Practice portal › Moving Charges and Magnetism › Motion of a Charge in a Magnetic Field
Asked in AIPMT 1997 · Circular motion: radius, period and energy
Given: velocity due East, field vertically upward, so ⃗v⊥⃗B.
Idea: a perpendicular entry gives a force of constant size always at right angles to the motion — the definition of uniform circular motion.
F=qvB sin 90°=qvB, constant in size while the speed is constant.
That force is perpendicular to ⃗v, so it does no work and the speed stays fixed.
It is also perpendicular to ⃗B, so the motion stays in the horizontal plane.
A constant-magnitude force always perpendicular to the velocity bends the path into a circle of radius r=(mv)/(qB).
So the particle moves in a circle at uniform speed, in the horizontal plane.
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