Irodov 3.164: Current Lines Bend Like Field Lines – Amit & Deepali Sinha Foundation

Problem Irodov 3.164

A steady direct current crosses the boundary between two conducting media of conductivities $\sigma_1$ and $\sigma_2$ . Let $\alpha_1$ and $\alpha_2$ be the angles that a current line makes with the normal to the boundary in medium 1 and in medium 2. Show that the current lines refract according to $\tan\alpha_2/\tan\alpha_1 = \sigma_2/\sigma_1$ .

The boundary between the media of conductivities σ_1 (above) and σ_2 (below), its normal, and a current line arriving in medium 1 at the angle α_1 to the normal. — The boundary between the media of conductivities $\sigma_1$ (above) and $\sigma_2$ (below), its normal, and a current line arriving in medium $1$ at the angle $\alpha_1$ to the normal.

Given

conductivities $\sigma_1$ , $\sigma_2$
steady current
angles $\alpha_1$ , $\alpha_2$ to the normal

Find

the law

\tan\alpha_2/\tan\alpha_1 = \sigma_2/\sigma_1

Solution

Step 1What the steady state keeps continuous

Two conditions hold at the boundary. The current is steady, so no charge accumulates on the boundary and the current flowing into a flat pillbox straddling it equals the current flowing out: the normal component of $\vect j$ is continuous. The field of direct current is conservative, so its line integral round a thin rectangle straddling the boundary vanishes: the tangential component of $\vect E$ is continuous.

\begin{gathered} j_{1n} = j_{2n}\\ E_{1t} = E_{2t} \end{gathered} \tag{1}

Step 2The tangential current from the field

Ohm’s law in local form, $\vect j = \sigma\vect E$ , holds in each medium. Applied to the tangential components and combined with the second condition of (1):

\frac{j_{1t}}{\sigma_1} = \frac{j_{2t}}{\sigma_2} \tag{2}

Step 3The angles of the current lines

The current line refracted at the boundary, drawn for σ_2 = 2σ_1 and α_1 = 30°: it leaves at α_2 = 49°, further from the normal, in the better conductor. — The current line refracted at the boundary, drawn for $\sigma_2 = 2\sigma_1$ and $\alpha_1 = 30^\circ$ : it leaves at $\alpha_2 = 49^\circ$ , further from the normal, in the better conductor.

A current line runs along $\vect j$ , so the tangent of its angle to the normal is the ratio of tangential to normal components in each medium:

\begin{gathered}\tan\alpha_1 = \frac{j_{1t}}{j_{1n}}\\ \tan\alpha_2 = \frac{j_{2t}}{j_{2n}}\end{gathered}

Dividing the two, the normal components cancel by the first condition of (1):

\frac{\tan\alpha_2}{\tan\alpha_1} = \frac{j_{2t}}{j_{1t}}\cdot\frac{j_{1n}}{j_{2n}} = \frac{j_{2t}}{j_{1t}}

By (2) the ratio of the tangential components is the ratio of the conductivities, which gives the law to be demonstrated:

\frac{\tan\alpha_2}{\tan\alpha_1} = \frac{\sigma_2}{\sigma_1}

(3)

Check

Equal media, $\sigma_1 = \sigma_2$ , give $\alpha_2 = \alpha_1$ : the line goes straight through. A line along the normal, $\alpha_1 = 0$ , stays along it, $\alpha_2 = 0$ . For a nearly insulating medium 1, $\sigma_1 \to 0$ with $\alpha_2$ finite, $\tan\alpha_1 \to 0$ : in a poor conductor next to a good one the current lines meet the boundary almost along the normal, as field lines meet the surface of a metal.

Remark

The law has exactly the form of the refraction of $\vect E$ lines at a dielectric boundary, with $\varepsilon_2/\varepsilon_1$ in place of $\sigma_2/\sigma_1$ , because steady currents and electrostatics share their boundary conditions. This is why electrostatic fields can be modelled in an electrolytic tank: electrodes in a uniform conducting medium carry the same potential pattern as electrodes of the same shape in a uniform dielectric.