Question

A slightly divergent beam of charged particles, accelerated by a potential difference V propagates from a point A long the axis of a solenoid. The beam is focussed at a point at a distance $l$ from A by two magnetic induction value $B_1$ and $B_2$ one after another. Find specific charge q/m of the particles.

Answer 1

$\frac{q}{m} = \frac{8\pi^2 V}{l^2 (B_2-B_1)^2}$

Answer 2

1. Particle Velocity: Charged particles accelerated by a potential difference $V$ gain kinetic energy $qV$. If $v$ is their velocity and $m$ their mass, then $\frac{1}{2}mv^2 = qV$, which gives $v = \sqrt{\frac{2qV}{m}}$.

2. Helical Motion in Solenoid: A slightly divergent beam implies particles have velocity components parallel ($v_{||}$) and perpendicular ($v_{\perp}$) to the solenoid's axis. The magnetic field $B$ acts along the axis. The motion is helical with time period $T = \frac{2\pi m}{qB}$ and pitch $p = v_{||}T$. For a "slightly divergent" beam, we approximate $v_{||} \approx v$. Thus, $p \approx vT = \sqrt{\frac{2qV}{m}} \frac{2\pi m}{qB}$.

3. Focusing Condition: A beam is focused at a distance $l$ if $l$ is an integer multiple of the pitch $p$. So, $l = n \times p$, where $n$ is an integer. $l = n \sqrt{\frac{2qV}{m}} \frac{2\pi m}{qB}$.

4. Two Successive Focusings: The problem states the beam is focused at distance $l$ by $B_1$ and then by $B_2$. This means:

   • With $B_1$, focusing occurs at $l$. Let this correspond to $n_1$ turns: $l = n_1 p_1 = n_1 \sqrt{\frac{2qV}{m}} \frac{2\pi m}{qB_1}$.
   • With $B_2$, focusing occurs at $l$. Let this correspond to $n_2$ turns: $l = n_2 p_2 = n_2 \sqrt{\frac{2qV}{m}} \frac{2\pi m}{qB_2}$. For "consecutive" focusing, $n_1$ and $n_2$ are consecutive integers. Let $n_1 = n$ and $n_2 = n+1$.

5. Relating B and n: From the focusing conditions: $l B_1 = n_1 \sqrt{\frac{2qV}{m}} \frac{2\pi m}{q}$ $l B_2 = n_2 \sqrt{\frac{2qV}{m}} \frac{2\pi m}{q}$ Let $C = \sqrt{\frac{2qV}{m}} \frac{2\pi m}{q}$. Then $l B_1 = n_1 C$ and $l B_2 = n_2 C$. This gives $\frac{B_1}{n_1} = \frac{C}{l}$ and $\frac{B_2}{n_2} = \frac{C}{l}$. Therefore, $\frac{B_1}{n_1} = \frac{B_2}{n_2}$, or $\frac{B_2}{B_1} = \frac{n_2}{n_1}$. With $n_1=n$ and $n_2=n+1$, we get $\frac{B_2}{B_1} = \frac{n+1}{n}$, which implies $n B_2 = (n+1)B_1 \implies n(B_2-B_1) = B_1 \implies n = \frac{B_1}{B_2-B_1}$.

6. Finding Specific Charge (q/m): Square the expression for $C$: $C^2 = \left(\sqrt{\frac{2qV}{m}} \frac{2\pi m}{q}\right)^2 = \frac{2qV}{m} \frac{4\pi^2 m^2}{q^2} = \frac{8\pi^2 V m}{q}$. From $l B_1 = n_1 C$, we have $C = \frac{l B_1}{n_1}$. Squaring this: $C^2 = \frac{l^2 B_1^2}{n_1^2}$. Equating the two expressions for $C^2$: $\frac{8\pi^2 V m}{q} = \frac{l^2 B_1^2}{n_1^2}$ $\frac{q}{m} = \frac{8\pi^2 V n_1^2}{l^2 B_1^2}$.

7. Final Result: Substitute $n_1 = \frac{B_1}{B_2-B_1}$: $\frac{q}{m} = \frac{8\pi^2 V}{l^2 B_1^2} \left(\frac{B_1}{B_2-B_1}\right)^2 = \frac{8\pi^2 V}{l^2 B_1^2} \frac{B_1^2}{(B_2-B_1)^2} = \frac{8\pi^2 V}{l^2 (B_2-B_1)^2}$.