Why is the emf over an inductor equal to that of the battery? In a hypothetical circuit with only an inductor and a DC voltage source (no resistance), why is the voltage across the inductor the same as the source voltage?
I get that the charges coming from the battery need to lose their energy, so do so over the inductor, but surely when the voltage across the battery and inductor are equal, there is no net voltage acting on the current, and so no change in current and so no back emf?
Thanks :)
 A: An inductor is basically a passive element that could store the electrical energy in it's magnetic field. Ideally the inductor has no resistance even though it could offer a reactance to time varying currents. Now, in your hypothetical circuit, we can divide the situation into three stages:   
When you switch on the current, the current through the inductor rises from zero to a particular value. So, there is a change in the current through the inductor, which in turn, by its property of self-inductance will try to oppose that change by creating a back emf which appears across the inductor. Suppose the inductance of the coil is $L$. Then the applied emf  
$$V=L\frac{di}{dt}=-V_{ind}$$  
where $V_{ind}$ is the voltage induced in the coil. The induced emf is utilized to oppose the change in the current.
After some time, the current in the circuit becomes stable. Then the energy stored in the magnetic field of the inductor is completely utilized and is zero so that the induced emf across the inductor vanishes. This means  
$$V_{ind}=0\implies i=\textrm{constant}$$  
as expected. However, one could not expect $V=0$. It's because the emf applied is completely utilized for the flow of current through the circuit. If the circuit has no resistance, then a maximum current flows through the circuit. The inductor then behaves like a connecting wire. In such a case, one should not expect any voltage across the inductor. 
The last case is when you turn off the current. The current through the inductor falls from a non-zero value to zero. The inductor opposes this change in the circuit by converting the magnetic energy stored in the inductor into electrical energy to support the current flow. So there is a delay we can observe for the current to become zero.   
You can expect a voltage drop to appear across the inductor only when there is a change in the current in the circuit. When the current becomes stable, there is no voltage drop across the inductor. As a matter of conservation of energy, one may write, in general, at any instant $t$, the applied emf (assuming zero resistance in the circuit):  
$$V=L\frac{di}{dt}$$  
If the circuit has an effective resistance $R$, then the above equation becomes  
$$\underbrace{V}_{\textrm{applied emf}}=\underbrace{L\frac{di}{dt}}_{\textrm{voltage across inductor}}+\underbrace{iR}_{\textrm{voltage across resistor}}$$
A: An inductor is, under the simplest interpretation, just a coil of wire. The only time it has an effect on the circuit different than just a wire element is when there is changing current flow through it. If you connect a DC voltage source to an inductor, you will not have any changing current flow. Therefore, you are basically short circuiting your battery. In this case, we can no longer approximate the wire (including the wire of the inductor) as having zero resistance. 
The voltage of the charged particles flowing through the circuit will drop uniformly along the length of the circuit (assuming uniform wire composition). You would obviously have a very high amount of current since this is essentially a short circuit (but since there is no change, the inductor doesn't generate any magnetic fields, back EMF, etc).
A: The reason why the voltage across the source and the voltage across the load are the same is because of Kirchhoff’s voltage law. It doesn’t actually matter what type of source and what type of load it is. The voltages across the elements of any two-element circuit must always be equal. 
