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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
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B
2VmeB2
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C
2VmeB
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D
2Vme2B
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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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