On-Axis B Field of a Finite Cylindrical Current Layer: 04-BS-9, December 2013, Q2

Question 2: On-Axis B Field of a Finite Cylindrical Current Layer (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.

Quantity	Value
Layer radius $a$	$0.5$ cm $=5\times10^{-3}$ m
Layer length $L$	$2$ cm $=0.02$ m
Total encircling current $I$	$1$ mA $=1\times10^{-3}$ A

Find. The magnetic flux density $B$ on the axis, at the midpoint of the layer.

Cylindrical current layer of radius a, length L, total current 1 mA. B evaluated on-axis at the midpoint.

Approach. Model the layer as a stack of circular current loops (Biot–Savart for a loop on its own axis), integrate along the axis over the layer length using the supplied integral aid, and evaluate the closed form at the centre.

Field of one elemental loop. A ring of current at axial position $z'$ (measured from the centre) contributes, at the centre ($z=0$): $$dB=\frac{\mu_0\,n I}{2}\cdot\frac{a^2}{(a^2+z'^2)^{3/2}}\,dz',\qquad n=\frac{N}{L}\ \text{(turns per unit length)}$$
Integrate over the layer using the aid. With $z'=au$, $dz'=a\,du$, the aid integral gives $\int (1+u^2)^{-3/2}du=u(1+u^2)^{-1/2}$, so $$\int_{-L/2}^{L/2}\frac{dz'}{(a^2+z'^2)^{3/2}}=\frac{1}{a^2}\left[\frac{u}{\sqrt{1+u^2}}\right]_{-L/2a}^{L/2a}=\frac{L}{a^2\sqrt{a^2+(L/2)^2}}$$ Carrying this through the prefactor collapses $nIL=NI=I_{\text{total}}$, giving the closed form $$B=\frac{\mu_0\,I_{\text{total}}\,L}{2a\sqrt{a^2+(L/2)^2}}=\frac{\mu_0\,I_{\text{total}}}{\sqrt{L^2+4a^2}}$$
Substitute the numbers. $$B=\frac{(4\pi\times10^{-7})(1\times10^{-3})}{\sqrt{(0.02)^2+4(0.005)^2}}=\frac{(4\pi\times10^{-7})(1\times10^{-3})}{0.02236}$$ $$B=\boxed{5.62\times10^{-8}\ \text{T}=56.2\ \text{nT}}$$

Quantity	Result
$B$ at centre, on-axis	$5.62\times10^{-8}$ T $=56.2$ nT

Concept & Theory

Topic. On-axis field of a finite solenoidal current layer, derived from the Biot–Savart law for a loop.

Key relations. Single-loop axial field $B=\mu_0 I a^2/2(a^2+z^2)^{3/2}$; finite-solenoid centre field $B=\mu_0 N I/\sqrt{L^2+4a^2}$; long-solenoid limit $B\to\mu_0 n I$ as $L\to\infty$.

Why this works. Each turn of the current layer is a circular loop whose own on-axis field is a standard Biot–Savart result; superposing (integrating) every turn along the layer's length gives the finite-solenoid result. The aid integral is exactly the piece needed to carry out that integration in closed form.

Common pitfall. Using the infinite-solenoid approximation $B=\mu_0 nI$ directly — here $a/L=0.25$ is not small, so the finite-length correction factor $L/\sqrt{L^2+4a^2}\approx0.894$ matters (it would overstate $B$ by about 12% if ignored).

Source. Sadiku, Elements of Electromagnetics, Ch. 7 (Biot–Savart Law, magnetic field of a solenoid).

Question 2 of 8: On-Axis B Field of a Finite Cylindrical Current Layer

Question 2: On-Axis B Field of a Finite Cylindrical Current Layer (20 marks)