Rotating a superconducting loop

By the definition $V=L \frac {di}{dt}$ , where $V$ is the voltage induced across the circuit, $L$ is the inductance (Henries), and $\frac {di}{dt}$ (the time rate of change of current with respect to time), we know how to relate one of the quantities we do know $L$ to current $i$ . However, let's get rid of voltage.

By Faraday's law of induction, we can relate an electromotive force (or voltage) $E$ to the time rate of change of electric flux $\Phi$ in the following way $E=-\frac {d\Phi}{dt}$ .

Because the coil here is a superconductor, there is no resistance. Thus, $V=E$ . And then we can set these two equations equal to each other.

$L \frac {di}{dt}=\frac {d\Phi}{dt}$ (the negative sign can be eliminated)

Eliminate $dt$ .

$L di = d\Phi$

Then we can integrate, pulling out our inductance $L$ because it remains constant.

$L i = \Phi$ and $i=\frac {\Phi}{L}$ follows.

Returning to the diagram, we see that initially the flux $\Phi = B \cdot A$ was 0, because there are no field lines passing through the closed loop circuit. Or mathematically, $B \cdot A = 0$ , where $A$ is the area vector directed perpendicular to the loop.

In the second case, $B \cdot A \neq 0$ , and actually $B \cdot A =$ , after a little geometry, $0.01 \times 0.01 \times \cos(60)$

$i = \frac {\Phi}{L} \Rightarrow 50 A$

We have dϕ = L x dI where ϕ = magnetic flux

                                          and  I = current

now initially ϕ = I = 0

finally ϕ = B . A

         = BA cosθ          where  θ = 60

so dϕ = 5 x 10^(-5) Weber

so dI = dϕ/L

   =  { 5 x 10^(-5)} / {10^(-6)} Amp

   =  50 Amp.

which is final current as initial current was zero.

The emf is given by V=d(flux) /dt, and the self inductance is given by V=L dI/dt. Equating the two we get d(flux) =L dI The total change in flux is ABsin30 = 5x10E-5 Tm^2. Hence I=(5x10E-5)/(1E-6) = 50A

Since the loop is superconducting $R=0$ we have that $\mathcal{E}=I R=0.$ On other hand, from Faraday's Law we know that $\mathcal{E}=-\frac{d\Phi}{dt}.$
Therefore the magnetic flux through the loop cannot change i.e. $\frac{d\Phi}{dt}=0$ . In other words, $\Phi_{i}= \Phi_{f}.$ Since initially, the plane of the loop is parallel to magnetic field we have that $\Phi_{i}=0$ When the loop is rotated, the magnetic flux is $\Phi_{f}= L I_{f}+ B A \cos(90^{\circ}-30^{\circ})$ Setting the final flux equal to the initial flux we obtain
$|I_{f}|=\frac{1}{2} \frac{BA}{L}= 50 A.$ Note that the negative sign of the current indicates its direction which can be also determined from Lenz's Law.