A charged particle of mass m and charge q travels on a circular path of radius r that is perpendicular to a magnetic field B . The time taken by the particle to complete one revolution is

  • Option 1)

        \frac{2\pi qB}{m}

  • Option 2)

    \frac{2\pi m}{qB}

  • Option 3)

    \frac{2\pi mq}{B}

  • Option 4)

    \frac{2\pi mq}{B}

 

Answers (1)

As we learnt in

Radius of charged particle -

r=\frac{mv}{qB}=\frac{P}{qB}=\frac{\sqrt{}2mk}{qB}=\frac{1}{B}\sqrt{}\frac{2mV}{q}

-

 

 

Time period of charged particle -

T=\frac{2\pi m}{qB}

- wherein

independent of speed of particle 

 

 

 

 

T=\frac{2\pi }{\omega }=\frac{2\pi r}{v}..........(i)

\because   centripetal force = magnetic force

\therefore \; \; \frac{mv^{2}}{r}=qvB\Rightarrow v=\frac{qBr}{m}..............(ii)

From\; \; (i)\; and \; (ii)

\therefore \; \; \; T=\frac{2\pi r\times m}{qBr}=\frac{2\pi m}{qB}

 


Option 1)

    \frac{2\pi qB}{m}

Incorrect option

Option 2)

\frac{2\pi m}{qB}

Correct option

Option 3)

\frac{2\pi mq}{B}

Incorrect option

Option 4)

\frac{2\pi mq}{B}

Incorrect option

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