Three particle each with a charge of 595 mu C and a speed of

Three particle, each with a charge of 5.95 mu C and a speed of 395.0 m/s, enter a uniform magnetic field whose magnitude is 0.625 T. What is the magnitude of the magnetic force on particle a? What is the magnitude of the magnetic force on particle b? What is the magnitude of the magnetic force on particle c?

Solution

A .

force is calculated as follows:

     F = q v X B = q v B sin = 5.95 e -6 * 395 * 0.625 sin 30 = 7.344e-4 N

B .

magnetic force:

    F = q v B sin = 5.95 e -6 * 395 * 0.625 sin 90 =1.4689e-3 N = 14.689e-4 N

C .

Magnetic force:

   F = q v B sin = 5.95 e -6 * 395 * 0.625* sin 150 = 7.34 e-4 N