Motional EMF Induced in a Moving Rod Crossing a Vertical Field: 04-BS-9, December 2018, Q5

Question 5: Motional EMF Induced in a Moving Rod Crossing a Vertical Field (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
Rod length $\ell$	$2$ m, aligned $30^\circ$ north of east
Velocity $v$	$50$ m/s, due north
Magnetic flux density $B$	$1\times10^{-5}$ T, vertically up

Find. The magnitude and polarity of the EMF induced between the rod's two tips.

Top-down (bird's-eye) view: rod at 30° north of east, moving north at 50 m/s through a vertically-up B field (out of this top-down page). The east-directed force qv×B drives positive charge to the rod's north-east tip.

Approach. Compute the motional-EMF force per unit charge $\vec v\times\vec B$, then take its component along the rod's own length vector (dot product) to get the potential difference between the tips.

Force per unit charge. With $v$ north and $B$ up, the right-hand rule gives $\vec v\times\vec B$ pointing EAST, magnitude $$|\vec v\times\vec B|=vB=(50)(1\times10^{-5})=5\times10^{-4}\ \text{V/m}$$
Component along the rod. The rod (length $2$ m) lies $30^\circ$ north of east, so its east-projected length is $\ell\cos30^\circ$: $$\text{EMF}=(\vec v\times\vec B)\cdot\vec\ell=(5\times10^{-4})(2)\cos30^\circ$$
Result and polarity. $$\boxed{\text{EMF}=8.660\times10^{-4}\ \text{V}=0.866\ \text{mV}}$$ Since the driving force $\vec v\times\vec B$ points east and the rod's own axis has a positive east component, positive charge is pushed toward the north-east tip — that tip is at the HIGHER (positive) potential.

Quantity	Result
$\|\vec v\times\vec B\|$	$5\times10^{-4}$ V/m
Induced EMF	$8.660\times10^{-4}$ V $=0.866$ mV
Polarity	north-east tip positive, south-west tip negative

Concept & Theory

Topic. Motional EMF — the Lorentz force on charge carriers inside a moving conductor.

Key relations. $\vec F=q\vec v\times\vec B$; $\text{EMF}=\displaystyle\int(\vec v\times\vec B)\cdot d\vec\ell$.

Why this works. Every free charge inside the moving rod feels the same $q\vec v\times\vec B$ force, equivalent to an internal battery whose EMF is the line integral of that force per unit charge along the rod. Only the component of $\vec v\times\vec B$ that lies ALONG the rod actually separates charge; the perpendicular component just presses charge against the rod's sides with no net EMF contribution.

Common pitfall. Using the full rod length instead of its projection onto the $\vec v\times\vec B$ direction (forgetting the $\cos30^\circ$ factor), or getting the polarity backwards by not tracking, via the right-hand rule, which physical direction $\vec v\times\vec B$ actually points.

Source. Sadiku, Elements of Electromagnetics, Ch. 9 (Faraday's law, motional EMF); Hayt & Buck, Ch. 10.

Question 5 of 8: Motional EMF Induced in a Moving Rod Crossing a Vertical Field

Question 5: Motional EMF Induced in a Moving Rod Crossing a Vertical Field (20 marks)