Past exam of the physics 2 paper of the University of Oxford / 2019 / 8

For a circular coil, use the Biot-Savart lawAt an axial position $x$, every wire element is at distance $r=\sqrt{a^2+x^2}$. The transverse magnetic field contributions cancel by rotational symmetry. The axial part of $d\mathbf{\ell }\times \mathbf{r}$ is $ad\ell$, and $\oint d\ell =2\pi a$. Multiplying by the $N$ turns givesThe electric current's sense fixes the axial direction by the right-hand rule.

Put the two coil centres at $z=\pm a/2$. By superposition, their axial magnetic field isBoth magnetic field contributions point the same way. Evaluation givesThe corresponding coefficients are $0.715542$ and $0.676777$. Each individual magnetic field falls as one moves away from its own coil; their sum is much flatter near the midpoint.

For a single coil, differentiating $(a^2+x^2{)}^{-3/2}$ twice produces $3(4x^2-a^2)(a^2+x^2{)}^{-7/2}$. At $x=a/2$ this vanishes. Thus the pair has $B^{\prime }(0)=0$ by reflection symmetry and $B^{\prime \prime }(0)=0$ by its chosen separation; all odd derivatives vanish there. In fact,This Helmholtz coil arrangement produces a nearly uniform central magnetic field, useful for controlled charged particle in a uniform magnetic field and magnetic field calibration. The small orbit used below lies within that central region.

The electron follows a semicircle under the transverse Lorentz force. Parallel, oppositely directed incoming and outgoing paths are separated by a diameter, so the orbit radius is $r=1.00\mathrm{cm}$. Acceleration from rest through the electric potential difference $V$ gives kinetic energy $m_e{v}^2/2=eV$, where $e>0$ is the elementary electric charge. Equating the Lorentz force magnitude to the required centripetal force givesThe magnetic field must be perpendicular to the plane of the turn, with its sign chosen for the electron's negative electric charge. In the illustration the electron enters rightwards at the upper endpoint, so a magnetic field into the page bends it downwards.

The nonrelativistic estimate has $v\simeq 0.198c$. If the small special relativity correction is retained, use $pc=\sqrt{K(K+2m_e{c}^2)}$ with $K=eV$ and $B=p/(er)$. This gives $B\simeq 3.389\times 10^{-2}\mathrm{T}$, about $0.49\%$ larger. The leading answer is adequate at the precision implicit in the classical treatment.