22-Elec-A7 Electromagnetics · May 2018

Question 3 of 8: EMF Induced in a Bar Rotating in a Vertical Field

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

Notes on this paper

Paper format. National Examinations, May 2018 — 16-Elec-A7, Electromagnetics. Closed book, three hours, approved Casio or Sharp calculator only. Eight questions, all of equal value; the paper states that any five constitute a complete paper, so each question carries 20 of the 100 marks that are actually graded. All eight are solved here, because this set is a study resource rather than an examination script.

Aids given on the paper. $\varepsilon_0 = 8.85\times10^{-12}\ \text{F/m}$, $\mu_0 = 4\pi\times10^{-7}\ \text{H/m}$, $e = 1.6\times10^{-19}\ \text{C}$. Those two constants fix $c = 1/\sqrt{\mu_0\varepsilon_0} = 2.9986\times10^{8}\ \text{m/s}$ and $\eta_0 = \sqrt{\mu_0/\varepsilon_0} = 376.82\ \Omega$ — which is why Question 1 quotes a 377 Ω line: it is free space itself, rendered as a cable.

Reference texts. D. M. Pozar, Microwave Engineering, 4th ed. (transmission-line transients, stubs, waveguide modes); M. N. O. Sadiku, Elements of Electromagnetics, 7th ed. (Biot–Savart, motional EMF, plane waves); W. H. Hayt and J. A. Buck, Engineering Electromagnetics, 9th ed.; F. T. Ulaby and U. Ravaioli, Fundamentals of Applied Electromagnetics, 8th ed.; C. A. Balanis, Antenna Theory: Analysis and Design, 4th ed. (short-element fields and the reactive near-field boundary).

Constants used throughout: the paper's own aid list is taken as authoritative, so $c = 2.9986\times10^{8}$ m/s and $\eta_0 = 376.82\ \Omega$. Question 1 supplies its own propagation velocity ($3\times10^{8}$ m/s) and that stated value is used there. Question 5 is the one place where the difference between $2.9986\times10^{8}$ and $3.00\times10^{8}$ changes the wording of the answer, and both readings are worked through explicitly.

Question 3: EMF Induced in a Bar Rotating in 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. A rigid conducting bar swept round a vertical axis through one of its ends, inside a uniform vertical field.

Given data
Quantity	Symbol	Value
Bar length (also the radius swept)	$L$	3 m
Rotation rate	$N$	600 rev/min
Flux density, vertical and uniform	$B$	$10^{-5}$ T
Axis	—	vertical, through one end
Bar orientation	—	horizontal, so $B$ is normal to its plane of motion

Find. The steady voltage appearing between the pivot end and the free end of the bar.

The bar sweeps a disc of radius L in a vertical field. Each element moves at v = ωr, so the motional field grows linearly from zero at the pivot to its largest value at the free end.

Approach. This is a homopolar (Faraday-disc) generator: integrate the motional field $\mathbf{v}\times\mathbf{B}$ along the bar, then confirm the result independently from the rate at which the bar sweeps area.

Convert the rotation rate. $$\omega = \frac{2\pi N}{60} = \frac{2\pi (600)}{60} = 62.83\ \text{rad/s}.$$
Write the motional field at radius $r$. The element at radius $r$ moves tangentially at $v = \omega r$, and $\mathbf{v}$ is horizontal while $\mathbf{B}$ is vertical, so the two are perpendicular and $$|\mathbf{v}\times\mathbf{B}| = \omega r B ,$$ directed along the bar. The field is therefore triangular: zero at the pivot, largest at the free end.
Integrate along the bar. $$\mathcal{E} = \int_{0}^{L}\omega B r\,dr = \tfrac{1}{2}\,\omega B L^{2}.$$ Substituting, $$\mathcal{E} = \tfrac{1}{2}(62.83)(10^{-5})(3)^{2} \;\Rightarrow\; \boxed{\mathcal{E} = 2.83\ \text{mV}} .$$
Check it a second way, from swept area. In one second the bar sweeps an angle $\omega$, covering an area $\tfrac{1}{2}\omega L^{2} = 282.7\ \text{m}^2$ of the horizontal plane. With $B$ normal to that plane the flux cut per second is $B(\tfrac{1}{2}\omega L^{2}) = 2.83$ mV, which reproduces the integral exactly. The two routes agree because they are the same statement of Faraday’s law.
Note the character of the answer. The voltage is direct, not alternating. Nothing about the geometry changes as the bar turns — the field is uniform, the bar stays horizontal, and the swept area grows at a constant rate — so the EMF is a constant 2.83 mV for as long as the rotation is maintained. Contact must be made with a slip ring at the rim and a brush on the shaft to use it.
Give the polarity. The force on a positive carrier is $q\mathbf{v}\times\mathbf{B}$, which for upward $B$ and anticlockwise rotation (seen from above) points radially outward. The free end is then the positive terminal; reversing either the field direction or the sense of rotation reverses it. The magnitude is unaffected either way.

Check: the paper states neither the sense of rotation nor which way the vertical field points, so the polarity above is quoted for the “field up, rotation anticlockwise viewed from above” case and flips with either. The magnitude 2.83 mV is independent of both choices.

Final results
Quantity asked	Result
Angular velocity	62.83 rad/s
Motional field at the free end	$\omega L B = 1.885\times10^{-3}$ V/m
Area swept per second	282.7 m$^{2}$/s
Induced EMF between the ends	2.83 mV (DC)
Polarity	free end positive for $B$ up, anticlockwise rotation

Topic. Motional EMF in a rotating conductor — the homopolar generator.

Key relations. $\mathcal{E} = \int (\mathbf{v}\times\mathbf{B})\cdot d\mathbf{l}$; for a bar of length $L$ turning at $\omega$ in a normal field, $\mathcal{E} = \tfrac{1}{2}\omega B L^{2}$; equivalently $\mathcal{E} = B\,dA/dt$ with $dA/dt = \tfrac{1}{2}\omega L^{2}$.

Why this works. Every free carrier in the bar is being dragged sideways through the field, so it feels a force $q\mathbf{v}\times\mathbf{B}$ along the bar. Charge piles up at the ends until the electrostatic field it creates exactly cancels that magnetic push, and the potential difference at that balance point is the line integral of the motional field. Because $v$ rises linearly with radius, the integrand is a straight line and the integral is simply the mean value $\tfrac{1}{2}\omega LB$ times the length $L$.

Common pitfall. Using $\mathcal{E} = BLv$ with the tip speed $v = \omega L$, which doubles the answer by pretending every point of the bar moves as fast as the tip. The other common error is expecting a sinusoid: a bar rotating in a uniform field normal to its plane of motion gives DC, not AC — the sinusoid appears only when a closed loop rotates so that the flux through it varies.

Source. Sadiku, Elements of Electromagnetics, 7th ed., §9.3 (motional EMF); Hayt and Buck, Engineering Electromagnetics, 9th ed., §10.1–10.2.

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