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A point charge of 5.0 × 10–7 C moves to the right at 2.6 × 105 m/s in a magnetic field that is directed into the screen and has a field strength of 1.8 × 10–2 T. What is the magnitude of the magnetic force acting on the charge? A. 0 N B. 2.3 × 10^–3 N C. 23 N D.2.3 × 10^11 N

The magnetic field 0.02 m from a wire is 0.1 T. What is the magnitude of the magnetic field 0.01 m from the same wire? A. 0.01 T B. 0.05 T C. 0.1 T D. 0.2 T

1 Answer

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1.) The formula for magnetic force is F = qvBsinθ.
Substituting, we have F=(5.0 × 10^–7 C) (2.6 × 10^5 m/s) (1.8 × 10^–2 T)(sin90°) [90° because point charge is perpendicular to magnetic field]

F = 2.34 x 10^-3 N
Answer is B.

2.) The distance is reduced by 1/2, so B is twice or 0.2 T. Answer is D.
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