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An electron (mass m) is accelerated through a potential difference of 'V' and then it enters in a magnetic field of induction 'B' normal to the lines. The radius of the circular path is (e = electronic charge)
An electron (mass m) is accelerated through a potential difference of 'V' and then it enters in a magnetic field of induction 'B' normal to the lines. The radius of the circular path is (e = electronic charge)
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Logic Tip: A useful standard formula to memorize for charged particles moving in magnetic fields is $R = \frac{\sqrt{2mK{qB}$, where $K$ is kinetic energy. Since $K = qV$ (where $q=e$), it becomes $\frac{\sqrt{2meV{eB} = \sqrt{\frac{2mV}{eB^2$.
Updated On: Apr 28, 2026
- $\sqrt{\frac{2eV}{m$
- $\sqrt{\frac{2Vm}{eB^2$
- $\sqrt{\frac{2Vm}{eB$
- $\sqrt{\frac{2Vm}{e^2B$
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The Correct Option is B
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