Exercise 4.1.19: Rogowski coil

Let $s$ be the arc length along the coil's axis (the closed curve $C$). In the element $ds$ there are $nds$ turns, each of area $A$ small enough that $B$ is uniform over it. The turns are perpendicular to $C$, so each links the flux $B\cdot dS={\mu }_{0}H\cdot a_{t}A$ with $a_t$ the tangent of $C$. The total flux linkage is

by Ampère's law. Only the enclosed current enters: the shape of $C$ and the position of the conductor inside it are irrelevant.

${\mu }_{0}nA=4\pi \times 10^{-7}\times 2000\times 10^{-4}=0.251$ μH (the mutual inductance between conductor and coil). For $i=500\sqrt{2}\sin \omega t$, $(di/dt{)}_{\max }=\omega \sqrt{2}{I}_{\mathrm{rms}}=314\times 707=2.22\times 10^5$ A/s, hence

An integrator restores a signal proportional to $i(t)$ itself.

A conductor outside $C$ contributes to $H$ at every point of the coil, but ${\oint }_{C}H\cdot dl=0$ for it: its contributions cancel around the loop, and the coil does not see it.

Units: ${\mu }_{0}nAi$ is (H/m)(1/m)(m${}^2$)(A) $=$ H A $=$ Wb. The coil contains no iron, so it neither saturates nor retains magnetisation: the same coil reads $10$ A and a $100$ kA fault current. The return lead is brought back through the middle of the winding so that the loop as a whole encloses no area of its own and does not pick up the flux through the big loop. A clamp meter is the same principle with a split iron core and a Hall sensor (Application 4.1.3).