A particle carrying charge 4 `microCoulombs moves at constant velocity through crossed electric and magnetic fields (i.e., the fields are mutually perpendicular). The direction of motion of the charge is perpendicular to both fields. The electric field has strength E = 95000 N / C and the magnetic field has strength B = .97 Tesla, and the effects of the two fields on the charges are in opposite directions.
The velocity we are to test is v = E / B = 95000 N / C / ( .97 Tesla) = 95000 N / C / ( .97 N / ( (C / s) * m)) = 97930 m/s.
The electric field exerts a force F = q * E independent of particle velocity. We obtain
The magnetic field exerts a force F = q v B which does depend on the velocity:
These forces are seen to be equal in magnitude.
The forces of the two fields on the 16 `microCoulomb charge will both have magnitude 152 * 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.
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.
"