106MT2Pr-solutions

Question 1

The circuit with the capacitor removed

The first step in the Thévenin equivalent circuit approach is to remove the capacitor, and calculate the open circuit voltage between a and b.

$I_{1}={\frac {100{\textrm {V}}}{20+30{\textrm {\Omega }}}}=2{\textrm {A}}$ $I_{2}={\frac {100{\textrm {V}}}{10+90{\textrm {\Omega }}}}=1{\textrm {A}}$

$V=20I_{1}-10I_{2}=(20{\textrm {\Omega }})(2{\textrm {A}})-(10{\textrm {\Omega }})(1{\textrm {A}})=30V$

Question 2

(a) The normal component of ${\vec {B}}$ must be zero.

(b) To find the field outside the superconductor, we place an "image dipole" inside the superconductor a depth $d$ below the surface, oriented at an angle ($\pi -\theta$) from vertical. This automatically satisfies the boundary condition. The field outside the superconductor is the field due to the original dipole and the image dipole.

(c) Taking the $y$-axis normal to $\mu$, parallel to the superconducting sheet, and into the page:

${\vec {\tau }}=-{\frac {\mu ^{2}\sin 2\theta }{64\pi \epsilon _{0}c^{2}d^{3}}}{\hat {y}}$

(d) Stable equilibria: $\theta =\pi /2$, $\theta =3\pi /2$. Unstable equilibria: $\theta =0$, $\theta =\pi$

(e) $F=-{\frac {3\mu ^{2}\sin 2\theta }{64\pi \epsilon _{0}c^{2}d^{4}}}(3+\cos 2\theta )$ away from the superconductor.