An electron, initially at rest, is accelerated uniformly through a potential difference over a distance of 0.02 meters. If the electron reaches a final velocity of 4.0 x 106 m/s, what is the magnitude of the magnetic force it experiences when it subsequently enters a uniform magnetic field of 0.5 Tesla, perpendicular to its velocity?
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The answer
Correct answer: B. 3.2 x 10-13 N
Tested concept: Magnetic force on moving charges
Explanation
The magnetic force (F_B) on a charged particle moving in a magnetic field is given by the formula F_B = qvBsinθ. For an electron, the charge (q) is approximately 1.6 x 10-19 C. The velocity (v) is given as 4.0 x 106 m/s. The magnetic field strength (B) is 0.5 T. The angle (θ) between the velocity and the magnetic field is 90° (perpendicular), so sin(90°) = 1. F_B = (1.6 x 10-19 C) (4.0 x 106 m/s) (0.5 T) 1 F_B = (1.6 4.0 * 0.5) x 10^(-19 + 6) N F_B = (3.2) x 10-13 N. Thus, the electron experiences a magnetic force of 3.2 x 10-13 Newtons. Distractor 1 (1.6 x 10-13 N): This result would occur if the magnetic field strength was incorrectly taken as 0.25 T instead of 0.5 T, or if the charge was incorrectly taken as half the electron charge. Distractor 3 (6.4 x 10-13 N): This result could arise from incorrectly doubling the magnetic field strength or the electron's charge in the calculation. Distractor 4 (4.8 x 10-13 N): This result could arise from an arithmetic error, such as using a magnetic field strength of 0.75 T instead of 0.5 T.
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