Question

An electron accelerated through a potential difference V in a uniform tranverse magnetic field moves in a circle of radius R. If the accelerating potential is increased to 2V, the electron in the same magnetic field will follow a circle of radius:

• A. R
• B. √2 R
• C. 2R
• D. 4R

Answer 1

Answer: (B)

√2 R

Answer 2

Explanation of the solution:

1. Energy gain: When an electron is accelerated through a potential difference $V$, its kinetic energy ($KE$) increases by an amount equal to the work done by the electric field, which is $eV$. $KE = eV$

2. Velocity of the electron: The kinetic energy is also given by $\frac{1}{2}mv^2$, where $m$ is the mass of the electron and $v$ is its velocity. $\frac{1}{2}mv^2 = eV$ Solving for $v$: $v = \sqrt{\frac{2eV}{m}}$

3. Motion in a magnetic field: When an electron moves with velocity $v$ perpendicular to a uniform magnetic field $B$, the magnetic force $F_B = evB$ provides the necessary centripetal force $F_c = \frac{mv^2}{R}$ for it to move in a circle of radius $R$. $evB = \frac{mv^2}{R}$

4. Radius in terms of velocity: From the force balance, we can express the radius $R$: $R = \frac{mv}{eB}$

5. Radius in terms of potential difference: Substitute the expression for $v$ from step 2 into the equation for $R$ from step 4: $R = \frac{m}{eB} \sqrt{\frac{2eV}{m}}$ To simplify, bring $m$ and $e$ inside the square root: $R = \frac{1}{B} \sqrt{\frac{m^2}{e^2} \frac{2eV}{m}}$ $R = \frac{1}{B} \sqrt{\frac{2mV}{e}}$

6. Relationship between R and V: From the derived formula, we can see that the radius $R$ is directly proportional to the square root of the accelerating potential $V$ (assuming $m$, $e$, and $B$ are constant): $R \propto \sqrt{V}$

7. Calculate the new radius: Let $R_1$ be the initial radius when the potential is $V_1 = V$. Let $R_2$ be the new radius when the potential is $V_2 = 2V$. $\frac{R_2}{R_1} = \frac{\sqrt{V_2}}{\sqrt{V_1}}$ $\frac{R_2}{R_1} = \sqrt{\frac{2V}{V}}$ $\frac{R_2}{R_1} = \sqrt{2}$ $R_2 = \sqrt{2}R_1$

Therefore, if the accelerating potential is increased to $2V$, the electron will follow a circle of radius $\sqrt{2}R$.