Axial Resistance, Current Density and Power of a Solid Cylindrical Resistor: 17-Phys-A3 Electromagnetics, Undated paper, Q6

Question 6: Axial Resistance, Current Density and Power of a Solid Cylindrical Resistor (22 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. A solid cylindrical conductor of radius $a=1.25$ mm, length $L=6.5$ mm, conductivity $\sigma=60.2$ S/m, with flat circular end faces at $z=0$ (terminal A) and $z=L$ (terminal B) held as equipotentials — i.e. an ordinary axial-current wire resistor, current flowing uniformly along $\hat a_z$ between the two end caps.

Find. $R_{AB}$; the current density $\mathbf J$ and total current $I$; the dissipated power $P$.

Approach. With the end faces as equipotentials and no $\rho$- or $\phi$-dependence, $V(z)$ is linear (1-D Laplace equation), giving a uniform axial $\mathbf J=\sigma\mathbf E$; then $R=L/(\sigma A)$, and $\mathbf J,I,P$ scale with whatever potential difference $V_{AB}$ is actually applied across the terminals (not stated numerically in the problem), so parts (b)/(c) are reported per volt of $V_{AB}$.

Part (a) — Resistance. Cross-sectional area $A=\pi a^2=\pi(1.25\times10^{-3})^2=4.909\times10^{-6}$ m². $$R_{AB}=\frac{L}{\sigma A}=\frac{6.5\times10^{-3}}{(60.2)(4.909\times10^{-6})}=\boxed{22.0\ \Omega.}$$
Part (b) — Current density and total current. With uniform axial field $E_z=V_{AB}/L$, $$\boxed{\mathbf J=\sigma\frac{V_{AB}}{L}\,\hat a_z=(9262\ \text{S/m}^2)\,V_{AB}\,\hat a_z\ \text{A/m}^2}$$ (uniform over the whole cross-section $0\le\rho\le1.25$ mm), and $$\boxed{I=JA=\frac{V_{AB}}{R_{AB}}=(0.04546\ \text{S})\,V_{AB}\ \text{A}}$$ (Ohm's law, as it must be, since $J$ was built from $E=V_{AB}/L$ in the first place).
Part (c) — Power dissipated (Joule's law). $$P=\int\mathbf J\cdot\mathbf E\,dV = I^2R_{AB}=\boxed{\frac{V_{AB}^2}{R_{AB}}=(0.04546\ \text{W/V}^2)\,V_{AB}^2\ \text{W}.}$$

Final results
Quantity	Value
Resistance, $R_{AB}$	22.0 Ω
Current density, $\mathbf J$	$9262\,V_{AB}\,\hat a_z$ A/m²
Total current, $I$	$V_{AB}/22.0$ A
Power, $P$	$V_{AB}^2/22.0$ W

Check: the printed problem gives the resistor's geometry and conductivity but never states a numeric applied voltage or current across A–B, so parts (b)/(c) are reported as exact coefficients times the (unstated) potential difference $V_{AB}$ — substituting any assumed $V_{AB}$ (e.g. 1 V) recovers a single number immediately.

Topic: DC resistance and current density of a uniform conducting rod (axial-current resistor).

Key relations: $R=L/(\sigma A)$; $\mathbf J=\sigma\mathbf E$; $I=JA=V/R$; $P=I^2R=V^2/R$.

Why this works: With both end faces held as equipotentials and no angular or radial dependence imposed on the geometry, the potential can only vary linearly along $z$ (the 1-D Laplace solution), which makes $\mathbf E$, hence $\mathbf J$, exactly uniform over the whole cross-section — the same idealisation behind the everyday $R=\rho_{res}L/A$ wire formula ($\rho_{res}=1/\sigma$ here).

Common pitfall: Mistaking this axial geometry (current flows along the cylinder's LENGTH between flat end caps) for a RADIAL/coaxial resistor (current flowing between inner and outer curved surfaces), which would need the very different $R=\ln(b/a)/(2\pi\sigma L)$ form — the two are easy to conflate from the figure's oblique 3-D sketch.

Source: Sadiku, Elements of Electromagnetics, Ch. 5 (resistance, current density, Joule's law).

Question 6 of 10: Axial Resistance, Current Density and Power of a Solid Cylindrical Resistor

Question 6: Axial Resistance, Current Density and Power of a Solid Cylindrical Resistor (22 marks)