# A proton is accelerated through a potential difference $V$, subjected to a uniform magnetic field acting normal to the velocity of the proton. If the potential difference is doubled, how will the radius of the circular path described by the proton in the magnetic field change?

The radius of the Circular path by the proton in the magnetic field change can be described as:-

$r=\frac{1}{B}\sqrt{\frac{2mV}{q}}$

Where     $B$     is the magnetic field

$V$    is the Potential difference

Therefore

$r\propto \sqrt{V}$

let r be the radius when the potential difference is V and r' be the radius when the potential difference is 2V

$\\\frac{{r}'}{r}= \sqrt{\frac{2V}{V}}=\sqrt{2}\\\\r'=\sqrt{2}r$

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