Question

A homogeneous beam of proton (mass $m$, charge $e$) having cross sectional radius $R$ is accelerated through potential difference $V$ and finally it carries equivalent current $I$. The potential difference between the surface and axis of the beam is

• A. $\frac{2I}{\pi \epsilon_0}\sqrt{\frac{m}{15eV}}$
• B. $\frac{I}{\pi \epsilon_0}\sqrt{\frac{m}{21eV}}$
• C. $\frac{2I}{\pi \epsilon_0}\sqrt{\frac{m}{20eV}}$
• D. $\frac{I}{\pi \epsilon_0}\sqrt{\frac{m}{32eV}}$

Answer 1

Answer: (D)

$\frac{I}{\pi \epsilon_0}\sqrt{\frac{m}{32eV}}$

Answer 2

To determine the potential difference between the surface and axis of the proton beam, we follow these steps:

1. Determine the velocity of protons ($v$):

   When protons (charge $e$, mass $m$) are accelerated through a potential difference $V$, their kinetic energy is $eV$. Using the kinetic energy formula: $KE = \frac{1}{2}mv^2 = eV$ Solving for $v$: $v = \sqrt{\frac{2eV}{m}}$

2. Relate current ($I$) to charge density ($\rho$):

   The current $I$ carried by the beam is given by the product of charge density, cross-sectional area, and velocity. The cross-sectional area of the beam is $A = \pi R^2$. $I = \rho A v = \rho (\pi R^2) v$ From this, the charge density $\rho$ is: $\rho = \frac{I}{\pi R^2 v}$

3. Calculate the electric field ($E$) inside the beam:

   The beam is a uniformly charged cylinder. To find the electric field inside, we use Gauss's Law. Consider a cylindrical Gaussian surface of radius $r$ ($r < R$) and length $L$ coaxial with the beam. The charge enclosed by this Gaussian surface is $Q_{enc} = \rho (\pi r^2 L)$. Gauss's Law states $\oint \vec{E} \cdot d\vec{A} = \frac{Q_{enc}}{\epsilon_0}$. Due to symmetry, the electric field $\vec{E}$ is radial and its magnitude is constant on the Gaussian surface. So, $E (2\pi r L) = \frac{\rho (\pi r^2 L)}{\epsilon_0}$ $E = \frac{\rho r}{2\epsilon_0}$

4. Calculate the potential difference ($\Delta V$) between the surface and axis:

   The potential difference between the surface ($r=R$) and the axis ($r=0$) is given by the negative integral of the electric field from the axis to the surface. Since the beam is positively charged, the potential at the axis will be higher than at the surface. We are looking for the magnitude of the potential difference. $\Delta V = |V_{surface} - V_{axis}| = \left|-\int_{0}^{R} E dr\right|$ $\Delta V = \left|-\int_{0}^{R} \frac{\rho r}{2\epsilon_0} dr\right|$ $\Delta V = \left|-\frac{\rho}{2\epsilon_0} \left[ \frac{r^2}{2} \right]_{0}^{R}\right|$ $\Delta V = \left|-\frac{\rho R^2}{4\epsilon_0}\right|$ $\Delta V = \frac{\rho R^2}{4\epsilon_0}$

5. Substitute $\rho$ and $v$ into the expression for $\Delta V$:

   Substitute the expression for $\rho$ from step 2: $\Delta V = \frac{1}{4\epsilon_0} \left(\frac{I}{\pi R^2 v}\right) R^2$ $\Delta V = \frac{I}{4\pi \epsilon_0 v}$

   Now, substitute the expression for $v$ from step 1: $\Delta V = \frac{I}{4\pi \epsilon_0 \sqrt{\frac{2eV}{m}}}$ To simplify and match the given options, we can bring the factor of 4 into the square root. Since $4 = \sqrt{16}$: $\Delta V = \frac{I}{\pi \epsilon_0} \frac{1}{4} \sqrt{\frac{m}{2eV}}$ $\Delta V = \frac{I}{\pi \epsilon_0} \sqrt{\frac{1}{16} \frac{m}{2eV}}$ $\Delta V = \frac{I}{\pi \epsilon_0} \sqrt{\frac{m}{16 \times 2eV}}$ $\Delta V = \frac{I}{\pi \epsilon_0} \sqrt{\frac{m}{32eV}}$

Comparing this result with the given options, it matches option D.