Question 5 of 7: Faraday induction in a circular loop
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. 16-Elec-A7 Electromagnetics, December 2019 — 3 hours, closed book, approved calculator only. Seven questions of equal value (20 marks each); the paper states that five completed questions constitute a complete paper. All seven are solved here, because the set is intended as a study resource.
Reference texts (22-Elec-A7 Electromagnetics). F. T. Ulaby & U. Ravaioli, Fundamentals of Applied Electromagnetics, 8th ed.; M. N. O. Sadiku, Elements of Electromagnetics, 7th ed.; W. H. Hayt & J. A. Buck, Engineering Electromagnetics, 9th ed.; D. M. Pozar, Microwave Engineering, 4th ed.; C. A. Balanis, Antenna Theory: Analysis and Design, 4th ed.
Constants used throughout. $c = 3.00\times 10^{8}\ \text{m/s}$, $\varepsilon_0 = 8.854\times 10^{-12}\ \text{F/m}$, $\mu_0 = 4\pi\times 10^{-7}\ \text{H/m}$, $\eta_0 = 376.7\ \Omega$. Where a boundary case could turn on the choice between $c = 3\times 10^{8}$ and $c = 2.998\times 10^{8}$ m/s, both readings are stated.
Figure reading. In Figure 1(b) the maxima sit at z = 0, 1.5 and 3.0 m and the minima at z = 0.75 and 2.25 m, so the wavelength is 3.00 m. Three independent checks confirm that reading — the quarter-wave maximum-to-minimum spacing, the fact that the stated 3.75 m line is then exactly 1.25 wavelengths, and the fact that z = l then lands on a voltage minimum where the quarter-wave transformer result must give $Z_{\text{in}} = Z_0/S$.
Question 5: Faraday induction in a circular loop (20 marks)
Given. A rigid, stationary circular loop of radius 40 cm lies in the $x$–$y$ plane, immersed in a spatially uniform but exponentially growing magnetic flux density.
Given data
Quantity
Symbol
Value
Flux density
$\mathbf{B}(t)$
$0.2e^{5t}\,\hat{\mathbf{a}}_z\ \text{T}$
Loop radius
$\rho$
$0.40\ \text{m}$
Loop plane / normal
—
$x$–$y$ plane, normal $\hat{\mathbf{a}}_z$
Loop resistance
$R$
$10\ \Omega$
Alternative field (part e)
$\mathbf{B}(t)$
$0.2e^{5t}\,\hat{\mathbf{a}}_x\ \text{T}$
Find. The induced emf, the curl of E, the loop current with its physical direction, a sketch, and the emf when the field is re-oriented along $\hat{\mathbf{a}}_x$.
Approach. Compute the flux linkage, differentiate it for Faraday's law, use the differential form for the curl, divide by $R$ for the current, and settle the direction with Lenz's law.
Compute the loop area and the flux linkage. $$A = \pi\rho^{2} = \pi(0.40)^{2} = 0.5027\ \text{m}^{2}.$$ Because B is uniform and parallel to the loop normal, the surface integral is trivial: $$\Phi(t) = \int_S \mathbf{B}\cdot d\mathbf{S} = B_z(t)\,A = (0.2)(0.5027)e^{5t} = 0.1005e^{5t}\ \text{Wb}.$$
Apply Faraday's law. With the loop circulation taken in the right-hand sense about $+\hat{\mathbf{a}}_z$ (i.e. counter-clockwise seen from $+z$), $$\text{emf} = -\frac{d\Phi}{dt} = -(5)(0.1005)e^{5t} \;\Longrightarrow\; \boxed{\text{emf}(t) = -0.503\,e^{5t}\ \text{V}}$$ i.e. a magnitude of $0.503e^{5t}$ V, with the minus sign recording that it drives current the other way round the reference circulation.
Take the curl of E directly. The differential (point) form of Faraday's law needs no integration at all: $$\nabla\times\mathbf{E} = -\frac{\partial \mathbf{B}}{\partial t} = -(0.2)(5)e^{5t}\,\hat{\mathbf{a}}_z \;\Longrightarrow\; \boxed{\nabla\times\mathbf{E} = -1.00\,e^{5t}\, \hat{\mathbf{a}}_z\ \text{V/m}^{2}}$$ This is consistent with part (a) by Stokes' theorem: $\oint\mathbf{E}\cdot d\mathbf{l} = \int(\nabla\times\mathbf{E})\cdot d\mathbf{S} = (-1.00e^{5t})(0.5027) = -0.503e^{5t}$ V, exactly the emf found above.
Get the current from Ohm's law. The loop is a closed conductor of total resistance 10 Ω, so $$i(t) = \frac{\text{emf}(t)}{R} = \frac{-0.503e^{5t}}{10} \;\Longrightarrow\; \boxed{i(t) = -50.3\,e^{5t}\ \text{mA}}$$ relative to the counter-clockwise reference — that is, a current of magnitude $50.3e^{5t}$ mA circulating clockwise when viewed from the $+z$ direction.
Confirm the direction with Lenz's law. The upward flux is increasing, so the induced current must generate a downward ($-\hat{\mathbf{a}}_z$) magnetic moment inside the loop to oppose the change. By the right-hand rule that requires clockwise circulation seen from $+z$, which is exactly the sign obtained algebraically. The sketch below shows this.
Re-orient the field (part e). With $\mathbf{B} = 0.2e^{5t}\,\hat{\mathbf{a}}_x$, the field now lies in the plane of the loop, so it is everywhere perpendicular to the loop normal: $$\Phi = \int_S \mathbf{B}\cdot d\mathbf{S} = 0.2e^{5t}\,(\hat{\mathbf{a}}_x\cdot\hat{\mathbf{a}}_z)A = 0 \quad\text{for all } t,$$ and therefore $$\boxed{\text{emf}(t) = 0\ \text{V}}.$$ No flux threads the loop at any instant, so however fast the field grows, nothing is induced. (Note this is a statement about the net linkage: the local $\nabla\times\mathbf{E}$ is still non-zero, pointing along $-\hat{\mathbf{a}}_x$, but its flux through this particular surface vanishes.)
Part (d) — the loop viewed from +z. B points out of the page and is growing, so the induced current runs clockwise (seen from +z) to oppose it, per Lenz's law.
Check: modelling assumptions. The loop is taken as rigid and stationary and the field as spatially uniform, so the motional term $\oint(\mathbf{v}\times\mathbf{B})\cdot d\mathbf{l}$ vanishes identically and only the transformer term survives. Self-inductance is neglected, as the question's resistance-only data implies. The exponential growth $e^{5t}$ is of course unbounded; the results are the formal answer over whatever interval the model is asserted to hold.
