Past exam of the physics B course of the University of Cambridge / 2022 / Paper 1 / Section C

The Poynting vector is the local electromagnetic energy flux density:Here $\mathbf{E}$ is the electric field and $\mathbf{H}$ the magnetic field intensity. For an oriented area element $d\mathbf{S}$, $\mathbf{N}\cdot d\mathbf{S}$ is the electromagnetic power crossing that element in its positive normal direction. Thus ${\int }_S\mathbf{N}\cdot d\mathbf{S}$ is outward power through a closed surface, in watts. The Poynting vector has units $\mathrm{W}{\mathrm{m}}^{-2}$, and its direction describes local electromagnetic energy transport.

The unit relative permittivity and vacuum-valued magnetic permeability imply $\mathbf{D}={ϵ}_0\mathbf{E}$ and $\mathbf{B}={\mu }_0\mathbf{H}$. The electromagnetic energy per unit volume isThe two contributions are the stored electric and magnetic energies. Although called $U$ in this paper, this quantity is an energy density; the total electromagnetic energy is ${\int }_{V}UdV$.

Use the divergence identity for a cross product, followed by Faraday's law of induction and the Ampère-Maxwell equation:With the electromagnetic energy density of part (b), these giveIntegrate over a fixed volume $V$ and apply the divergence theorem, choosing $d\mathbf{S}$ outward:This is Poynting's theorem. The outward electromagnetic energy flux equals minus the power transferred to matter and minus the rate at which electromagnetic energy accumulates inside. For the dissipative electrical conductor in this question, the matter term is Ohmic heating.

Take the electric current along $+z$, and use cylindrical coordinates. A uniform current in a straight homogeneous wire hasThe local Ohm's law $\mathbf{J}=\sigma \mathbf{E}$ gives the longitudinal electric fieldFor a circular Amperian loop of radius $r\le a$, the enclosed electric current is $I_{\mathrm{enc}}=I{r}^2/a^2$. The magnetostatic Ampère law gives $2\pi r{H}_{\varphi }=I_{\mathrm{enc}}$, henceSince $\hat{\mathbf{z}}\times \hat{\mathbf{\varphi }}=-\hat{\mathbf{r}}$, the Poynting vector isIt is zero at the centre and points radially inward elsewhere. Its units are $\mathrm{W}{\mathrm{m}}^{-2}$. These are the interior fields in the long-wire approximation, neglecting end and contact effects.

The Ohmic heating per unit volume is $\mathbf{E}\cdot \mathbf{J}=J^2/\sigma$, uniform in this wire. Multiplying by its volume gives the dissipated powerwhere the electrical resistance is $R_{\mathrm{wire}}=L/(\sigma \pi a^2)$. The result is in watts; over a time interval $\Delta t$ the dissipated energy is $P_{\mathrm{loss}}\Delta t$.

On the curved wire surface, the outward normal is $\hat{\mathbf{r}}$ and the Poynting vector has the constant radial value $N_r(a)=-I^2/(2\sigma {\pi }^2{a}^3)$. Its outward flux isThe Poynting vector is radial, so it has zero normal component on the two end faces in this approximation. This is therefore also the outward flux through the complete wire surface. Equivalently, positive electromagnetic power $P_{\mathrm{loss}}$ flows inward.

The fields are steady, so the stored electromagnetic energy does not change. Poynting's theorem consequently requires the incoming power to equal the Ohmic heating in part (e), exactly as found. The energy reaches the wire through the surrounding electromagnetic field. A local check is $\nabla \cdot \mathbf{N}=r^{-1}{\partial }_r(r{N}_r)=-I^2/(\sigma {\pi }^2{a}^4)=-\mathbf{E}\cdot \mathbf{J}$.

Let the unit normal $\mathbf{n}$ point from material 1 to material 2, and let their relative magnetic permeabilities be ${\mu }_1,{\mu }_2$, so ${\mathbf{B}}_i={\mu }_0{\mu }_i{\mathbf{H}}_i$ for linear isotropic media. Apply Gauss's law for magnetism to an infinitesimal Gaussian pillbox straddling the interface. The side contribution vanishes as its height tends to zero, leavingThe normal magnetic flux density is continuous.

