A particle that has a 8.2 microcoulumb charge moves with a velocity of magnitude

Question

A particle that has a 8.2 microcoulumb charge moves with a velocity of magnitude 5.0 x 105 m/s. When the velocity points along the +x axis, the particle experiences no magnetic force, although there is a magnetic field present. The maximum possible magnetic force that the charge could experience has a magnitude of 0.68 N. Find the magnitude and direction of the magnetic field. Note that there are two possible answers for the direction of the field. Make sure you draw a diagrams indicating charges, velocities, forces, and magnetic fields.

Explanation / Answer

charge q = 8.2 x10^-6 coulumbs

velocity v = 5 x10^5 m/sec

force F= 0.68 N

when the charge is moving X-direction the maximum force will occur hen the magnetic field is perpendicular to motion

F = q vB  

SO B = F/qv   =0.68/4.1 = 0.168 T

and the direction is either y or z direction

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