to the kinetic energy. The equation is

V x e = 1/2 x m x v² where V is potential, e electron charge, m electron mass, and v is velocity. Solve for the velocity.

2) When entering the magnetic field, the electron experiences a force perpendicular to the plane formed by the velocity and magnetic field vectors. If the velocity vector and the magnetic field vectors form an angle of 90 degrees, the resultant force on the electron causes the electron to describe a circular path. The relevant equation is F = e x v x B where B is the magnetic field strength in Teslas.

The force F is equal to the centripetal force m x v² / r, where r is the radius of curvature given in the problem. By setting the centripetal force equal to the force by the magnetic field, you can solve for B as the other variables are known.

3) As the force on the electron by the magnetic field is at right angles to the velocity vector, the speed of the electron shouldn't be affected as it leaves the magnetic field, so it stays the same.