A magnet produces a 0.40 T field between its poles, directed horizontally to the east. A dust particle with charge q = -8.0×10^-18 C is moving vertically downwards with a speed of 0.30 cm/s in this field. Whilst it is in the magnetic field, the dust particle is also in an electric field of strength 1.00×10^-2 V/m pointing to the north.
(a) What is the magnitude and direction of the magnetic force on the dust particle?
(b) What is the magnitude and direction of the electric force on the dust particle?
(c) What is the magnitude of the net force on the dust particle?
F=(k|q1q2|) / r^2
The Attempt at a Solution
a) Magnitude: F=|q|vB = (-8 x 10^-18 C)(-0.003 m/s)(0.4 T) = 9.6 x 10^-21 N
Direction: Since the velocity is south and the force must be perpendicular to the velocity, the force must lie in a plane perpendicular to the north/south axis.
b) I know it has to do with "1.00 x 10^-2 V/m" but all I could do with it was:
Magnitude: B=E/V --> (1.00 x 10^-2 V/m) / (0.003 m/s) = 3.3(repeating) T
c) Would the net force be the sum of answers "a" and "b" or would it be:
Net force = sqrt(a^2 + b^2)
a: being the force in a)
b: being the force in b)