22-Elec-A7 Electromagnetics · December 2018

Question 5 of 8: Terminal voltage of a rod crossing a slab of magnetic field

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format. National Examinations, December 2018, 16-Elec-A7 Electromagnetics. Three hours, closed book, one approved Casio or Sharp calculator. Eight questions, all of equal value; the rubric says any five constitute a complete paper, so a candidate answers five — but all eight are worked here, because the set is a study resource. Aids printed on the paper: $\varepsilon_0 = 8.85\times10^{-12}$ F/m, $\mu_0 = 4\pi\times10^{-7}$ H/m, $e = 1.6\times10^{-19}$ C.

Reference texts. M. N. O. Sadiku, Elements of Electromagnetics, 7th ed.; W. H. Hayt and J. A. Buck, Engineering Electromagnetics, 9th ed.; F. T. Ulaby and U. Ravaioli, Fundamentals of Applied Electromagnetics, 8th ed.; D. M. Pozar, Microwave Engineering, 4th ed.; C. A. Balanis, Antenna Theory: Analysis and Design, 4th ed.

Constants used throughout. Question 6 hands us $\omega = 2\pi\times10^{10}$ rad/s together with $k = (2\pi/3)\ \text{cm}^{-1}$, so the setter’s phase velocity is $\omega/k = 3.000\times10^{8}$ m/s exactly; Question 3’s cut-off arithmetic only closes on round numbers under the same choice. Accordingly $c = 3\times10^{8}$ m/s and $\eta_0 = 120\pi = 376.99\ \Omega$ are used here. The paper’s own aid list would give $c = 2.99863\times10^{8}$ m/s and $\eta_0 = 376.83\ \Omega$ — a 0.05 % shift that changes no answer except at one deliberate boundary case, flagged in Question 3.

Question 5: Terminal voltage of a rod crossing a slab of magnetic 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. A rod carried bodily westward, at an angle to its own direction of travel, through a slab of uniform vertical field.

Given data
Quantity	Symbol	Value
Rod length	$L$	2 m
Rod orientation	—	horizontal, NW–SE
Velocity	$v$	30 m/s, due west
Field magnitude inside the wall	$B$	$10^{-5}$ T
Field direction	—	vertically up
Wall orientation and thickness	—	vertical, north–south, 30 m

Find. The voltage of the NW tip relative to the SE tip as a function of time, plotted from first contact to final exit.

Plan view, drawn with the motion running left to right and a compass rose carrying the true orientation. The rod meets the wall face obliquely, so it enters over a finite time.

Approach. Evaluate the motional force per unit charge, project it along the rod to get the peak EMF, then use the rod’s extent along the direction of travel to set the entry and exit ramps.

