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  • A proton of mass m moving with a speed v ( , velocity of light in vacuum ) completes a circular orbit in time T in a uniform magnetic fiel

A proton of mass m moving with a speed v ( \ll c, velocity of light in vacuum ) completes a circular orbit in time T in a uniform magnetic field. If the speed of proton is increased to \sqrt{2 V}. What will be the time needed to complete the circular orbit?

Option: 1

\sqrt{2} T


Option: 2

T


Option: 3

\frac{T}{\sqrt{2}}


Option: 4

\frac{T}{2}


Answers (1)

best_answer

The magnetic force on a proton of mass m moving along a circular path in a uniform magnetic field, 

\vec{F}_m=q(\vec{V} \times \vec{B})

\therefore F_m=q v B \sin 90^{\circ}

[ B = magnetic field, q = charge of proton ] =qVB since the proton is moving along a circular path of radius r, \frac{m v^2}{r}=q \vee B \therefore V=\frac{q B r}{m}

Note the time taken by the proton to make one complete revolution is \frac{2 \pi r}{v}

Hence, T=\frac{2 \pi r}{v} \text { or, } T=\frac{2 \pi r \times m}{q B r} \text { or, } T=\frac{2 \pi m}{q B}

Since T does not depend on the velocity of the proton, so if the velocity changes to \sqrt{2} v then T will remain unchanged. 

Posted by

vinayak

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