Question 9 of 10: On-axis field and self-flux of a circular current loop

Question 9: On-axis field and self-flux of a circular current loop

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 single circular turn of radius 3 m lying in the $xy$-plane, carrying 1 A in the $+\mathbf{a}_\phi$ sense.

Find. $\mathbf{H}(z)$ on the axis and its value at $z = 0.5$ m; and the flux linking the loop's own area under a uniform-field assumption.

[Figure not reproduced: Figure 5 (redrawn). A 1 A anticlockwise loop of radius 3 m; on its axis H is purely axial and points along +a_z. See the official exam paper.]

Approach. Derive the on-axis field from the Biot–Savart law, using symmetry to cancel the transverse contributions, then multiply the loop-centre flux density by the loop area for part (b).

1. Set up the Biot–Savart integral. For a current element on the loop at azimuth $\phi'$, $$d\vec{H} = \frac{I\,d\vec{l}\times\hat{a}_R}{4\pi R^2}, \qquad d\vec{l} = b\,d\phi'\,\hat{a}_\phi, \qquad \vec{R} = z\hat{a}_z - b\hat{a}_\rho$$ so $R = \sqrt{b^2 + z^2}$ is the same for every element — the field point lies on the axis, so every element is equidistant. That is what makes the integral elementary.
2. Use symmetry to kill the transverse part. The cross product $\hat{a}_\phi\times(z\hat{a}_z - b\hat{a}_\rho)$ has one component along $+\hat{a}_\rho$ (proportional to $z$) and one along $+\hat{a}_z$ (proportional to $b$). Integrating around the full turn, the radial parts of diametrically opposite elements point in opposite directions and cancel exactly, so only the axial part survives: $$dH_z = \frac{I b\,d\phi'}{4\pi R^2}\cdot\frac{b}{R} = \frac{I b^2\,d\phi'}{4\pi (b^2 + z^2)^{3/2}}$$
3. (a) Integrate over the turn. The integrand is independent of $\phi'$, so the integral just multiplies by $2\pi$: $$\boxed{\vec{H}(0,0,z) = \frac{I b^2}{2\left(b^2 + z^2\right)^{3/2}}\hat{a}_z\ \text{A/m}}$$ The direction is $+\hat{a}_z$ for current in $+\hat{a}_\phi$, which is the right-hand rule. Two limits check the form: at $z = 0$ it reduces to the familiar loop-centre value $I/2b = 0.1667$ A/m, and for $z \gg b$ it becomes $Ib^2/2z^3$, the dipole law.
4. Evaluate at $z = 0.5$ m. With $b^2 = 9$ and $b^2 + z^2 = 9.25$ m$^2$, $(9.25)^{3/2} = 28.133$: $$H_z = \frac{(1)(9)}{2(28.133)} = 0.15996\ \text{A/m} \;\Rightarrow\; \boxed{\vec{H} = 0.1600\,\hat{a}_z\ \text{A/m}, \quad \vec{B} = \mu_0\vec{H} = 0.2010\,\hat{a}_z\ \mu\text{T}}$$ The field has dropped only 4 % from its centre value, because 0.5 m is a sixth of the loop radius — still well inside the near-uniform region. A direct numerical Biot–Savart integration around the loop returns the same 0.15996 A/m and confirms the $+\hat{a}_z$ sense.
5. (b) Flux through the loop's own area. The loop area lies in the plane $z = 0$, so the field to use for the stated uniform-field assumption is the centre value $$B_0 = \mu_0\frac{I}{2b} = \frac{4\pi\times10^{-7}}{2(3)} = 0.20944\ \mu\text{T}$$ and with $A = \pi b^2 = \pi(9) = 28.274$ m$^2$, $$\boxed{\Phi = B_0 A = (2.0944\times10^{-7})(28.274) = 5.922\ \mu\text{Wb}}$$ Because the loop carries 1 A, this doubles as an estimate of its self-inductance, $L \approx \Phi/I = 5.92\ \mu$H.
6. Comment on the assumption. The uniform-field idealisation is what makes part (b) a one-line calculation, but the true field is not constant over the disc: it grows without bound as the wire is approached, which is why the exact self-inductance of a circular loop needs a wire radius and a logarithm. If instead the field at the part-(a) point $z = 0.5$ m is used, $\Phi = (2.0100\times10^{-7})(28.274) = 5.683\ \mu$Wb — 4 % lower, which brackets the sensitivity of the estimate.