Set 54 Problem #16

Set 54 Problem number 16

Problem

Show that the charge of 9 `microCoulombs passes without change of direction through a pair of crossed electric and magnetic fields, the fields being mutually perpendicular and also perpendicular to the velocity, if its velocity is v = E / B, assuming electric field strength E = 39000 N / C and magnetic field strength B = .98 Tesla. Assume that the effects of the two fields on the charges are in opposite directions. Show that the same is the case if the charge is 15 `microCoulombs, and explain why the change in charge has no effect on the result.

Solution

The velocity we are to test is v = E / B = 39000 N / C / ( .98 Tesla) = 39000 N / C / ( .98 N / ( (C / s) * m)) = 39790 m/s.

The electric field exerts a force F = q * E independent of particle velocity. We obtain

force exerted by electrical field = 9 `microCoulomb * 39000 N / C = 35.1 * 10^-2 N, in the direction of the field.

The magnetic field exerts a force F = q v B which does depend on the velocity:

force exerted by magnetic field = 9 `microCoulomb * 39790 m/s * .98 Tesla = 35.09 * 10^-2 N.

These forces are seen to be equal in magnitude.

Since the force exerted by the magnetic field is perpendicular to the field and to the particle velocity, it must be in the direction of the electric field or opposite to this direction.
By the assumption of the problem the forces are opposite in direction.
The forces are thus equal and opposite and cancel one another so that the particle experiences no net force from the to fields and hence passes through the two fields undeflected.

The forces of the two fields on the 15 `microCoulomb charge will both have magnitude 58.5 * 10^-2 N, so that this charge will also pass undeflected through the two fields.

More generally we see that sense both forces are proportional to the charge, any charged particle will pass undeflected through the two fields.

Generalized Solution

Generally if v = E / B, the force exerted by the electric field on a charge q is q E and the force exerted by the magnetic field has magnitude q v B = q ( E / B ) * b = q E. If the fields are appropriately oriented the forces will be equal and opposite, and any charge q moving at this velocity will pass through the fields without a change in direction.