Faraday's Law or Kirchoff's Loop rule? 
$$\mathbf{B}=\begin{cases}
B_{0}t ;t \in [0,T]\\
B_{0}T ;t>T
\end{cases}$$
Assume negligible resistance for the connecting wire and other necessary constants.
Should I use Faraday's Law or Kirchoff's Voltage Law? I tried both of them end up with different answers. Also I assumed the area of the loop to be A in the case of Faraday's law. And what's the justification of the method? I think we should use Faraday's here as there is a changing magnetic flux.
 A: Kirchoff's Voltage Law is just a restatement that the path integral of $\vec{E}$ around any closed loop is equal to zero.  This is because the voltage drop across any element in a circuit is just the path integral from one side of the element to the other.
In situations where there is a changing magnetic field, however, we have
$$
\oint \vec{E} \cdot d\vec{s} = - \frac{d \Phi}{dt}.
$$
(This can be proven using Stokes' theorem and the differential form of Faraday's Law.)  Thus, Kirchoff's Law cannot be expected to hold in such situations.
A: This is an interesting problem.  I believe all you need to do is calculate the additional emf produced by the B field and add it as a second source in the loop (paying attendtion to sign).  As i see it, the loop already has a voltage source, ε, which is a battery (based on the symbol).  So, the answer would be to use both.
Use Faraday to calculate the emf induced by the B field.  Then use kirchoff to find the I that flows as a result of these 2 voltage sources.
A: Now since here there is a changing magnetic flux, application of Faraday's law would be advised. And also intuitively makes sense to use this law as when there is a change in $\phi_{B}$, an emf is produced in the loop that may increase or decrease the current flowing depending on the direction of magnetic field.
Now using faraday's law,
$$∮E⃗ ⋅ds⃗ =−\frac{dΦ}{dt}$$
Going clockwise in the loop. Hence taking the area vector from right hand thumb rule, i.e. into the plane.
The integral evaluates to:
$-ε+iR_{1}+iR_{2} ---(1)$
$\phi_{B}=\mathbf{B}\cdot\mathbf{A}=-B_{0}At ; t \in [0,T]$
$\frac{dΦ}{dt}=-B_{0}A\ ; \ t \in [0,T] ---(2)$
$(1)=(2)$
$-ε+iR_{1}+iR_{2}=-(-B_{0}A)$
Rearranging:
$$i=\frac{B_{0}A+ε}{R_{1}+R_{2}}$$
A: There are two independent sources of emf in this circuit, one being the time-changing B field, the other the battery.  Use simple superposition.  Careful with the direction of the current induced by the B field.
I am new on this site and need time to adapt to the differences between LaTex and MathJax to write any formulas.
