Magnetic Field Keeping an Injected Electron on a Straight Path: 04-BS-9, May 2016, Q6

Question 6: Magnetic Field Keeping an Injected Electron on a Straight Path (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

Quantity	Value
Accelerating potential $V_{\text{acc}}$	$10^4$ V
Capacitor voltage $V_{\text{cap}}$	$100$ V
Plate separation $d$	$1$ mm $=10^{-3}$ m
Electron charge/mass	$-1.6\times10^{-19}$ C, $9.1\times10^{-31}$ kg

Find. The magnitude and direction of $\vec B$ so the electron travels straight through, undeflected.

Local right-handed frame: $\hat x$ along the electron's velocity, $\hat y$ along the capacitor's $E$ field between the plates, $\hat z=\hat x\times\hat y$; $\vec B$ must lie along $\pm\hat z$ (in the plane of the plates, perpendicular to $v$).

Approach. Find the electron's speed from the accelerating potential (energy conservation), the capacitor's field from its voltage and gap, then require the magnetic force to exactly cancel the electric force — the same velocity-selector balance as a straight-through mass/velocity filter.

Electron speed after acceleration. $eV_{\text{acc}}=\tfrac12 m v^2$: $$v=\sqrt{\frac{2eV_{\text{acc}}}{m}}=\sqrt{\frac{2(1.6\times10^{-19})(10^4)}{9.1\times10^{-31}}}$$ $$v=\boxed{5.930\times10^7\ \text{m/s}}$$
Field between the capacitor plates. $$E_{\text{cap}}=\frac{V_{\text{cap}}}{d}=\frac{100}{10^{-3}}$$ $$E_{\text{cap}}=\boxed{1.0\times10^5\ \text{V/m}}$$
Balance electric and magnetic forces, $evB=eE_{\text{cap}}$. In the local frame of the figure ($\hat x=v$-direction, $\hat y=E$-direction), $\vec F_B=q\vec v\times\vec B$ cancels $\vec F_E=q\vec E$ only when $\vec B$ is along $\pm\hat z$ (perpendicular to both $v$ and $E$, i.e. lying in the plane of the plates); working through the electron's negative charge fixes the sign so that $\vec B$ along $+\hat z$ gives $\vec F_B$ opposing $\vec F_E$: $$B=\frac{E_{\text{cap}}}{v}=\frac{1.0\times10^5}{5.930\times10^7}$$ $$B=\boxed{1.686\times10^{-3}\ \text{T, along }+\hat z\ \text{(in the plane of the plates, perpendicular to }v\text{)}}$$

Quantity	Result
Electron speed $v$	$5.930\times10^7$ m/s
Capacitor field $E_{\text{cap}}$	$1.0\times10^5$ V/m
Required $B$	$1.686\times10^{-3}$ T, $\perp$ to both $v$ and $E_{\text{cap}}$

Check: the source does not fix absolute compass/plate-polarity directions for this setup, so the direction of $\vec B$ is reported here in a local right-handed frame tied to the stated velocity and field directions rather than a compass bearing.

Concept & Theory

Topic. Velocity-selector balance of electric and magnetic (Lorentz) forces on an accelerated charge.

Key relations. Energy conservation $eV=\tfrac12 mv^2$; uniform-plate field $E=V/d$; force balance $qvB=qE\Rightarrow B=E/v$.

Why this works. $\vec F_B=q\vec v\times\vec B$ is always perpendicular to both $\vec v$ and $\vec B$; requiring it to exactly cancel the fixed-direction $\vec F_E=q\vec E$ (perpendicular to $v$) forces $\vec B$ to be perpendicular to both $v$ and $E$ — a unique direction (up to sign) once $v$ and $E$'s directions are fixed, with the sign resolved by the charge's sign.

Common pitfall. Reusing the accelerating potential's field instead of the capacitor's own $E=V_{\text{cap}}/d$ (they are two separate stages: one sets $v$, the other sets the deflecting force); or forgetting the electron's negative charge when resolving the direction of $\vec B$.

Source. Sadiku, Elements of Electromagnetics, Ch. 8 (Lorentz force, motion of charged particles in combined $E$ and $B$ fields).

Question 6 of 8: Magnetic Field Keeping an Injected Electron on a Straight Path

Question 6: Magnetic Field Keeping an Injected Electron on a Straight Path (20 marks)