Past exam of the physics course of the year 1 of the natural sciences course of the University of Cambridge / 2024 / Paper 1 / Section B / 10

In magnetostatics, Ampère's circuital law isIt relates magnetic circulation around a closed curve to the steady current through a spanning surface. It is especially useful when symmetry makes the tangential field constant on a chosen path: an infinite straight wire gives $B={\mu }_{0}I/(2\pi r)$, while an ideal long solenoid gives approximately $B={\mu }_{0}nI$ inside.

The Biot-Savart law gives the field contribution from the actual current geometry:The cross product determines the direction by the right-hand rule. Integrating around a circular loop of radius $a$ gives $B={\mu }_{0}I/(2a)$ at its centre. A finite wire segment can also be integrated directly, even though it lacks enough symmetry for an Ampère-loop shortcut. The full steady-current circuit includes a return path; a segment calculation specifies that segment's contribution.

Place the wire on the $y$ axis, its endpoints at $-a$ and $b$, and the observation point at $(-d,0)$. With current along $+y$, $d{\mathbf{\ell }}^{\prime }\times \mathbf{r}$ points out of the page. Its magnitude givesIn the question's angle convention, $\cos \beta =b/\sqrt{d^2+b^2}$ and $\cos \alpha =-a/\sqrt{d^2+a^2}$. ThereforeThe minus sign in $\cos \alpha$ matters because $\alpha$ is the exterior obtuse angle.

For a regular hexagon of side $a$, the perpendicular centre-to-side distance is $d=\sqrt{3}a/2$. Each side subtends endpoint angles with $\cos \beta =1/2$ and $\cos \alpha =-1/2$. Its centre field is consequently ${\mu }_{0}I/(4\pi d)$.

All six contributions have the same normal direction, soThe direction follows the current's right-hand rule. Equal side lengths alone would not fix an arbitrary distorted hexagon's field; the usual regular-coil geometry is used here.

The regular hexagon has area $A=6(\sqrt{3}{a}^2/4)=3\sqrt{3}{a}^2/2$. Choose $t=0$ with its positive normal parallel to the imposed field. The external magnetic flux is $\Phi =B_{0}A\cos \omega t$, so Faraday's law of induction givesTaking the stated resistance to determine the loop response,A different initial orientation adds a phase. The sign is with respect to the chosen loop orientation and obeys Lenz's law; reversal of that orientation reverses the signed current. This expression neglects self-inductance, as in the resistive-loop model.