Question:
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
Solution and Explanation
Concept:
When an electron is accelerated through a potential difference $V$, the electrical potential energy is converted into kinetic energy ($\frac{1}{2}mv^2 = eV$). When this moving electron enters a uniform magnetic field perpendicularly, it experiences a magnetic Lorentz force that acts as the centripetal force, making it move in a circular path ($evB = \frac{mv^2}{R}$).
Step 1: Determine the velocity of the electron after acceleration.
Equate the kinetic energy gained to the electrical work done: $$\frac{1}{2}mv^2 = eV$$ Solve for the velocity $v$: $$v^2 = \frac{2eV}{m} \implies v = \sqrt{\frac{2eV}{m$$
Step 2: Determine the radius of the circular path in the magnetic field.
Equate the magnetic force to the centripetal force: $$evB = \frac{mv^2}{R}$$ Solve for the radius $R$: $$R = \frac{mv}{eB}$$
Step 3: Substitute the velocity expression into the radius equation.
$$R = \frac{m}{eB} \left(\sqrt{\frac{2eV}{m\right)$$ To simplify, bring the outside terms $m$ and $e$ inside the square root by squaring them: $$R = \frac{1}{B} \sqrt{\frac{m^2 \cdot 2eV}{e^2 \cdot m$$ Cancel common factors of $m$ and $e$: $$R = \frac{1}{B} \sqrt{\frac{2mV}{e$$ Bring the $B$ inside the square root as well to match the options: $$R = \sqrt{\frac{2mV}{eB^2$$
When an electron is accelerated through a potential difference $V$, the electrical potential energy is converted into kinetic energy ($\frac{1}{2}mv^2 = eV$). When this moving electron enters a uniform magnetic field perpendicularly, it experiences a magnetic Lorentz force that acts as the centripetal force, making it move in a circular path ($evB = \frac{mv^2}{R}$).
Step 1: Determine the velocity of the electron after acceleration.
Equate the kinetic energy gained to the electrical work done: $$\frac{1}{2}mv^2 = eV$$ Solve for the velocity $v$: $$v^2 = \frac{2eV}{m} \implies v = \sqrt{\frac{2eV}{m$$
Step 2: Determine the radius of the circular path in the magnetic field.
Equate the magnetic force to the centripetal force: $$evB = \frac{mv^2}{R}$$ Solve for the radius $R$: $$R = \frac{mv}{eB}$$
Step 3: Substitute the velocity expression into the radius equation.
$$R = \frac{m}{eB} \left(\sqrt{\frac{2eV}{m\right)$$ To simplify, bring the outside terms $m$ and $e$ inside the square root by squaring them: $$R = \frac{1}{B} \sqrt{\frac{m^2 \cdot 2eV}{e^2 \cdot m$$ Cancel common factors of $m$ and $e$: $$R = \frac{1}{B} \sqrt{\frac{2mV}{e$$ Bring the $B$ inside the square root as well to match the options: $$R = \sqrt{\frac{2mV}{eB^2$$
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