Past exam of the physics course of the year 1 of the natural sciences course of the University of Cambridge / 2022 / Section B / 10

For an infinite straight wire in vacuum, Ampère's circuital law around a circle of radius $R$ gives $B(2\pi R)={\mu }_{0}I$. ThereforeHere ${\mu }_0$ is the vacuum permeability and the azimuthal direction follows the right-hand rule around the electric current. The magnetic field is measured in $\mathrm{T}$.

Choose the electric current counterclockwise as viewed from $+z$; reversing it reverses the magnetic field. Parameterize the circular wire byThe cross product in the Biot-Savart law isConsequently the transverse terms integrate to zero over $0\le \varphi \le 2\pi$, while the axial term is constant:ThusThe direction is $+z$ on both halves of the axis for this electric current orientation, including $z<0$; it is not “away from the plane” on both sides. The right-hand rule gives the same direction.

Read the original diagram as a clockwise arc carrying electric current from its upper endpoint to its lower endpoint, with tangent semi-infinite leads extending to the right. Their crossing is insulated, so there is no extra junction. Choose $+z$ out of the page. At the centre the arc produces a negative axial magnetic field, and each straight lead a positive one, by the right-hand rule.

For the arc, every element is at distance $R$, with $d\mathbf{l}$ perpendicular to its displacement to the centre. The Biot-Savart law therefore givesFor each straight lead, write $s\ge 0$ for distance along the tangent measured from its endpoint. Its perpendicular distance from the centre is $R$, soThe integral equals $1/R$ because an antiderivative is $s/[R\sqrt{R^2+s^2}]$. Adding both leads and the arc,Using the full infinite-wire magnetic field for either lead would incorrectly double its contribution.

To calculate magnetic-field cancellation of an arc and tangent leads away from the plane, put $a^2=R^2+h^2$. Each point on the arc is distance $a$ from $(0,0,h)$, and the axial component of $d\mathbf{l}\times \mathbf{s}$ still has magnitude $R^{2}d\varphi$. Thus its axial magnetic field isFor either tangent lead the axial numerator remains $Rds$, while the cubed distance is $(R^2+h^2+s^2{)}^{3/2}$. HenceAdding both leads givesFor $ϵ=h/R\ll 1$, a binomial series yieldsThe exact condition for cancellation of the axial component isFor example, keeping $\theta =2$ would leave a small positive axial magnetic field proportional to $h^2/R^2$. Both contributions weaken with height, but the arc factor is $(1+{ϵ}^2{)}^{-3/2}$ whereas the lead factor is $(1+{ϵ}^2{)}^{-1}$. The arc therefore loses a greater fraction of its axial magnetic field. Increasing its angle compensates for this loss. Physically, the arc is all at the same finite radius, whereas distant elements of the semi-infinite leads undergo a smaller fractional change in distance on moving a little out of the plane.

The cancellation here concerns $B_z$ only; off the plane it does not assert that every component of the magnetic field vanishes. The small-height result also stays within the tangent-lead geometry shown in the original paper.