Velocity Selector — Magnetic Field Cancelling an Electric Deflecting Force: 04-BS-9, December 2018, Q8

Question 8: Velocity Selector — Magnetic Field Cancelling an Electric Deflecting Force (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 $V_{\text{acc}}$	$1\times10^4$ V
Beam direction	horizontal, east
Deflecting electric field $E$	$1\times10^6$ V/m, pointing down

Find. The magnitude and direction of the magnetic flux density $B$ that exactly cancels the electric deflecting force.

Crossed-field (velocity selector) geometry: electron velocity east, deflecting E field down (giving an UPWARD electric force on the negative electron), balancing magnetic field north (so v×B points up, cancelling F_E).

Approach. Find the electron's speed from the accelerating potential (work–energy theorem), then require the magnetic force to exactly cancel the electric force — the velocity-selector condition $\vec v\times\vec B=-\vec E$ — to solve for $B$.

Speed from the accelerating voltage. $$eV_{\text{acc}}=\tfrac12m_ev^2\ \Rightarrow\ v=\sqrt{\frac{2eV_{\text{acc}}}{m_e}}$$ $$v=\sqrt{\frac{2(1.6\times10^{-19})(1\times10^4)}{9.11\times10^{-31}}}$$ $$\boxed{v=5.927\times10^{7}\ \text{m/s, east}}$$
Force-balance condition. Net force must vanish: $q\vec E+q\vec v\times\vec B=0\ \Rightarrow\ \vec v\times\vec B=-\vec E$. The electron is negative, so the downward $\vec E$ pushes it UP ($\vec F_E=(-e)\vec E$ points up); the magnetic force must therefore act DOWNWARD to cancel it. Because $\vec F_B=(-e)\,\vec v\times\vec B$, a downward force on the electron requires $\vec v\times\vec B$ itself to point straight UP (equivalently $\vec v\times\vec B=-\vec E$), and by the right-hand rule with $v$ east this requires $\vec B$ to point NORTH (east $\times$ north $=$ up).
Magnitude, from $vB=E$. $$B=\frac{E}{v}=\frac{1\times10^{6}}{5.927\times10^{7}}$$ $$\boxed{B=1.687\times10^{-2}\ \text{T}=16.87\ \text{mT, pointing north}}$$

Classical kinematics is the exam-level method here and is what the answer above uses. At $10^4$ V the electron reaches about $0.2c$, so a relativistic check ($\gamma=1+eV/m_ec^2=1.0195$) gives $v\approx5.84\times10^{7}$ m/s and $B\approx1.71\times10^{-2}$ T, about 1.5% higher. The direction (north) is unchanged, and the classical value is the expected answer on this paper.

Quantity	Result
Electron speed $v$	$5.927\times10^{7}$ m/s
Magnetic flux density $B$	$1.687\times10^{-2}$ T $=16.87$ mT
Direction of $B$	north

Concept & Theory

Topic. Velocity selector — balance of electric and magnetic (Lorentz) forces on a charged particle.

Key relations. Work–energy theorem $eV=\tfrac12mv^2$; Lorentz force $\vec F=q(\vec E+\vec v\times\vec B)$; equilibrium condition $\vec v\times\vec B=-\vec E$.

Why this works. A charge moving through crossed, mutually perpendicular $E$ and $B$ fields experiences electric and magnetic forces that can be made to exactly cancel for one particular speed — here $v$ is fixed by the accelerating voltage, so $B$ is the quantity solved for. This is the operating principle of a mass spectrometer's velocity selector.

Common pitfall. Forgetting the electron's negative charge and getting the electric-force direction backwards (and hence the required $B$ direction backwards too); or substituting the DEFLECTING field $E$ into the accelerating-voltage speed formula instead of using energy conservation with $V_{\text{acc}}$.

Source. Sadiku, Elements of Electromagnetics, Ch. 8 (Lorentz force, charged-particle motion in fields); Hayt & Buck, Ch. 8.

Question 8 of 8: Velocity Selector — Magnetic Field Cancelling an Electric Deflecting Force

Question 8: Velocity Selector — Magnetic Field Cancelling an Electric Deflecting Force (20 marks)