= Barkhausen pulse area
{c}
{title2=$\int e\,dt=-\Delta\lambda$}
Write the <magnetic flux linkage> of the stationary <pickup coil> as $\lambda=\sum_{n=1}^N\Phi_n$, in <SI> units of <weber (unit)>. If every turn links the same <magnetic flux> $\Phi$, then $\lambda=N\Phi$; the turn count is dimensionless. With a fixed winding and terminal convention, <Faraday's law of induction> gives the induced <electromotive force>
$$
e(t)=-\frac{d\lambda}{dt},\qquad
\int_{t_a}^{t_b}e(t)\,dt=-[\lambda(t_b)-\lambda(t_a)].
$$
This is the <fundamental theorem of calculus> applied to <magnetic flux linkage>. Its units are $\mathrm{V\,s}=\mathrm{Wb}$, not <volts> alone. Reversing the leads reverses the sign. In a high-input-<electrical resistance> voltage measurement the input approximately follows this <electromotive force>; coil loading and the <frequency response> of the electronics require separate calibration.
Separate direct drive linkage from material linkage: $\lambda=\lambda_{\rm drive}+\lambda_{\rm material}$. Even an empty <pickup coil> has $-d\lambda_{\rm drive}/dt$. In an ideal long, uniformly coupled specimen with negligible <demagnetizing field> correction and <eddy current> shielding, a useful material approximation is
$$
\lambda_{\rm material}=\mu_0 N A_s M,\qquad
\int e_{\rm material}\,dt=-\mu_0 N A_s\Delta M.
$$
Here $A_s$ is specimen cross-sectional <area>, not the larger coil aperture; $\mu_0$ is <vacuum permeability> and $M$ the axial <magnetization> sampled by the coil. The units check directly: $(\mathrm{V\,s/(A\,m)})(\mathrm{m^2})(\mathrm{A/m})=\mathrm{V\,s}$. Finite geometry and nonuniform <magnetization> generally require a calibrated coupling in place of $\mu_0NA_s$. The exact linkage identity survives that failure of the uniform approximation.
As an original illustrative calculation, stipulate $N=100$, $A_s=10^{-6}\,\mathrm{m^2}$, $\Delta M=10^3\,\mathrm{A/m}$ and rounded $\mu_0=4\pi\times10^{-7}\,\mathrm{V\,s/(A\,m)}$. Then
$$
\Delta\lambda=4\pi\times10^{-8}\,\mathrm{Wb}
\approx0.125664\,\mu\mathrm{Wb},\qquad
\int e_{\rm material}\,dt\approx-0.125664\,\mathrm{mV\,ms}.
$$
A uniform change over $0.10\,\mathrm{ms}$ would give a rectangular pulse of $-1.25664\,\mathrm{mV}$; over $0.20\,\mathrm{ms}$ it gives $-0.628319\,\mathrm{mV}$ with the same area. Thus pulse height mixes change size with change speed. These numbers describe an ideal model, not observed pulses or material constants.
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