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Asked by Balakrishna Last Modified
Answer When a charged particle moves with velocity vv perpendicular to a uniform magnetic field BB, it experiences a magnetic force (FBFB) perpendicular to both vv and BB. According to the Lorentz force law, the magnetic force (FBFB) acting on the particle is given by:
FB=qv×BFB=qv×B
Since the force is always perpendicular to the velocity, it changes the direction of the velocity but not its magnitude. This results in the particle moving along a circular path.
To find the radius of the circular path, we equate the magnetic force (FBFB) to the centripetal force (FcFc) required to keep the particle in circular motion:
qvB=mv2rqvB=rmv2
Where:
From this equation, we can solve for the radius rr of the circular path:
r=mvqBr=qBmv
Therefore, the expression for the radius rr of the circular path is given by:
r=mvqBr=qBmv
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