Question

An electron of mass m is accelerated through a potential difference of V and then it enters a magnetic field of induction B normal to the lines of force. Then the radius of the circular path is

• A. $\sqrt{\frac{2eV}{m}}$
• B. $\sqrt{\frac{2Vm}{e{B}^2}}$
• C. $\sqrt{\frac{2Vm}{eB}}$
• D. $\sqrt{\frac{2Vm}{e^{2}B}}$

Answer 1

Answer: (B)

$\sqrt{\frac{2Vm}{e{B}^2}}$

$KE=eV=\frac{1}{2}{m}_e{V}^2$
and $R=\frac{m_{e}V}{eB}$
$=\frac{m_e}{eB}\times \sqrt{\frac{eV\times 2}{m_e}}$
$=\sqrt{\frac{2V{m}_e}{e{B}^2}}$