106MT2Pr-solutions

Question 1

(a)

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.

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

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

The circuit with Voltage set to zero

The second step is to calculate the equivalent resistance $R_{0}$ by reducing the voltage to zero and calculating the resistance between a and b

$R_{0}={\frac {(20)(30)}{20+30}}\Omega +{\frac {(10)(90)}{10+90}}\Omega =21{\textrm {Ohms}}$

The Thévenin equivalent circuit is then a 30V source connected in series with a 21 Ohms resistor.

Thévenin equivalent circuit

So the current at t=0 seconds is simply

$I=V/R={\frac {30{\textrm {V}}}{21\Omega }}=1.43{\textrm {A}}$

(b)

V(t)

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.