What is the magnetic force on a particle that has 0.000500 C of charge and is moving at 2.50 10m/s to the right through a magnetic field that is 4.20 T and pointing away from
you? Specify both magnitude and direction in your answer.

1.) F = 5.3×10^-3 N

2.) Positive y - direction

Explanation:

The parameters given are:

Charge q = 0.0005C

Velocity V = 2.5010 m/s

Magnetic field B = 4.2 T

Magnetic force F = BVqsinØ

F = BVq

since Ø = 90 degree

Substitute all the parameters into the formula

F = 4.2 × 2.5010 × 0.0005

Therefore, the magnetic force on a particle is F = 5.3 × 10^-3 N

2.) According to Fleming's left hand rule, the direction of the magnetic force will be perpendicular to the magnetic field which moving upward of the screen.

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