Question

As shown in the figure, in the x-y plane, an electron source continuously emits N electrons per second with an average velocity of v along the positive x-direction, forming an electron beam of width 2b. The electrons are evenly distributed along the y-axis and are symmetrical about the x-axis. The electron beam is directed into a uniform magnetic field with a radius R and centered at the origin O in the x-y plane. After passing through the magnetic field, the electrons are deflected and uniformly emitted from point P. Directly below the magnetic field region, there is a pair of parallel metal plates K and A perpendicular to the y-axis, with the distance between point P and plate K being d. There is a small hole of width 2$\ell$ between the plates, symmetrical about the y-axis. Plate K is grounded, and a small adjustable voltage $U_{AK}$ is applied between plates A and K. All electrons passing through the hole are collected and conducted by plate A, thus forming an electric current. It is known that $b = \frac{\sqrt{3}}{2}R, d=\ell$, the mass of the electron is m, the charge of the electron is -e, and the interaction between electrons can be neglected.

• A. The magnitude of magnetic intensity B is $\frac{mv}{eR}$.
• B. The range of the angle between the electron stream and negative y-axis when it is emitted from point P is -60° ≤ 0 ≤ 60°.
• C. The number of $e^-$ that pass through the small hole per second when $U_{AK}=0$ is 0.75 N.
• D. The potential ($U_K-U_A$) at which current becomes zero is $\frac{mv^2}{e}$.

Answer 1

Answer: (A), (B), (C)

A, B, C

Answer 2

Analysis of Option (A):

Electrons move with velocity $v$ perpendicular to the uniform magnetic field $B$. The magnetic force provides the centripetal force for circular motion. The radius of curvature of the electron's path is given by:

$r_c = \frac{mv}{eB}$

The problem states that the magnetic field has a radius $R$ and is centered at the origin, and electrons are deflected to point P $(0, -R)$. This implies that the radius of curvature of the electron's path is $R$.

Therefore, $R = \frac{mv}{eB}$, which gives $B = \frac{mv}{eR}$. So, Option (A) is correct.

Analysis of Option (B):

The problem states that the electrons are "uniformly emitted from point P" and provides a beam width $2b = \sqrt{3}R$. For all electrons from a beam of width $2b$ to be focused onto a single point P after passing through a uniform magnetic field, the magnetic field must act as a focusing lens. While a simple uniform circular field does not naturally focus all parallel incident rays to a single point in this manner, such problems often imply a specific magnetic field configuration or an idealization where this focusing occurs. Assuming such a focusing mechanism, the problem states the range of angles. If this is the case, then Option (B) is correct.

Analysis of Option (C):

Assuming Option (B) is correct, electrons are emitted from point P $(0, -R)$ with velocities whose angles $\theta$ with the negative y-axis range from -60° to 60°. The velocity components are $v_x = v \sin\theta$ and $v_y = -v \cos\theta$ (since electrons are moving generally downwards). The electrons travel from P to plate K, which is at a distance $d$ below P. The hole in plate K has a width $2\ell$, symmetrical about the y-axis. We are given $d=\ell$.

The time taken for an electron to travel vertically from P to plate K is $t = \frac{d}{|v_y|} = \frac{d}{v \cos\theta}$. During this time, the horizontal distance traveled is $x = v_x t = (v \sin\theta) \frac{d}{v \cos\theta} = d \tan\theta$.

For an electron to pass through the hole, its horizontal position $x$ must satisfy $|x| \leq \ell$. Since $d=\ell$, we have $|\ell \tan\theta| \leq \ell$, which simplifies to $|\tan\theta| \leq 1$. This condition means $-45^\circ \leq \theta \leq 45^\circ$.

The problem states that electrons are "uniformly emitted" within the range of angles from -60° to 60°. The total angular range is $60^\circ - (-60^\circ) = 120^\circ$. The angular range for electrons passing through the hole is $45^\circ - (-45^\circ) = 90^\circ$.

The fraction of electrons passing through the hole is $\frac{90^\circ}{120^\circ} = \frac{3}{4} = 0.75$. Since $N$ electrons are emitted per second, the number of electrons passing through the hole per second is $0.75 N$. So, Option (C) is correct, assuming Option (B) is correct.

Analysis of Option (D):

Current becomes zero when no electrons reach plate A. Plate K is grounded ($U_K=0$). Electrons pass through the hole in plate K with kinetic energy $KE_K$. Since they are emitted from P with speed $v$ and travel to K (both at zero potential relative to the source), $KE_K = \frac{1}{2}mv^2$.

As electrons move from K to A, their potential energy changes. The change in potential energy is $\Delta PE = (-e)U_A - (-e)U_K = -e(U_A-U_K) = -eU_{AK}$.

By conservation of energy, $KE_K + PE_K = KE_A + PE_A$. $\frac{1}{2}mv^2 + 0 = KE_A + (-e)U_A$. For the current to become zero, electrons must be stopped before reaching plate A, meaning their kinetic energy $KE_A$ becomes zero at or before reaching A.

So, $\frac{1}{2}mv^2 = -eU_A$. This implies $U_A = -\frac{mv^2}{2e}$. The question asks for the potential $(U_K-U_A)$.

$U_K-U_A = 0 - (-\frac{mv^2}{2e}) = \frac{mv^2}{2e}$. Option (D) states $\frac{mv^2}{e}$, which is twice the calculated value. So, Option (D) is incorrect.

Based on the analysis, Options A, B, and C are correct, assuming the problem implies a specific focusing magnetic field for B and C to hold.