Question

A particle that has an 7.3-μC charge moves with a velocity of magnitude 4 × 105 m/s along the +x axis. It experiences no magnetic force, although there is a magnetic field present. The maximum possible magnetic force that the charge with the given speed could experience has a magnitude of 0.320 N. Find the magnitude and direction of the magnetic field. Note that there are two possible answers for the direction of the field.

Answer 1

If the particle's velocity were perpendicular to the magnetic field, the magnetic force would be given by:

F = qvB

F is the magnetic force, q is the charge, v is the velocity, and B is the magnetic field strength.

Given values:

F = 0.320N

q = 7.3×10⁻⁶C

v = 4×10⁵m/s

Plug in the values and solve for B:

0.320 = (7.3×10⁻⁶)(4×10⁵)B

B = 0.110T

The magnetic force depends on the cross product between the particle's velocity vector v and the magnetic field vector B. If v and B are parallel then the cross product is 0. v points in the +x direction, therefore B must point in either the +x or -x direction.