Displacement Current from the Circulation of H: 22-Elec-A7 Electromagnetics, December 2013, Q6

Question 6: Displacement Current from the Circulation of H (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. The average tangential magnetic field around the rim of a 1 m square horizontal contour in vacuum is known, and Ampère's law in vacuum links it to the displacement current threading the square.

Given data
Quantity	Symbol	Value
Square side	$s$	1 m
Contour length	$L$	4 m
Enclosed area	$A$	1 m$^{2}$
Average tangential magnetic field	$\langle H_t\rangle$	$10^{-4}$ A/m rms
Frequency	$f$	1 MHz
Medium	—	vacuum (no conduction current)

Find. The rms spatial-average of $\partial E_z/\partial t$ over the square.

Figure 6.1 — The horizontal square contour. H circulates around the rim; the vertical displacement current threads the enclosed area.

Approach. Integrate the supplied curl equation over the square and apply Stokes' theorem, turning the local curl into a circulation of H around the rim and a flux of $\epsilon_0\,\partial\mathbf{E}/\partial t$ through the area.

Integral form of the supplied law. Integrating $\nabla\times\mathbf{H} = \epsilon_0\,\partial\mathbf{E}/\partial t$ over the square and applying Stokes' theorem,$$\oint_{C}\mathbf{H}\cdot d\boldsymbol{\ell} = \epsilon_0\int_{A}\frac{\partial \mathbf{E}}{\partial t}\cdot d\mathbf{A}.$$For a horizontal contour the area element is vertical, so only the vertical component $E_z$ contributes — exactly the quantity the question asks for.
Left-hand side: circulation of H. The stated figure is the spatial average of the tangential component along the rim, so the line integral is that average times the perimeter:$$\oint_{C}\mathbf{H}\cdot d\boldsymbol{\ell} = \langle H_t\rangle\,L = (10^{-4})(4) = 4.00\times10^{-4}\ \text{A (rms)}.$$This is the total enclosed current — here entirely displacement current, since vacuum carries no conduction current.
Right-hand side: displacement current. Likewise the surface integral is the spatial average of the integrand times the area,$$\epsilon_0\left\langle\frac{\partial E_z}{\partial t}\right\rangle A = (8.85\times10^{-12})\left\langle\frac{\partial E_z}{\partial t}\right\rangle (1.00).$$
Solve for the required average. Equating the two sides,$$\left\langle\frac{\partial E_z}{\partial t}\right\rangle = \frac{\langle H_t\rangle L}{\epsilon_0 A} = \frac{4.00\times10^{-4}}{(8.85\times10^{-12})(1.00)},$$$$\boxed{\left\langle \partial E_z/\partial t \right\rangle = 4.52\times10^{7}\ \text{V}\,\text{m}^{-1}\text{s}^{-1}\ \text{(rms)}}$$
Consistency check using the frequency. The 1 MHz figure is not needed for the answer, but it lets the result be sanity checked: for a sinusoidal field $\langle\partial E_z/\partial t\rangle = \omega\langle E_z\rangle$, so$$\langle E_z\rangle = \frac{4.52\times10^{7}}{2\pi\times10^{6}} = 7.19\ \text{V/m (rms)}.$$A field of a few volts per metre alongside a tenth of a milliamp per metre of H is entirely reasonable for a near-field region at 1 MHz (their ratio, about $7\times10^{4}\ \Omega$, is far above $\eta_0 = 377\ \Omega$, which simply says the 1 m loop is a tiny fraction of the 300 m wavelength and the region is capacitive rather than radiating).

Final Results
Quantity	Symbol	Value
Circulation of H (enclosed current)	$\oint\mathbf{H}\cdot d\ell$	$4.00\times10^{-4}$ A rms
Enclosed area	$A$	1.00 m$^{2}$
Average vertical field derivative	$\langle\partial E_z/\partial t\rangle$	$4.52\times10^{7}$ V m$^{-1}$ s$^{-1}$ rms
Implied vertical field at 1 MHz	$\langle E_z\rangle$	7.19 V/m rms

Topic. Ampère–Maxwell law in integral form; displacement current.

Key relations. $\nabla\times\mathbf{H} = \mathbf{J} + \epsilon_0\,\partial\mathbf{E}/\partial t$; Stokes' theorem $\oint_C\mathbf{H}\cdot d\boldsymbol{\ell} = \int_A(\nabla\times\mathbf{H})\cdot d\mathbf{A}$; for a sinusoid $\partial E/\partial t \to j\omega E$.

Why this works. Maxwell's addition of the displacement-current term means a changing electric field magnetises its surroundings exactly as a real current would. Stokes' theorem converts the local statement into a bookkeeping identity between something measured on a rim and something happening over the enclosed surface, so a magnetic measurement around a loop is a direct measurement of the total current — conduction plus displacement — through it. Because the contour here is horizontal its normal is vertical, which is why only the vertical component of E appears.

Common pitfall. Using the area instead of the perimeter on the left (or vice versa) is the commonest error, and it changes the answer by a factor of four here. It is also tempting to divide by $\omega$ somewhere: the question asks for the derivative itself, so the frequency belongs only in the optional cross-check.

Source. Sadiku, Elements of Electromagnetics, 7th ed., §9.4 (displacement current) and §9.5 (Maxwell's equations in integral form); Hayt & Buck, 9th ed., §10.4.

Question 6 of 8: Displacement Current from the Circulation of H

Question 6: Displacement Current from the Circulation of H (20 marks)