Question

When an electron accelerated from rest through a potential difference \( V_1 \) enters a uniform magnetic field, the maximum force on it is F. If the potential difference is changed to \( V_2 \), then the maximum force on the electron in the same magnetic field is 4F, then \( \frac{V_1}{V_2} \) =

• A. 1 : 4
• B. 4 : 1
• C. 2 : 1
• D. 1 : 16

Hint:

Remember the energy conservation for accelerating a charged particle: $qV = \frac{1}{2}mv^2$. The maximum magnetic force on a charged particle is $F = qvB$. Combine these two relations to see how force depends on potential difference, typically as $F \propto \sqrt{V}$.

Answer 1

The Correct Option is D

Step 1: Relate potential difference to kinetic energy and velocity
When an electron (charge $e$, mass $m_e$) is accelerated from rest through a potential difference $V$, its potential energy is converted into kinetic energy. \[ e V = \frac{1}{2} m_e v^2 \] From this, we can express the velocity $v$ as: \[ v = \sqrt{\frac{2eV}{m_e}} \quad \cdots (1) \] Step 2: Relate velocity to the maximum magnetic force
When a charged particle moves in a uniform magnetic field $B$, the magnetic force $F$ on it is given by: \[ F = qvB \sin\theta \] For an electron ($q=e$), the force is $F = evB \sin\theta$. The maximum force occurs when $\sin\theta = 1$ (i.e., the velocity is perpendicular to the magnetic field), so: \[ F_{\text{max}} = evB \quad \cdots (2) \] Step 3: Combine the relationships for the first case
Substitute equation (1) into equation (2): \[ F = eB \sqrt{\frac{2eV_1}{m_e}} \quad \cdots (3) \] Here, $F$ is the maximum force when the potential difference is $V_1$. Step 4: Combine the relationships for the second case
When the potential difference is changed to $V_2$, the maximum force becomes $4F$. Let the new force be $F' = 4F$. \[ 4F = eB \sqrt{\frac{2eV_2}{m_e}} \quad \cdots (4) \] Step 5: Find the ratio \( \frac{V_1}{V_2} \)
Divide equation (3) by equation (4): \[ \frac{F}{4F} = \frac{eB \sqrt{\frac{2eV_1}{m_e}}}{eB \sqrt{\frac{2eV_2}{m_e}}} \] \[ \frac{1}{4} = \sqrt{\frac{V_1}{V_2}} \] To remove the square root, square both sides: \[ \left(\frac{1}{4}\right)^2 = \frac{V_1}{V_2} \] \[ \frac{1}{16} = \frac{V_1}{V_2} \] So, the ratio $\frac{V_1}{V_2}$ is $1:16$. Step 6: Analyze Options
\begin{itemize} \item Option (1): 1 : 4. Incorrect. \item Option (2): 4 : 1. Incorrect. \item Option (3): 2 : 1. Incorrect. \item Option (4): 1 : 16. Correct, as it matches our calculated ratio. \end{itemize}