Next use a narrow rectangular Amperian loop crossing the interface. The two short-side contributions vanish in the zero-height limit. The magnetostatic Ampère law for magnetic field intensity giveswhere ${\mathbf{K}}_f$ is any free surface current density. Thus, if there is no free surface current density on the interface, the tangential magnetic field intensity is continuous. In that case the component relations areThe distinction between free and bound electric currents matters: bound currents due to magnetization are already incorporated in the definition of magnetic field intensity. A change of magnetic permeability alone does not make the tangential magnetic flux density continuous.

The starting point is Gauss's law for magnetism, $\nabla \cdot \mathbf{B}=0$: a flux tube carries the same magnetic flux through successive transverse sections. To identify that tube approximately with the physical core, one must also make a negligible-leakage approximation. The large core magnetic permeability makes the core route much less reluctant than competing paths through the surrounding air; the small gap further reduces the incentive for a long external return path. The gentle taper, $s\gg R-r$, lets the field follow the changing section with only small transverse corrections.

For two cuts and the intervening lateral surface, the divergence theorem gives ${\Phi }_2-{\Phi }_1=-{\Phi }_{\mathrm{side}}$. Neglecting lateral leakage givesThe normal magnetic flux density is continuous at each core–air interface by part (a), so this same magnetic flux crosses the gap. To use the gap's geometrical area $\pi r^2$ in the later formulas, the magnetic fringing must also be negligible, normally requiring $h\ll r$. This is the small-gap interpretation of the paper's approximation. Large magnetic permeability alone does not eliminate leakage or magnetic fringing for arbitrary gap geometry.

For the closed untapered torus, apply the magnetostatic Ampère law to a circular path through the core. It links all $N$ turns, each carrying electric current $I$:Since $L\gg R$, variation of the path length and of the tangential magnetic field intensity across the section is negligible. Thus $H_{\mathrm{in}}L\simeq NI$, giving, within this thin-torus approximation,The units are $\mathrm{A}{\mathrm{m}}^{-1}$. The core magnetic flux density is $B_{\mathrm{in}}={\mu }_0\mu NI/L$ in the linear material model.

With uniform section and negligible magnetic fringing, constant magnetic flux gives equal magnetic flux densities in the core and gap:Here $L$ is the core arc length and $h$ the additional gap length. Applying the magnetostatic Ampère law around the complete magnetic circuit,Consequently,This tends to $NI/h$ when the gap dominates the magnetic reluctance, $\mu h\gg L$, and reduces to the no-gap result as $h\to 0$. The gap magnetic flux density is $B_{\mathrm{gap}}={\mu }_0{H}_{\mathrm{gap}}$.

Let the approximately conserved magnetic flux be $\Phi$. ThenThe length $L$ shown in the original diagram is the untapered core arc; the two tapers are separate lengths $s$ each. In the untapered core, flux conservation givesFor one taper use distance $u$ from its narrow end. The straight taper in the diagram has local radiusand magnetic field intensityIts contribution to the line integral in the magnetostatic Ampère law is found explicitly:Adding the gap, the untapered arc and both tapers givesThereforeThis uses a linear radius profile, approximately uniform magnetic flux density over each section, a linear constant magnetic permeability, and negligible leakage and magnetic fringing. For a non-linear taper profile $a(u)$ the honest generalization isThus an arbitrary smooth taper with the same end radii and length need not give the printed formula; the straight-sided geometry supplies the additional information.

Set $\alpha =r/R<1$. Comparing with the untapered expression in part (d), using the same $N,I,\mu ,h,L$ and large radius $R$,The stated limit $s/L\ll \alpha$ makes $2s\alpha \ll L{\alpha }^2$, soThe taper concentrates magnetic flux into a smaller pole-face area, increasing the gap magnetic field intensity and magnetic flux density for the same winding electric current. If $\mu h\ll L{\alpha }^2$, the enhancement approaches $1/{\alpha }^2=(R/r{)}^2$. If instead $\mu h\gg L$, the gap dominates the magnetic reluctance and the enhancement approaches unity: both fields approach $NI/h$.

The benefit is therefore strongest when the untapered core's contribution is significant, while the taper lengths contribute little. The same-core-length convention is the comparison defined by parts (d) and (e); if one instead fixes the full iron path length, the untapered comparator has length $L+2s$, a difference negligible in the stated limit. A smaller gap area can increase the field even though the total magnetic flux need not increase. Magnetic saturation and magnetic fringing, excluded from this model, limit the ideal concentration.