Irodov 3.260: Pressure and Shear on a Current Sheet – Amit & Deepali Sinha Foundation

Problem Irodov 3.260

A flat conducting sheet carrying a current is placed in an external uniform magnetic field, after which the induction just on one side of the sheet is $\vect B_1$ and just on the other side $\vect B_2$ , with $B_1 > B_2$ . Seen edge on, with side 1 on the left, the three cases are: (a) $\vect B_1$ and $\vect B_2$ both parallel to the sheet and pointing the same way (up); (b) both parallel to the sheet, $\vect B_1$ up and $\vect B_2$ down; (c) both crossing the sheet obliquely, $\vect B_1$ arriving from the left at about $55^\circ$ from the normal and pointing upward, $\vect B_2$ leaving to the right at about $12^\circ$ from the normal and pointing slightly downward. In each case find the magnetic force per unit area of the sheet and the direction of the current in it.

The current-carrying plane edge on, with the field B_1 on its left and B_2 on its right, denser on the left: (a) both upward; (b) B_1 upward, B_2 downward; (c) both crossing the plane, the upward component reversed on the right. — The current-carrying plane edge on, with the field $\vect B_1$ on its left and $\vect B_2$ on its right, denser on the left: (a) both upward; (b) $\vect B_1$ upward, $\vect B_2$ downward; (c) both crossing the plane, the upward component reversed on the right.

Given

$\vect B_1$ on the left, $\vect B_2$ on the right, $B_1 > B_2$
(a) both up along the plane
(b) $\vect B_1$ up, $\vect B_2$ down
(c) oblique, at about $55^\circ$ and $12^\circ$ from the normal

Find

the force per unit area and the direction of the current, in each case

Solution

Step 1Components at the plane

Let $\vect n$ be the unit normal to the plane pointing from side 1 to side 2 (to the right in the figure), $\vect t$ the unit vector along the plane pointing up in the figure, and $\vect k = \vect n\times\vect t$ the one pointing out of the page. Since $\nabla\cdot\vect B = 0$ , the flux into a thin pillbox straddling the plane is zero, so the normal component is the same on both sides:

B_{1n} = B_{2n} \equiv B_n \tag{1}

It is zero in cases (a) and (b). The tangential components $B_{1t}$ and $B_{2t}$ , taken along $\vect t$ , may differ: that difference is the current.

Step 2The current from the jump in the tangential field

Apply Ampere’s law to a rectangle in the plane of the page, of length $l$ along the sheet, straddling it, traversed counterclockwise (up on side 2, down on side 1). The short sides contribute nothing as the rectangle is flattened, and the current crossing it is $i_k\,l$ , where $i_k$ is the component of the surface current density along $\vect k$ :

(B_{2t} - B_{1t})\,l = \mu_0\,i_k\,l, \qquad i_k = \frac{B_{2t} - B_{1t}}{\mu_0} \tag{2}

Reading the tangential components off the three cases:

\begin{gathered} \text{(a)}\quad B_{2t} = B_2 < B_1 = B_{1t}\\ \text{(b)}\quad B_{2t} = -B_2 < 0 < B_1 = B_{1t}\\ \text{(c)}\quad B_{2t} < 0 < B_{1t} \end{gathered}

So $B_{2t} < B_{1t}$ and $i_k < 0$ in all three:

The current flows into the page in all three cases.

Step 3The field that pushes the plane

The sheet’s own field is $\mu_0 i/2$ on each side, parallel to it and reversed across it, and it exerts no net force on the sheet itself. So $\vect B_1$ and $\vect B_2$ are the external field plus and minus that self-field, and the external field at the plane is their average:

\vect B_0 = \frac{\vect B_1 + \vect B_2}{2} = B_n\,\vect n + \frac{B_{1t} + B_{2t}}{2}\,\vect t \tag{3}

Step 4The force per unit area

A strip of unit width and unit length carries the current $i_k\vect k$ across $\vect B_0$ , so the force per unit area is the cross product below. The unit vectors multiply as $\vect k\times\vect n = \vect t$ and $\vect k\times\vect t = -\vect n$ , so with (3) and then (2):

\vect f = i_k\,\vect k\times\vect B_0 = \frac{B_{2t} - B_{1t}}{\mu_0}\left(B_n\,\vect t - \frac{B_{1t} + B_{2t}}{2}\,\vect n\right) \tag{4}

Step 5The three cases

The normal part of (4) is $(B_{1t}^2 - B_{2t}^2)/(2\mu_0)$ , and by (1) the equal normal components cancel in the difference of squares:

B_{1t}^2 - B_{2t}^2 = (B_1^2 - B_n^2) - (B_2^2 - B_n^2) = B_1^2 - B_2^2

In (a) and (b) the field has no normal component, $B_n = 0$ , and only this normal part is left:

\vect f = \frac{B_1^2 - B_2^2}{2\mu_0}\,\vect n \quad\text{(a), (b)}

(5)

Its magnitude is $(B_1^2 - B_2^2)/(2\mu_0)$ , normal to the plane and directed to the right, from the stronger field toward the weaker. In (c) $B_n > 0$ and $B_{1t} - B_{2t} > 0$ , so the tangential part of (4) survives as well:

Case (c): the current in the plane runs into the page; the force f per unit area on it has the normal part f_n, toward the weaker field, and the tangential part f_t, down along the plane. — Case (c): the current in the plane runs into the page; the force $\vect f$ per unit area on it has the normal part $f_n$ , toward the weaker field, and the tangential part $f_t$ , down along the plane.

\vect f = \frac{B_1^2 - B_2^2}{2\mu_0}\,\vect n - \frac{B_n(B_{1t} - B_{2t})}{\mu_0}\,\vect t \quad\text{(c)}

(6)

The normal part is again $(B_1^2 - B_2^2)/(2\mu_0)$ to the right; the tangential part, of magnitude $B_n(B_{1t} - B_{2t})/\mu_0$ , points down along the plane. For the angles of the drawing the tangential part is about 1.6 times the normal one, so the force points well below the normal, as the figure shows.

Check

The force on the sheet is also the jump of the Maxwell stress across it, $\vect f = (\mathsf T_2 - \mathsf T_1)\cdot\vect n$ , where $\mathsf T\cdot\vect n = B_n\vect B/\mu_0 - B^2\vect n/(2\mu_0)$ on each side. This gives $(B_1^2 - B_2^2)\vect n/(2\mu_0) + B_n(\vect B_2 - \vect B_1)/\mu_0$ , and since $\vect B_2 - \vect B_1 = (B_{2t} - B_{1t})\vect t$ it is (6) term by term (the form that reduces to (5) when $B_n = 0$ ). A force normal to the sheet alone would fail this test whenever $B_n \ne 0$ : the current, along $\vect k$ , crosses the normal field $B_n\vect n$ too, and $i_k\vect k\times B_n\vect n$ lies in the plane.

Remark

The normal part is a difference of magnetic pressures, $B^2/(2\mu_0)$ on each face, and it depends only on the magnitudes because the normal components, equal on both sides, cancel in it. The tangential part is a shear: field lines kinked at the sheet pull it along, like a stretched string pulling sideways on a peg it bends round. A sheet held in a field that crosses it obliquely must therefore be braced against sliding as well as against being pushed.