Past exam of the physics 2 paper of the University of Oxford / 2017 / 11

For a long air-cored solenoid, Ampère's law applied to a rectangular loop running inside and outside the winding gives $B\ell \simeq {\mu }_{0}NI$, with the external magnetic field negligible away from the ends. Hence $B={\mu }_{0}nI$ along the winding axis, where $n=N/\ell$. Each turn links magnetic flux $BA$ with $A=\pi a^2$, so the total magnetic flux linkage is $NBA=LI$. It follows thatHere $n=2000{\mathrm{m}}^{-1}$, $a=0.010\mathrm{m}$ and $\ell =0.50\mathrm{m}$, so $N=1000$ andThe value uses the stated long-solenoid approximation and ignores end corrections and winding resistance.

The original diagram places $R_2$ in parallel with the inductor; $R_1$ connects their common node to the source only when the switch is in position 2. Let $v(t)$ be their shared voltage and $i_L(t)$ the downward inductor electric current. Since $i_L(0^{-})=0$ and a finite voltage cannot change it instantaneously, the inductor initially behaves as an open branch. ThusAt long times, an ideal inductor carrying constant electric current has $v=L{\dot{i}}_L=0$, shorting $R_2$. Therefore $I_2(\infty )=0$ and $i_L(\infty )=V_0/R_1=10\mathrm{A}$.

For the full transient, Kirchhoff's current law gives $(V_0-v)/R_1=i_L+v/R_2$ with $v=L{\dot{i}}_L$. Equivalently, the source seen by the inductor has the Thévenin equivalentSolving $L{\dot{i}}_L+R_{\mathrm{th}}{i}_L=V_{\mathrm{th}}$ givesThe relevant resistance is the parallel combination with the ideal voltage source suppressed, not $R_1+R_2$ and not $R_1$ alone.

During switch-on, the Joule heating in $R_2$ has instantaneous power $P_2=R_2[V_0/(R_1+R_2){]}^2{e}^{-2t/\tau }$. Integrating the finite transient givesAlthough the source continues supplying power to $R_1$ indefinitely, $R_2$ is eventually shorted and its heat integral converges. It would be incorrect to identify this heat with the final stored magnetic energy, because the source and $R_1$ also participate during switch-on.

Opening the switch removes the source and $R_1$ but leaves a closed path through the inductor and $R_2$. Set the new time origin at switch-off. Inductor current continuity gives $i_L(0)=V_0/R_1=10\mathrm{A}$. Since the two parallel branches are now isolated, Kirchhoff's current law gives $I_2=-i_L$, and $L{\dot{i}}_L=-R_2{i}_L$. ThusAll the initially stored magnetic energy is ultimately converted to Joule heating in $R_2$:Direct integration of $R_2{i}_L^2(t)$ gives the same result. The initial shared voltage is $v(0^{+})=-R_2{i}_L(0)=-1000\mathrm{V}$; its reversal drives the return current. The ideal circuit permits this transient voltage; real switching would also be limited by stray capacitance, breakdown and other parasitic effects.