An electron enters a magnetic field of 0.5 T perpendicular to its velocity. If its kinetic energy is 100 eV, find the radius of its circular path. (me=9.1×10−31 kg, e=1.6×10−19 C.)
Solution — Step by Step
KE=21mv2⟹v=m2⋅KE
KE=100×1.6×10−19=1.6×10−17 J.
v=9.1×10−312×1.6×10−17=3.52×1013≈5.93×106 m/s
qvB=rmv2⟹r=qBmv
r=1.6×10−19×0.59.1×10−31×5.93×106=8×10−205.4×10−24≈6.75×10−5 m
Final answer: r≈67.5μm
Why This Works
A charged particle in a perpendicular magnetic field feels a force always perpendicular to its velocity — exactly the condition for uniform circular motion. The Lorentz force does no work, so kinetic energy and speed stay constant; only direction changes.
The radius r=mv/qB tells you that heavier or faster particles curve less, while stronger fields curve them tighter. This is the principle behind cyclotrons, mass spectrometers, and old CRT TVs.
Alternative Method
Use the shortcut r=B1q2mV where V is the accelerating potential in volts (here V=100 V since KE=qV). Saves one step.
Common Mistake
Forgetting to convert eV to joules. 100 eV is not 100 J — it is 100×1.6×10−19 J. Lose a factor of 1019 and your radius becomes meters instead of micrometers. Always convert energy units before plugging into kinematic formulas.
For protons with the same KE, the radius scales as m. Proton radius would be roughly 1836 times larger than electron radius — about 2.9 mm here. Useful sanity check.
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