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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 F鈨�=(qvB,perpendiculartov鈨�) , B is the magnetic field strength. Because the force is always perpendicular to the particle鈥檚 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=2蟺mqB

Step by step solution

01

Given information

Given F鈨�=(qvB,perpendiculartov鈨�)

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=2蟺rv

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

T=(2蟺/v)(mv/qB)T=2蟺mqB

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