Homopolar Disc Machine — EMF, Torque and Output Power: 22-Mec-A5 Electrical and Electronics Engineering, December 2018, Q3

Question 3: Homopolar Disc Machine — EMF, Torque and Output Power (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 conducting disc of effective outer diameter $D$ and inner diameter $d$ rotates at $\omega$ rad/s about its own axis in a uniform vertical flux density $B$, while a radial current $I_2$ is passed through it between an inner and an outer ring brush. All quantities are symbolic; the front page supplies $1\ \text{hp} = 746\ \text{W}$ for the final conversion.

Find. The magnitude of the generated emf between the brushes, the electromagnetic torque acting on the rotor, and the output power expressed in horsepower.

Figure 3 — dc machine. The shaded ring is the elemental annulus of radius $r$ and radial width $dr$ suggested by the hint.

Approach. Both parts are single integrations over the same elemental annulus. For the emf, each element of the disc is a radial conductor moving through the field at the local tangential speed, so the motional-emf density $Bv$ is integrated along the radius. For the torque, the radial current in each element experiences a tangential force $B\,I_2\,dr$ whose moment about the axis is integrated over the same limits.

Write the motional emf of one annulus. A point at radius $r$ moves tangentially at $v = \omega r$, perpendicular to the vertical field, so the emf induced along the radial length $dr$ is $$de = B\,v\,dr = B\,\omega\,r\,dr.$$ Because $\mathbf{v} \times \mathbf{B}$ points radially everywhere, the contributions of successive annuli add directly in series between the two brushes.
Integrate between the brush radii. The brushes sit at $r = d/2$ and $r = D/2$, so $$e = \int_{d/2}^{D/2} B\,\omega\,r\,dr = B\,\omega\left[\frac{r^{2}}{2}\right]_{d/2}^{D/2} = \frac{B\,\omega}{2}\left(\frac{D^{2}}{4}-\frac{d^{2}}{4}\right),$$ which tidies to $$e = \boxed{\dfrac{B\,\omega\,(D^{2}-d^{2})}{8}}$$ The denominator 8 is simply the product of the two halvings: one from integrating $r\,dr$ and one from converting each diameter to a radius.
Write the tangential force on one annulus. The same element now carries the radial current $I_2$; a radial current in a vertical field produces a tangential force $$dF = B\,I_2\,dr,$$ and its moment about the shaft is $dT = r\,dF = B\,I_2\,r\,dr$.
Integrate for the torque. Over the same limits, $$T = \int_{d/2}^{D/2} B\,I_2\,r\,dr = \boxed{\dfrac{B\,I_2\,(D^{2}-d^{2})}{8}}$$ The torque and the emf therefore share the identical geometric factor $B(D^{2}-d^{2})/8$, which is the machine constant of this rotor.
Convert to output power. Mechanical power is torque times angular speed, $$P = T\,\omega = \frac{B\,I_2\,\omega\,(D^{2}-d^{2})}{8},$$ and dividing by the printed conversion gives $$\text{hp} = \boxed{\dfrac{B\,I_2\,\omega\,(D^{2}-d^{2})}{8 \times 746}}$$
Check by energy conversion. The electrical power crossing the brushes is $e\,I_2 = B\omega(D^{2}-d^{2})I_2/8$, which is exactly $T\omega$. The two independent integrations agree, confirming that the machine constant is common to both and that no factor of two has been lost in either.

Final results — Question 3
Quantity	Result
Generated emf between the brushes	$e = \dfrac{B\,\omega\,(D^{2}-d^{2})}{8}$
Electromagnetic torque on the rotor	$T = \dfrac{B\,I_2\,(D^{2}-d^{2})}{8}$
Mechanical (output) power	$P = T\omega = e\,I_2 = \dfrac{B\,I_2\,\omega\,(D^{2}-d^{2})}{8}$
Output in horsepower	$\text{hp} = \dfrac{B\,I_2\,\omega\,(D^{2}-d^{2})}{5968}$
Energy-conversion check	$T\omega = e\,I_2$ identically — satisfied

Check — idealisations. The field is taken as uniform over the whole annulus and normal to the disc, brush contact drop and disc resistance are neglected, and the current is assumed to flow purely radially (no circulating eddy paths). Real Faraday-disc machines lose an appreciable fraction of their output to brush drop precisely because they are low-voltage, very-high-current devices.

Concept & Theory

Topic. The homopolar (acyclic, Faraday-disc) machine and the equality of its motional and force-based machine constants.

Key relations. Motional emf $de = (\mathbf{v}\times\mathbf{B})\cdot d\boldsymbol{\ell} = B\omega r\,dr$; Lorentz force $dF = B I\,d\ell$; torque $dT = r\,dF$; energy balance $T\omega = e I$; and $1\ \text{hp} = 746\ \text{W}$.

Why this works. The disc machine is the one rotating machine with no commutation at all: the flux through any radial path never reverses, so the emf between inner and outer brushes is genuinely direct and needs no commutator segments. The integration is necessary because the tangential velocity is not uniform — it grows linearly with radius — so the outer parts of the disc contribute far more emf per unit length than the inner parts. That the same integral $\int r\,dr$ appears in both the emf and the torque is not a coincidence: it is the reciprocity that guarantees electrical and mechanical power balance in any lossless electromechanical converter, and it is the reason the check $T\omega = eI_2$ closes exactly.

Common pitfall. Substituting the diameters where the derivation calls for radii, which inflates the answer by a factor of four. Writing $e = B\omega(D^{2}-d^{2})/2$ (forgetting to convert diameters to radii) is the single most common error on this question; the safeguard is the $T\omega = eI_2$ check, which fails immediately if the factor is wrong in only one of the two derivations.

Source. Chapman, Electric Machinery Fundamentals, 5th ed., Ch. 1 (motional emf, induced force, the linear machine) as the direct antecedent of the rotating disc.

Question 3 of 8: Homopolar Disc Machine — EMF, Torque and Output Power

Question 3: Homopolar Disc Machine — EMF, Torque and Output Power (20 marks)