The electromagnetic flowmeter
The electromagnetic flowmeter is the inverse of the electromagnetic pump (Prob. 22-7). It operates as follows. See Fig. 22-16. A conducting fluid flows in a nonconducting tube between the poles of a magnet. Electrodes on either side of the tube and in contact with the fluid measure the $v \times B$ field, and thus the quantity of fluid that flows through the tube per second. This is a Hall effect, except that here ions of both signs move with the fluid in the same direction.
Faraday attempted to measure the velocity of the Thames River in this way in 1832. The magnetic field was, of course, that of the earth.
In the absence of turbulence, the fluid velocity in a tube of radius $R$ is of the form $v=v_{0}\left(1-r^{2} / R^{2}\right)$. The $v \times B$ field in the fluid is therefore not uniform. This gives rise to circulating currents with $\boldsymbol{J}=\sigma(-\boldsymbol{V} V+\boldsymbol{v} \times \boldsymbol{B})$. The potential $V_{0}$ results from the charges that accumulate on the electrodes.
(a) Sketch a cross section of the tube, showing qualitatively, by means of arrows of various lengths, the magnitude and direction of $v \times B$.
(b) Sketch another cross section, showing the lines of current flow. The current drawn by the electrodes is negligible.
(c) Neglect end effects by setting $\partial / \partial z=0$. Use the fact that $\nabla \cdot J=0$ to show that
$$
\boldsymbol{\nabla}^{2} V=B \frac{\partial v}{\partial y}=B \frac{\partial v}{\partial r} \sin \phi
$$
Since this Laplacian is equal to $-\rho / \epsilon_{0}$, the volume charge density $\rho$ is zero
on the axis, where $\partial v / \partial r=0$ and at $\phi=0$, and at $\pi$, See page 402 .
(d) Set $V=V^{\prime} \sin \phi$, where $V^{\prime}$ is independent of $\phi$, and show that
$$
\frac{d^{2} V^{\prime}}{d r^{2}}+\frac{1}{r} \frac{d V^{\prime}}{d r}-\frac{V^{\prime}}{r^{2}}=B \frac{d v}{d r}
$$
(e) You can solve this differential equation as follows: (1) express the left-hand side as a derivative, (2) integrate, (3) multiply both sides by $r,(4)$, express both sides as derivatives, (5) integrate. This will leave you with two constants of integration, one of which is easy to dispose of. You can find the value of the other by remembering that $J_{y}=0$ at $x=0, y=R$ if the voltmeter draws zero current.
Note that the output voltage is independent of the conductivity of the fluid, if one assumes that the voltmeter draws zero current.
Find the output voltage as a function of the volume $Q$ of fluid that flows, in one second.
(f) We have neglected cdge effects in the region where the fluid enters, and emerges from, the magnetic field. Sketch lines of current flow for these, two regions. These currents reduce the output voltage somewhat.