Imagine a Loop of perfectly conducting wire with a Resistor with resistance R. The north pole of a magnet goes straight towards the loop and thus the magnetic flux through the Loop changes with time which induces an emf. My question is: If for example the rate of change of the exterior Flux is constant 10 Webers/sec (assming that this flux goes perpendicular to the surface) and the Resistance of resistor in loop is 5 Ohms. Would the current produced be 2 Amps? Or would we also need to consider the induced emf that tries to oppose the exterior change in flux? (which would then yield a smaller net flux and thus a smaller current through the Loop)
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Or would we also need to consider the induced emf that tries to oppose the exterior change in flux?
Correct, you would have to consider this. The tendency of the loop to resist a change in its own current, because changing current induces a changing magnetic flux which induces a back emf, is known as the loop's inductance, denoted by $L$. The differential equation governing the current in the loop is then:
$$I = \frac{1}{R} \left[ \mathcal{E} - L \frac{\mathrm{d}I}{\mathrm{d}t} \right]$$
where $\mathcal{E}$ is the external emf due to the changing flux from the external magnet.
The value of $L$ is a function of the loop's geometry; a loop containing many tightly packed coils, for instance, would be much more efficient at generating a magnetic field that has a flux through itself, than a simple circular loop.
