Irodov 3.265: A Generator with No Moving Parts – Amit & Deepali Sinha Foundation

Problem Irodov 3.265

Two parallel plates, each of area $S$ and a distance $d$ apart, are placed in a stream of conducting liquid of resistivity $\rho$ , which flows parallel to the plates with constant speed $v$ . A uniform magnetic field of induction $B$ is applied parallel to the plates and at right angles to the flow. The plates are joined through an external resistance $R$ . What power is released in $R$ ? For what value of $R$ is this power largest, and what is the largest power?

The plates edge on, a distance d apart, in the stream of conducting liquid (resistivity ρ, speed v), with B parallel to the plates and across the stream (its sense is not given; drawn into the page). The leads are joined through R. — The plates edge on, a distance $d$ apart, in the stream of conducting liquid (resistivity $\rho$ , speed $v$ ), with $\vect B$ parallel to the plates and across the stream (its sense is not given; drawn into the page). The leads are joined through $R$ .

Given

plate area $S$
gap $d$
resistivity $\rho$
speed $v$
field $B$ , $\vect B \perp \vect v$ , both parallel to the plates
resistance $R$
flow uniform and unchanged by the current (assumed)

Find

the power in

R

; the

R

that maximises it; the maximum power

Solution

Step 1The emf of the moving liquid

A positive carrier moving with the liquid: the magnetic force q v· B points across the gap, toward the upper plate for this sense of B, so the upper plate charges positive. — A positive carrier moving with the liquid: the magnetic force $q\vect v\times\vect B$ points across the gap, toward the upper plate for this sense of $\vect B$ , so the upper plate charges positive.

A carrier of charge $q$ moving with the liquid feels the magnetic force $q\vect v\times\vect B$ . Since $\vect v$ and $\vect B$ are perpendicular and both parallel to the plates, this force is perpendicular to the plates and has the same magnitude $qvB$ everywhere in the gap. Its work per unit charge carried across the gap is the emf of the liquid layer:

\mathcal{E} = \int_0^d \left|\vect v\times\vect B\right| dx = \int_0^d vB\,dx

\mathcal{E} = vBd

(1)

Step 2The circuit

The liquid between the plates is the source, and it has an internal resistance: a column of length $d$ and cross-section $S$ of resistivity $\rho$ ,

r = \frac{\rho d}{S} \tag{2}

In series with $R$ , this source drives the current

I = \frac{\mathcal{E}}{R + r}

and the power released in $R$ is

P = I^2R = \frac{\mathcal{E}^2 R}{(R + r)^2} \tag{3}

Substituting (1) and (2):

P = \frac{v^2B^2d^2R}{(R + \rho d/S)^2}

(4)

Step 3The best load

The power vanishes for $R \to 0$ and for $R \to \infty$ , so it has a maximum between. Differentiating (3) with respect to $R$ :

\frac{dP}{dR} = \mathcal{E}^2\,\frac{(R + r)^2 - 2R(R + r)}{(R + r)^4} = \mathcal{E}^2\,\frac{r - R}{(R + r)^3}

This is zero when the load equals the internal resistance (2):

R = r = \frac{\rho d}{S}

(5)

Putting this load into (3), and then (1) and (2):

P_{\mathrm{max}} = \frac{\mathcal{E}^2 r}{(2r)^2} = \frac{\mathcal{E}^2}{4r} = \frac{v^2B^2d^2}{4}\cdot\frac{S}{\rho d}

P_{\mathrm{max}} = \frac{v^2B^2 S d}{4\rho}

(6)

Check

Units: $vBd$ is in volts (a tesla is V $\cdot$ s/m $^2$ ), and $\rho d/S$ is in ohms, so (4) is V $^2/\Omega$ , a power. Energy: the current $I$ crosses the liquid, length $d$ , at right angles to $\vect B$ , so the liquid feels the force $IdB$ , which opposes the flow by the right-hand rule. The stream does work against it at the rate $IdBv = I\mathcal{E}$ , exactly the power released in $R$ and in the liquid together. Reversing $\vect B$ swaps the plates’ polarity and leaves every answer unchanged, since only $B^2$ enters.

Remark

This is the magnetohydrodynamic generator: the moving conductor is the liquid or an ionised gas itself, so there is no rotor and no brushes. The largest power, (6), is the volume $Sd$ of the working liquid times $v^2B^2/4\rho$ , so a practical machine needs a fast stream, a strong field and a well-conducting medium; at the matched load half of the converted power is lost as heat in the liquid.