Irodov 3.261: A Pump with No Moving Parts – Amit & Deepali Sinha Foundation

Problem Irodov 3.261

An electromagnetic pump moves molten metal along a pipe of rectangular cross-section. A section of the pipe, filled with the metal, lies in a uniform magnetic field $B = 0.10$ T perpendicular to the pipe’s axis; the height of the channel measured along $\vect B$ is $a = 2.0$ cm. A current $I = 100$ A is passed through the metal across the pipe, perpendicular both to $\vect B$ and to the axis. Find the gauge pressure the pump produces.

The pipe section in perspective: the field B across it, the current I through the metal across the pipe, perpendicular to B and to the axis, the metal moving along the axis toward the viewer, and the channel's height a along B. — The pipe section in perspective: the field $\vect B$ across it, the current $I$ through the metal across the pipe, perpendicular to $\vect B$ and to the axis, the metal moving along the axis toward the viewer, and the channel’s height $a$ along $\vect B$ .

Given

$B = 0.10$ T
$I = 100$ A
height along $\vect B$ : $a = 2.0$ cm
friction in the section neglected (assumed)

Find

the gauge pressure

\Delta p

Solution

Step 1The force on the metal

Let $b$ be the width of the channel along the current. The current crosses the metal over the length $b$ , perpendicular to $\vect B$ , so by $\vect F = I\,\vect l\times\vect B$ the metal carrying it is pushed with the force

F = I\,b\,B \tag{1}

The Ampere force F on the current-carrying metal, along the pipe's axis, with the channel's width b along the current and its height a along B. — The Ampere force $\vect F$ on the current-carrying metal, along the pipe’s axis, with the channel’s width $b$ along the current and its height $a$ along $\vect B$ .

Its direction is that of $\vect l\times\vect B$ : with the current across the pipe and $\vect B$ perpendicular to both the current and the axis, it lies along the axis, toward the viewer in the figure, which is the way the metal is driven. It does not matter how the current spreads along the axis: every part of it crosses the full width $b$ in the same field, so the total is (1).

Step 2The pressure it builds

The force acts on the metal in the section, and the metal passes it on as a pressure difference between the two ends of the section, spread over the cross-section $ab$ . With friction neglected (assumed) and the same cross-section at both ends, the force balance of the metal in the section gives

\Delta p\,ab = I\,b\,B

and $b$ cancels. Numerically,

\Delta p = \frac{(100\text{ A})(0.10\text{ T})}{0.020\text{ m}}

\Delta p = \frac{IB}{a} = 0.50\ \text{kPa}

(2)

Check

Units: $\text{A}\cdot\text{T}/\text{m} = (\text{A}\cdot\text{T}\cdot\text{m})/\text{m}^2 = \text{N}/\text{m}^2$ , a pressure. Equivalently, $IB/a$ is the force density $jB$ , with $j = I/(a\,l)$ the current density for a section of length $l$ , times the length $l$ over which it acts: $(I/(al))\,B\,l = IB/a$ , as found.

Remark

The pressure, about 0.5 percent of an atmosphere here, rises in proportion to the current and the field and does not depend on the length of the active section. Such pumps, with no moving parts and nothing to seal, drive the liquid sodium coolant of fast reactors. Run backwards, with the metal driven through the field, the same arrangement becomes a generator, the principle of magnetohydrodynamic power conversion.