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
2eVm
B
2VmeB2
C
2VmeB
D
2Vme2B

Solution

## The correct option is A $$\sqrt{\dfrac{2Vm}{eB^{2}}}$$$$KE=eV=\dfrac{1}{2}m_eV^2$$and $$R=\dfrac{m_eV}{eB}$$$$=\dfrac{m_e}{eB}\times\sqrt{\dfrac{eV\times 2}{m_e}}$$$$=\sqrt{\dfrac{2Vm_e}{eB^2}}$$Physics

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