What is the value of the opposing voltage created by the inductor?
Surely it can't be L di/dt because then the opposing voltage and
source voltage would be equal thus making current to be zero, right?

$$\boxed{\text{It's like trying to solve this: }\dfrac{0}{0}}$$
When you apply a steady supply voltage (for example) across a perfect inductor, the back emf equals the applied voltage and, it remains equal to that applied voltage for the length of time that the applied voltage is connected.
On the face of it that appears to prevent current flow into the inductor (which is what I think you are alluding to). This of course appears problematic hence, I believe, this is the reason for your question.
For instance, how can current flow into an inductor when applied voltage and back-emf are equal; we know from the inductor formula that for an applied fixed voltage, V, the rate of change of current is: -
$$\dfrac{di}{dt} = \dfrac{V}{L}$$
But, this seems to be at odds with the back emf exactly equalling the applied emf. But, consider what is happening; the back emf and the applied voltage are both constant values and, they are across a zero ohm impedance hence, the current cannot be defined other than saying: -
$$\boxed{\text{The current is }\dfrac{0}{0}}$$
Why did I say that the impedance is zero? Answer: it's an inductor and, the back-emf is induced in series with that inductor hence with the applied voltage and back-emf voltage being equal, there is 0 volts across the &quot;true&quot; inductance and hence no spectral content across that inductor and, it therefore has to be represented by 0 ohms.
In other words, you cannot use the value of applied voltage and back-emf to calculate the current through the inductor. The current rising or falling delivers the back-emf and that's it.

What if the applied voltage is AC - well the difference in voltage between applied and back-emf is still zero and it appears across an impedance of zero ohms so: -
$$\boxed{\text{The current is still }\dfrac{0}{0}}$$