Question 5 of 8: On-Axis Field of an Atomic-Scale Current Loop
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams — December 2016 — 04-BS-9 Basic Electromagnetics. Three-hour, closed-book exam (approved Casio/Sharp calculator only). Aids given: $\varepsilon_0=8.85\times10^{-12}$ F/m, $\mu_0=4\pi\times10^{-7}$ H/m, $e=1.6\times10^{-19}$ C. Format: eight questions offered; any five constitute a complete paper and only the first five appearing in the answer book are marked. All eight are solved below for completeness.
Reference texts: Sadiku, Elements of Electromagnetics / Hayt & Buck, Engineering Electromagnetics — Coulomb's law and superposition, Gauss's law (integral and point form), Biot–Savart law and the field of straight conductors and current loops, Faraday's law and motional EMF, dipole fields, capacitance and dielectric breakdown; Young & Freedman, University Physics with Modern Physics — reflection geometry.
Question 5: On-Axis Field of an Atomic-Scale Current Loop (20 marks)
Find. The magnitude and direction of $\vec B$ at the field point.
Loop of radius $a$ carries current $I$ circulating clockwise viewed from above; the field point sits on the axis at height $z=a$.
Approach. Apply the standard on-axis field of a circular loop with $z=a$; determine the SIGN by the right-hand rule applied to the stated (clockwise-from-above) current sense.
Magnitude, on-axis loop field with $z=a$.
$$B=\frac{\mu_0Ia^2}{2(a^2+z^2)^{3/2}}=\frac{\mu_0Ia^2}{2(2a^2)^{3/2}}=\frac{\mu_0I}{4\sqrt2\,a}$$
$$B=\frac{(4\pi\times10^{-7})(0.01)}{4\sqrt2\,(10^{-10})}=\boxed{22.21\ \text{T}}$$
Direction by the right-hand rule. Curling the right-hand fingers in the current's sense as seen from above (clockwise) points the thumb DOWNWARD, i.e. the loop's magnetic moment $\vec m$ points down through the loop; the on-axis field is directed along $\vec m$ at every point on the axis (both above and below the loop, not just far away):
$$\boxed{\vec B=22.21\ \text{T, directed DOWNWARD (toward the loop), along the axis}}$$
Quantity
Result
Magnitude $B$
$22.21$ T
Direction
downward along the axis (toward the loop), set by the clockwise-from-above current
Check: a loop radius and field height of $10^{-10}$ m (atomic scale) with a macroscopic $10$ mA current gives an enormous field ($22.2$ T, stronger than any practical laboratory magnet); this is a direct, expected consequence of the given atomic-scale geometry (consistent with Questions 1, 2, and 7 on this same paper, which are also set at Ångstrom scale), not an arithmetic error.