Answer to Question #212260 in Electricity and Magnetism for Ani

Question #212260

Derive the expression of the trajectory when a charged particle of 

charge q is subjected to a magnetic field perpendicular to its plane. 

From here comment on the trajectory of the particle.


1
Expert's answer
2021-07-01T15:30:19-0400

Force that acts on the particle:


F=qvB.F=qvB.

Since this force is perpendicular to the velocity vector, the particle experiences centripetal acceleration, which determines the net force:


F=mv2R.F=m\frac{v^2}{R}.

Equiate:

qvB=mv2R, qB=mvR. R=mvqBqvB=m\frac{v^2}{R},\\\space\\ qB=\frac{mv}{R}.\\\space\\ R=\frac{mv}{qB}

The trajectory is a circle with radius R.


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