Practice portal › Moving Charges and Magnetism › Motion of a Charge in a Magnetic Field
Asked in GUJCET 2007 · Circular motion: radius, period and energy
Given: ⃗v=(2̂i+3̂j) m s⁻¹ and ⃗B=4̂k T.
Idea: the magnetic force is ⃗F=q(⃗v×⃗B), always perpendicular to both ⃗v and ⃗B.
Here ⃗v lies in the xy plane and ⃗B is along z, so ⃗v⊥⃗B and the force is qvB, not zero.
Being perpendicular to ⃗v, the force does no work: the speed √2²+3²=√13 m s⁻¹ cannot change.
What it does change is the direction of ⃗v — the electron turns along a circle in the xy plane.
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