How can there will be current in purely inductive circuit? In a purely inductive circiut there are two emf's one is applied and other is induced 
applied emf and induced emf are equal and opposite, then how can there will be current in a purely inductive circuit.
induced emf is not a potential drop and it produce current in opposite direction of change in current.
im confused in understanding the concept, please help.
 A: 
applied emf and induced emf are equal and opposite, then how can there will be current in a purely inductive circuit.

This reasoning is not valid. Every circuit element has some sort of “induced emf” that is always equal and opposite to the applied voltage. A resistor makes a voltage proportional to the current through it. According to this reasoning the equal and opposite voltage should prevent the current through a resistor. This is not how circuit elements work. 
Circuit elements establish a relationship between the voltage across them and the current through them. For resistors the voltage is proportional to the current, for inductors the voltage is proportional to the change in current, and for capacitors the change in voltage is proportional to the current. In all cases, KVL requires that the voltage across the element be equal and opposite to the “applied” voltage, but that is irrelevant to the relationship between the voltage and the current within the circuit element. 
A: 
applied emf and induced emf are equal and opposite, then how can there
  will be current in a purely inductive circuit.

But the induced emf is non-zero only if the current through the inductor is changing, correct?
Stipulate that the applied voltage across an ideal inductor is constant, $V_L = V_{DC}$.  The induced emf is then constant and opposite to the applied voltage.
$$\mathcal{E} = -V_{DC}$$
Now, you're asking how there can be a current through the inductor when the above holds.  For a constant emf, the inductor current must be changing at a constant rate, correct?
$$\mathcal{E} = -L\frac{di_L}{dt}\Rightarrow i_L(t) = i_L(0) + \frac{V_{DC}}{L}t$$
In summary, it cannot be the case that (1) the applied emf and induced emf are equal and opposite and (2) the inductor current is zero always.
A: For an inductor $\mathcal E_{\rm back} = - L \frac {dI}{dt}$ ie the current must be changing if the inductor is producing a back emf.
The back emf is in opposition to whatever is causing the change in the current though the inductor.  
This means that the total emf in the circuit consisting of a cell with emf $\mathcal E$ and an inductor is the sum of the emf of the cell and the back emf due to the inductor.  
$\mathcal E + \mathcal E _{\rm back}=0$
The total emf in the circuit is zero but that does not mean that nothing is happening particularly remembering that the back emf due to the inductor is only there because the current is changing.  
Multiplying the equation by the current $I$ and rearranging the equation gives  
$\mathcal E\, I = - \mathcal E_{\rm back} \,I = L \dfrac{dI}{dt}I$ 
This equation is really a restatement of the law of conservation of energy.  
Power supplied by the cell = minus the power supplied by the inductor, 
or put another way,  
power supplied by the power supply = power absorbed by the inductor and stored in the magnetic field of the inductor.  
You can now do an integration to find the energy provided by the cell which is stored in the inductor.
Assume that the initial current through the inductor is zero.  
$\displaystyle \int_0^t \mathcal E \, I \ dt = \int _0^IL\, I \, dI= \frac 12 LI^2$ 
or solve the differential equation $\mathcal E = L \frac {dI}{dt}$ for the current $I$ assuming an initial current of zero which gives $I(t) = \frac {\mathcal E}{L} t$ 
It is the asymmetry between the two components in the cell and inductor circuit which is to be contrasted to the symmetry when connecting together two cells each of emf $\mathcal E$ so that the positive terminal of one cell is connected to the positive terminal of the other cell etc.  
In this case the emfs of the cells are there irrespective of what value the current is and whether the current is changing or not.  
Again the total emf in the circuit is zero but there is symmetry so there is no reason why one cell should be the source of electrical energy and the other the sink of electrical energy which results in the current in such a circuit being zero.  
A: Going along a path around the circuit, the potential differences must be given by
$$V_s+V_L=0$$
where $V_s$ is the potential of the source and $V_L$ is the potential across the inductor.
But we know for the inductor
$$V_L=-L\frac{\text d I}{\text d t}$$
So we have 
$$V_s=L\frac{\text d I}{\text d t}$$
If the source has the form $V_s(t)=V_0\sin(\omega t)$ then we can find, upon integration
$$I(t)=-\frac{V_0}{\omega L}\cos(\omega t)$$
So we see that we get an alternating current with our alternating potential.