The motional field. A charge carried with the rod feels a force per unit charge $\mathbf{v}\times\mathbf{B}$. With east, north and up as $\hat{x},\hat{y},\hat{z}$, the velocity is $(-30,0,0)$ and the field $(0,0,10^{-5})$, so $$\mathbf{v}\times\mathbf{B} = (-30,0,0)\times(0,0,10^{-5}) = (0,\ 3\times10^{-4},\ 0)\ \text{V/m} .$$ It points due NORTH and has magnitude $vB = 3\times10^{-4}$ V/m — note it is perpendicular to the motion, which is why a rod aligned east-west would read nothing at all.
Project it along the rod. Only the component of the rod lying along that northward push contributes. The unit vector from the SE tip to the NW tip is $(-1,1,0)/\sqrt{2}$, so $$V_{pk} = (\mathbf{v}\times\mathbf{B})\cdot\mathbf{L} = 3\times10^{-4} \times \frac{2}{\sqrt{2}} = 3\times10^{-4} \times 1.414 = \boxed{424\ \mu\text{V}} ,$$ positive, so the NW tip is the high terminal. Equivalently $V = B v L_{NS}$ with $L_{NS} = L\cos 45^\circ = 1.414$ m the rod’s north-south span.
Find how long entry takes. The wall face runs north-south, so the rod crosses it progressively over its own EAST-WEST span, $$L_{EW} = L\cos 45^\circ = 1.414\ \text{m}, \qquad t_{ramp} = \frac{L_{EW}}{v} = \frac{1.414}{30} = 47.1\ \text{ms}.$$ The NW tip is the western tip, so it breaks the plane first and the immersed length grows linearly — hence a linear ramp in voltage, not a step.
Find the flat top. The rod is fully inside from $t = 47.1$ ms until its leading (NW) tip reaches the far face, $$t = \frac{30\ \text{m}}{30\ \text{m/s}} = 1.000\ \text{s},$$ and throughout that interval the EMF is the full 424 $\mu$V. This is worth saying explicitly: a rod is not a loop. A closed loop fully immersed in a uniform field reads zero, because its two transverse sides cancel; a rod has only one side, so its EMF persists.
Find the exit. Leaving mirrors entering: the immersed length falls linearly from 2 m to zero over another 47.1 ms, so the trace returns to zero at $$t = 1.000 + 0.047 = \boxed{1.047\ \text{s}} .$$
Describe the trace. The result is a single positive trapezoid: a 47.1 ms linear rise to 424 $\mu$V, a 953 ms flat top, and a 47.1 ms linear fall, total duration 1.047 s. It never changes sign, because the field direction and the velocity are both constant throughout. Its area, $V_{pk}(t_{total} - t_{ramp}) = 424\ \mu\text{V} \times 1.000\ \text{s} = 424\ \mu\text{Wb}$, is exactly $B$ times the swept area projected on the horizontal — the integral form of Faraday’s law, and a free check on the whole plot.

The answer: voltage of the NW tip with respect to the SE tip. A trapezoid, positive throughout, 424 μV peak, 1.047 s wide.

Check: the 30 m thickness is read as the wall’s extent along the rod’s path, i.e. measured east-west, perpendicular to the north-south wall — the only reading under which a ‘30 m thick’ north-south wall is a well-posed obstacle for a westward traveller. Read any other way the crossing time is undetermined. The peak voltage does not depend on this choice; only the 1.000 s flat-top duration does.

Final results
Quantity asked	Result
Motional field	$\|\mathbf{v}\times\mathbf{B}\| = 3\times10^{-4}$ V/m, due north
Peak voltage (NW tip positive)	$V_{pk} = 424\ \mu\text{V}$
Rise time (rod entering)	47.1 ms
Flat top (fully immersed)	952.9 ms, from 47.1 ms to 1.000 s
Fall time (rod leaving)	47.1 ms, ending at 1.047 s
Waveform	a single positive trapezoid, total width 1.047 s

Topic. Motional EMF and the difference between a rod and a loop.

Key relations. $\mathbf{f} = \mathbf{v}\times\mathbf{B}$ (force per unit charge); $\;V = \int (\mathbf{v}\times\mathbf{B})\cdot d\boldsymbol{\ell}$; $\;V = BvL_\perp$ for mutually perpendicular $B$, $v$ and the projected length; $\;\oint \mathbf{E}\cdot d\boldsymbol{\ell} = -d\Phi/dt$.

Why this works. Charges inside a conductor moving through a magnetic field are pushed sideways until the electrostatic field they build up balances the magnetic push, and the terminal voltage is that balance integrated along the conductor. Because the push is $\mathbf{v}\times\mathbf{B}$, it is perpendicular to both the motion and the field, and only the part of the rod lying along that direction contributes. The geometry then does the rest: the rod's north-south span sets how large the voltage is, while its east-west span sets how long the rod takes to cross the boundary and therefore how steep the ramps are. The same rod at $45^\circ$ supplies both numbers, which is the neat part of this question.

Common pitfall. The commonest error is to use the full 2 m length and report 600 $\mu$V, forgetting the projection. The second is to reason from Faraday's law about a loop and conclude the voltage vanishes while the rod is fully immersed; that cancellation needs a return conductor, which an isolated rod does not have.

Source. Sadiku, Elements of Electromagnetics, 7th ed., §9.2 (motional EMF); Hayt and Buck, Engineering Electromagnetics, 9th ed., §9.1.

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