A superconducting ring in a magnetic field

Initially, the current in the ring is zero. After swith off the field, the flux in the area of the ring will not change, because the ring is a superconductor. The flux in the area of the ring before the field be switched off is:

$\phi_0=B\cdot S_{ring}=B\cdot \pi a^2$

And $\phi_0=\phi_{self}=LI$ , so:

$I=\dfrac{B\cdot \pi a^2}{L}$ $I=\dfrac{0.5\cdot 10^{-3}\cdot \pi\cdot (50\cdot 10^{-3})^2}{4\cdot\pi\cdot 10^{-7}\cdot50\cdot10^{-3}\cdot\left(\ln\left(\frac{8\cdot50\cdot10^{-3}}{1\cdot10^{-3}}\right)-2\right)}$ $I=15.65\text{ A}$

The same Flux through the ring produced by the external field will remain inside the ring after removing the external field in condition T<Tc(critical Temp.)

a) Compute the total magnetic flux through the ring or [0.5 (1 E-3)]* [Pi r r] in this case 0.5 (1 E-3) Pi 0.05*0.05 Weber ,which according to the formula will equate as follows

[0.5 (1 E-3)][Pi 0.05 0.05]=[4 Pi (1 E-7) (0.05) (ln((80/1000)/(1/1000))-2)] [ I ]

Without going into arithmetic details I(current)=15.6584 Amp.

We have:

$\Phi=LI$ (1)

Differentiate both side of equation (1):

$\Rightarrow -d\Phi=LdI$

On the other hand, $\Phi=B.\pi.a^2$

$\Rightarrow -dB.\pi.a^2=LdI$

$\Rightarrow -\int_{0}^{B}dB.\pi.a^2=\int_{I}^{0}LdI$

$\Rightarrow B.\pi.a^2=LI$

$\Rightarrow I=\frac {B.\pi.a^2}{L}$

$\Rightarrow I=\frac {B.\pi.a}{\mu _0(ln(\frac {8a}{b})-2)}$

$\Rightarrow I\approx 15.658$

Since the resistivity is 0,there is no voltage drop across any segment of the wire when current is flowing through it.Thus the voltage remains constant throughout the wire and net emf is thus 0.now when the magnetic field is switched off, the magnetic field induction and the magnetic flux drop to zero in time dt. Thus the induced emf i.e change in flux can be represented as (area)*(dB/dt).let the rate of change of current now be equal to di/dt. Magnetic emf created by the self induction is (L(di/dt)).both the Emfs should cancel each other out.By writing di=i(final)-i(initial),we get the final current in the ring.

Ldi/dt=-(pi) a^2dB/dt Thus,Ldi=-(pi) a^2dB integrating both sides with lower limit of B&i as (.5*10^-3)T, 0A,resp. and upper one as 0T, iAresp. thus i=15.7A

The ring becomes superconducting as we cool it down. This implies that the induced E.M.F $\mathcal{E}=I R=0.$ On other hand, we know from Faraday's Law that $\mathcal{E}=-\frac{d\Phi}{dt}.$
Therefore the magnetic flux through the ring cannot change i.e. $\frac{d\Phi}{dt}=0$ . In other words, $\Phi_{i}= \Phi_{f}.$ The initial magnetic flux is given by $\Phi_{i}= B A= B \pi a^{2}$ whereas the final flux comes from the current in the ring and it's given by $\Phi_{f}= L I.$ Equating the fluxes, we obtain $I= \frac{B \pi a^{2}}{L}= \frac{\pi a B}{ \mu_{0} (\ln(\frac{8a}{b}-2))}=15.65 A.$