Magnetic Field Cancelling an Electric Force on a Moving Electron: 04-BS-9, December 2015, Q4

Question 4: Magnetic Field Cancelling an Electric Force on a Moving Electron (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
Electron velocity $v$ (east)	$6\times10^{4}$ m/s
Electric field $E$ (vertical, pointing down)	$10^{4}$ V/m
Electron charge	$q=-e=-1.6\times10^{-19}$ C

Find. The magnitude and direction of $\vec B$ so that $\vec F_B=-\vec F_E$.

Electron moving east in a downward $E$ field and an as-yet-unknown $B$; $B$ must point north for $\vec v\times\vec B$ (and hence $\vec F_B$) to balance $\vec F_E$.

Approach. Write both forces vectorially with $q=-e$; the magnetic-force direction $\vec v\times\vec B$ is perpendicular to both $\vec v$ and $\vec B$, so only a horizontal (north–south) $B$ can produce a vertical force when $\vec v$ is horizontal — solve that component directly.

Electric force on the electron. With $\vec E$ pointing down and $q=-e$, the force on the electron is opposite to $\vec E$, i.e. upward: $$F_E=eE=(1.6\times10^{-19})(10^{4})$$ $$F_E=\boxed{1.6\times10^{-15}\ \text{N, upward}}$$
Required magnetic force. For the magnetic force to cancel this, $\vec F_B$ must point downward with the same magnitude, $1.6\times10^{-15}$ N.
Solve for $B$ from $\vec F_B=q\vec v\times\vec B$. With $\vec v$ east and $\vec B$ chosen north (horizontal, perpendicular to $\vec v$), $\vec v\times\vec B$ points straight up, so $\vec F_B=(-e)(v\times B)$ points down — the needed sense. Equating magnitudes: $$evB=eE \implies B=\frac{E}{v}=\frac{10^{4}}{6\times10^{4}}$$ $$B=\boxed{0.1667\ \text{T} = \tfrac16\ \text{T, pointing north}}$$

Quantity	Result
$F_E$ on the electron	$1.6\times10^{-15}$ N, upward
Required $\vec B$	$0.1667$ T, pointing north

Concept & Theory

Topic. Balance of electric and magnetic (Lorentz) forces on a moving charge — the velocity-selector configuration.

Key relations. $\vec F=q(\vec E+\vec v\times\vec B)$; cancellation requires $q\vec v\times\vec B=-q\vec E$.

Why this works. $\vec F_B=q\vec v\times\vec B$ is always perpendicular to both $\vec v$ and $\vec B$; since $\vec v$ is horizontal (east) and the needed force is vertical, $\vec B$ must itself be horizontal and perpendicular to $\vec v$ — i.e. north–south — because any component of $\vec B$ along $\vec v$ contributes nothing to $\vec v\times\vec B$.

Common pitfall. Forgetting the electron's negative charge, which flips both the direction of $\vec F_E$ (opposite $\vec E$, not along it) and the sign convention needed when solving $\vec F_B=q\vec v\times\vec B$ for 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 4 of 8: Magnetic Field Cancelling an Electric Force on a Moving Electron

Question 4: Magnetic Field Cancelling an Electric Force on a Moving Electron (20 marks)