Irodov 3.259: Field Lines Push Like a Gas – Amit & Deepali Sinha Foundation

Problem Irodov 3.259

Two parallel planes carry surface currents that produce a uniform magnetic field of induction $B$ in the space between them and no field anywhere outside it. Find the magnetic force on each plane per unit of its area.

The two current-carrying planes edge on, with the uniform field B in the space between them and no field outside. — The two current-carrying planes edge on, with the uniform field $\vect B$ in the space between them and no field outside.

Given

uniform $B$ between the planes
$B = 0$ outside

Find

the force per unit area on each plane

Solution

Step 1The currents in the planes

Take a rectangular loop of length $l$ along one plane, straddling it, with one long side in the gap and the other outside. The field is $B$ along the inside side and zero along the outside one, and the short sides contribute nothing, because $\vect B$ is parallel to the plane and so perpendicular to them; so Ampere’s law gives the current per unit width $i$ of the plane:

B\,l = \mu_0\,i\,l, \qquad i = \frac{B}{\mu_0} \tag{1}

Both planes carry this current density. The field vanishes outside and doubles between them only if the two currents are opposite; the right-hand rule then fixes their senses relative to $\vect B$ .

The surface currents i and - i that make the field: each sheet gives μ_0 i/2 on both sides, the two add between the sheets and cancel outside. — The surface currents $\vect i$ and $-\vect i$ that make the field: each sheet gives $\mu_0 i/2$ on both sides, the two add between the sheets and cancel outside.

Step 2The field that acts on one plane

A uniform current sheet makes the field $\mu_0 i/2$ on each side of itself, parallel to the sheet and reversed across it. By symmetry its own field exerts no net force on it: the pushes on its two faces are equal and opposite. A plane is therefore pushed only by the field of the other plane, which at its position is, with (1),

B' = \frac{\mu_0 i}{2} = \frac{B}{2} \tag{2}

Step 3The force per unit area

The force per unit area on each sheet: f on one and - f on the other, both pointing out of the field region, so the planes repel. — The force per unit area on each sheet: $\vect f$ on one and $-\vect f$ on the other, both pointing out of the field region, so the planes repel.

The field $B'$ lies in the plane, perpendicular to the current, so a strip of the plane of unit width and unit length, carrying the current $i$ , feels the force $iB'$ . Per unit area, with (1) and (2):

f = i\,B' = \frac{B}{\mu_0}\cdot\frac{B}{2}

f = \frac{B^2}{2\mu_0}

(3)

The currents in the two planes are opposite, so the planes repel: each is pushed away from the field region, outward, normal to itself.

Check

Units: $\text{T}^2/(\text{H/m})$ is $\text{J/m}^3$ , an energy density, which is the same thing as $\text{N/m}^2$ , a pressure. Energy: hold the flux between the planes fixed, as superconducting planes would, and widen the gap $h$ by $dx$ . The flux per unit length, $\Phi = Bh$ , stays the same, so the field energy per unit area of plane, $(B^2/2\mu_0)h = \Phi^2/(2\mu_0 h)$ , falls by $\Phi^2\,dx/(2\mu_0 h^2) = (B^2/2\mu_0)\,dx$ , exactly the work done by the pressure (3).

Remark

The force per unit area equals the energy density of the field, $B^2/(2\mu_0)$ : a magnetic field pushes on the currents that confine it like a gas on the walls of its vessel. At $B = 1$ T this pressure is about $4\times10^5$ Pa, four atmospheres, which is why the windings of strong electromagnets must be mechanically braced.