Question

Electrons emitted with negligible speed from an electron gun are accelerated through a potential difference $V_0$ along the x-axis. These electrons emerge from a narrow hole into a uniform magnetic field of strength $B$ directed along $x-$axis, some electrons emerging at slightly divergent angles as shown in the figure. These paraxial electrons are refocused on the x-axis at a distance (Given charge of electron $=e$, mass of electron $=m$):

• A. $\sqrt{\frac{8{\pi }^{2}m{V}_0}{e{B}^2}}$
• B. $\sqrt{\frac{2{\pi }^{2}m{V}_0}{eB}}$
• C. $\sqrt{\frac{4{\pi }^{2}m{V}_0}{e{B}^2}}$
• D. $\sqrt{\frac{2{\pi }^{2}m{V}_0}{e{B}^2}}$

Answer 1

Answer: (A)

$\sqrt{\frac{8{\pi }^{2}m{V}_0}{e{B}^2}}$

The paraxial electrons will refocus on the x- axis after one revolution i.e. at a distance of one pitch of helix from the hole.
The time period of revolution $T=\frac{2\pi m}{qB}=\frac{2\pi m}{eB}$ is independent of the velocity of the electron.
As the electrons are only slightly divergent. $v_{||}\approx v$ where $v$ is the velocity of the electron emergent from the hole.
After accelerating through a potential $V_0$, Kinetic energy of the electron $KE=\frac{1}{2}m{v}^2=eV$
$v=\sqrt{\frac{2e{V}_0}{m}}$
Pitch of the helix will be equal to $v_{||}T$ as shown in the figure.
$v_{||}T=\sqrt{\frac{2e{V}_0}{m}}.\frac{2\pi m}{eB}=\sqrt{\frac{8{\pi }^{2}m{V}_0}{e{B}^2}}$