Mutual Inductance of Two Coaxial Loops of Very Different Size: 04-BS-9, May 2015, Q7

Question 7: Mutual Inductance of Two Coaxial Loops of Very Different Size (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
Large loop radius $b$	$5$ cm $=0.05$ m
Small loop radius $a$	$2$ mm $=0.002$ m
Axial separation $d$	$5$ cm $=0.05$ m

Find. The mutual inductance $M$ between the two loops.

Two coaxial loops, radii b (large) and a (small, a << b), planes 5 cm apart along the shared axis.

Approach. Since $a\ll b$ and $a\ll d$, treat the field the large loop produces at the small loop's location (on-axis, distance $d$) as uniform over the small loop's tiny area, exactly as in the small-loop-in-a-locally-uniform-field approximation used elsewhere in this subject.

On-axis field of the large loop, at the small loop's location (per unit current). $$B_{\text{axis}}(d)=\frac{\mu_0 b^2}{2(b^2+d^2)^{3/2}}=\frac{(4\pi\times10^{-7})(0.05)^2}{2\left[(0.05)^2+(0.05)^2\right]^{3/2}}$$ $$B_{\text{axis}}=\boxed{4.443\times10^{-6}\ \text{T per A}}$$
Flux through the small loop and mutual inductance. $$M=B_{\text{axis}}\cdot\pi a^2=(4.443\times10^{-6})\,\pi(0.002)^2$$ $$M=\boxed{5.583\times10^{-11}\ \text{H}=55.8\ \text{pH}}$$

Quantity	Result
$B_{\text{axis}}$ (per unit current)	$4.443\times10^{-6}$ T/A
$M$	$5.583\times10^{-11}$ H (55.8 pH)

Concept & Theory

Topic. Mutual inductance between coaxial loops of very different radius, using the small-loop approximation.

Key relations. On-axis field of a loop, $B(z)=\mu_0Ib^2/2(b^2+z^2)^{3/2}$; $M=\Phi_{\text{small}}/I$.

Why this works. Because $a\ll b,d$, the large loop's field is essentially uniform across the small loop's area, so the flux is simply that field times the small loop's area — avoiding the elliptic-integral mutual-inductance formula required for loops of comparable size. Reciprocity ($M_{12}=M_{21}$) makes it far simpler to compute the small loop's flux due to the large loop's field than the reverse.

Common pitfall. Using the centre-field formula ($z=0$) instead of the on-axis expression at $z=d=5$ cm, which would overstate $M$ by roughly a factor of $2\sqrt2$ here.

Source. Sadiku, Elements of Electromagnetics, Ch. 9 (Mutual inductance of coupled circuits).

Question 7 of 8: Mutual Inductance of Two Coaxial Loops of Very Different Size

Question 7: Mutual Inductance of Two Coaxial Loops of Very Different Size (20 marks)