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  • A proton of mass m and charge q is moving in a plane with kinetic energy E. If there exists a uniform magnetic field B, perpendicular to the plane of the motion, the proton will move in a circular

A proton of mass m and charge q is moving in a plane with kinetic energy E. If there exists a uniform magnetic field B, perpendicular to the plane of the motion, the proton will move in a circular path of radius.

Option: 1

\frac{2 E m}{q B}


Option: 2

\frac{\sqrt{2 E m}}{q B}


Option: 3

\frac{\sqrt{E m}}{2q B}


Option: 4

\frac{\sqrt{2 E q}}{m B}


Answers (1)

The kinetic energyof proton, 

E=\frac{1}{2} m v^2=\frac{1}{2} m \frac{q^2 B^2 r^2}{m^2}

=\frac{q^2 B^2 r^2}{2 m}\left[\because \text { velocity of the proton, } V=\frac{q B r}{m}\right]

\therefore \quad r^2=\frac{2 m E}{q^2 B^2}   or, r=\frac{\sqrt{2 E m}}{q B}

Posted by

Ramraj Saini

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