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Chapter 8: Q. 44 (page 200)

A charged particle of mass m moving with speed v in a plane perpendicular to a magnetic field experiences a force where q is the amount of charge and F⃗=(qvB,perpendiculartov⃗) , B is the magnetic field strength. Because the force is always perpendicular to the particle’s velocity, the particle undergoes uniform circular motion. Find an expression for the period of the motion. Gravity can be neglected.

Short Answer

Expert verified

Period is expressed as T=2πmqB

Step by step solution

01

Given information

Given F⃗=(qvB,perpendiculartov⃗)

02

Explanation

As the particle is moving in circular motion, so

(mv2)r=qvB

Expressed as r

r=(mv2)qvB=mvqB..........................(1)

Time period for this revolutions is

T=2πrv

Substitute value of r from equation(1), we get

T=(2π/v)(mv/qB)T=2πmqB

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