Magnetic force on a moving proton
A proton (q = 1.6 × 10⁻¹⁹ C) moves at 2 × 10⁶ m/s perpendicular to a 0.5 T magnetic field. What is the magnitude of the magnetic force?
Answer C: explanation for every choice
A
The force is zero only for motion parallel to the field.
B
This is half the correct value, as if sin θ = 0.5.
C
Correct. F = qvB sin 90° = (1.6 × 10⁻¹⁹)(2 × 10⁶)(0.5) = 1.6 × 10⁻¹³ N.
D
This leaves out the 0.5 T factor.
TakeawayF = qvB sin θ. Perpendicular motion gives sin θ = 1.
Equation usedMagnetic force on a moving charge
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