Question 6 of 8: Magnetic Field Above a Circulating Electron
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Examinations, May 2018 — 16-Elec-A7, Electromagnetics. Closed book, three hours, approved Casio or Sharp calculator only. Eight questions, all of equal value; the paper states that any five constitute a complete paper, so each question carries 20 of the 100 marks that are actually graded. All eight are solved here, because this set is a study resource rather than an examination script.
Aids given on the paper. $\varepsilon_0 = 8.85\times10^{-12}\ \text{F/m}$, $\mu_0 = 4\pi\times10^{-7}\ \text{H/m}$, $e = 1.6\times10^{-19}\ \text{C}$. Those two constants fix $c = 1/\sqrt{\mu_0\varepsilon_0} = 2.9986\times10^{8}\ \text{m/s}$ and $\eta_0 = \sqrt{\mu_0/\varepsilon_0} = 376.82\ \Omega$ — which is why Question 1 quotes a 377 Ω line: it is free space itself, rendered as a cable.
Reference texts. D. M. Pozar, Microwave Engineering, 4th ed. (transmission-line transients, stubs, waveguide modes); M. N. O. Sadiku, Elements of Electromagnetics, 7th ed. (Biot–Savart, motional EMF, plane waves); W. H. Hayt and J. A. Buck, Engineering Electromagnetics, 9th ed.; F. T. Ulaby and U. Ravaioli, Fundamentals of Applied Electromagnetics, 8th ed.; C. A. Balanis, Antenna Theory: Analysis and Design, 4th ed. (short-element fields and the reactive near-field boundary).
Constants used throughout: the paper's own aid list is taken as authoritative, so $c = 2.9986\times10^{8}$ m/s and $\eta_0 = 376.82\ \Omega$. Question 1 supplies its own propagation velocity ($3\times10^{8}$ m/s) and that stated value is used there. Question 5 is the one place where the difference between $2.9986\times10^{8}$ and $3.00\times10^{8}$ changes the wording of the answer, and both readings are worked through explicitly.
Question 6: Magnetic Field Above a Circulating Electron (20 marks)
Given. A single electron orbiting fast enough that it is indistinguishable from a steady current loop.
Given data
Quantity
Symbol
Value
Charge
$q$
$-1.6\times10^{-19}$ C
Orbit radius
$R$
$10^{-10}$ m (100 pm)
Circulation frequency
$f$
$2\times10^{15}$ Hz
Sense, viewed from above
—
clockwise
Field point
$h$
$10^{-10}$ m above the centre, on the axis
Find. The magnitude and direction of $\mathbf{B}$ at that axial point.
The orbit and the axial field point. Because the carrier is negative, the conventional current runs opposite to the electron, which puts B along the upward axis.
Approach. Replace the orbiting charge by the equivalent steady current $I = |q|f$, fix the current direction from the sign of the carrier, apply the on-axis loop formula, and finally justify the quasi-static treatment by comparing the orbit with a wavelength.
Convert the orbiting charge into a current. A single charge passing a fixed point on the orbit $f$ times per second constitutes a current $$I = |q|\,f = (1.6\times10^{-19})(2\times10^{15}) = 3.2\times10^{-4}\ \text{A} = 0.32\ \text{mA}.$$ Over any observation time long compared with the 0.5 fs orbital period this is an ordinary steady loop current.
Fix the direction of that current. The carrier is negative, so conventional current flows opposite to the electron's motion: the electron runs clockwise seen from above, therefore the current runs anticlockwise seen from above. Getting this backwards inverts the final answer, and it is the only place the sign of the charge enters.
Apply the on-axis loop field. For a circular loop of radius $R$ carrying $I$, at height $h$ on its axis, $$B = \frac{\mu_0 I R^{2}}{2\,(R^{2}+h^{2})^{3/2}} .$$ With $R = h = 10^{-10}$ m the denominator is $2(2\times10^{-20})^{3/2} = 5.657\times10^{-30}$.
Substitute. $$B = \frac{(4\pi\times10^{-7})(3.2\times10^{-4})(10^{-20})}{5.657\times10^{-30}} \;\Rightarrow\; \boxed{B = 0.711\ \text{T}} .$$ The direction follows the right-hand rule applied to the anticlockwise conventional current: $\mathbf{B}$ points vertically upward, along the axis away from the orbit plane.
Sanity-check the magnitude. Three quarters of a tesla from one electron looks alarming until the scale is noticed: the field point is 0.1 nm away, so $\mu_0 I/(2R)$ alone is 2.0 T at the centre and the $(1+h^2/R^2)^{3/2} = 2.83$ factor reduces it to 0.711 T. Independently, the loop's magnetic moment is $m = I\pi R^{2} = 1.005\times10^{-23}$ A·m$^{2}$, within about 8 % of the Bohr magneton $9.27\times10^{-24}$ A·m$^{2}$ — exactly what an atomic-scale orbit should give, so the model is behaving.
Justify the static formula at $2\times10^{15}$ Hz. The wavelength at that frequency is $\lambda = c/f = 150$ nm, while the field point is 0.1 nm away, so $$kh = \frac{2\pi h}{\lambda} = 4.2\times10^{-3} \ll 1 .$$ Retardation and radiation are utterly negligible over that separation, so Biot–Savart applies unmodified and the answer above is the field, not merely a low-frequency approximation to it.
The question asks for a single value of $\mathbf{B}$, so the time-averaged (loop) field is what is wanted. Strictly, a single point charge produces a field that also pulses at $2\times10^{15}$ Hz and its harmonics as the electron sweeps round; the average over one orbit is exactly the loop result quoted, and the pulsation is unobservable on any realistic instrument.
