# Which way does current flow if induced by a magnetic filed if two current are produced in opposite direction simultaneously?

I am studying how electric charge can be made to flow due to a change in the magnetic flux through a conductor, I understand if the magnetic flux changes there is an induced EMF in such a way to oppose the motion that caused it (Lenz's Law) however this confuses me as if this now induces a current then surely the force that is now created by this current (EMF) now creates another current that opposes the original current. So my question basically is which induced emf is dominant in this system?

I have a limited understanding of this myself but I think in a lot of cases the induced back EMF of the resulting current is assumed to be negligible compared to the original induced EMF.

If you really want to take into account the induced current opposing the original current then I believe you would have to account for the self-inductance of the conductor. The self-inductance tells you how the conductor opposes a change in the current flowing through it, which in this case is caused by a change in magnetic flux through the conductor, by the conductor itself creating a back EMF to oppose the change.

Here is an example to show how this could work:

Let the change in the magnetic flux caused by the initial magnetic field $$B_1$$ through a current loop's area in a small amount of time be $$\frac{d\phi_1}{dt} = \epsilon_1$$.

Let the loop have self-inductance $$L$$ and resistance $$R$$. The change in magnetic flux will cause a current that creates a magnetic field $$B_2$$ opposing the original magnetic field $$B_1$$. The current will have size $$I_1 = \frac{\epsilon_1}{R} = \frac{1}{R} \frac{d\phi_1}{dt}$$.

Now, this generated current $$I_1$$ produces its own magnetic field $$B_2$$ which generates its own change in magnetic flux through the current loop. This additional change in magnetic flux generates an opposing emf $$\epsilon_2$$ to the originally produced emf $$\epsilon_1$$ which in general is far smaller than $$\epsilon_1$$.

This opposing emf $$\epsilon_2 = L \frac{dI_1}{dt} = \frac{L}{R} \frac{d^2\phi_1}{dt^2}$$ so the opposing current is $$I_2 = \frac{\epsilon_2}{R} = \frac{L}{R^2} \frac{d^2\phi_1}{dt^2}$$

Thus, $$I_{net} = I_1-I_2 = \frac{1}{R} \frac{d\phi_1}{dt} - \frac{L}{R^2} \frac{d^2\phi_1}{dt^2}$$ (in the direction of the original current $$I_1$$).

However, I believe that if $$\frac{d^2\phi_1}{dt^2}$$ is sufficiently large the value of this could be negative and produce current in the opposite direction of the originally produced current $$I_1$$.

I'm not sure if the math or assumptions of this derivation are correct but I think the general idea behind it is sound. If there are any issues please let me know in the comments below.

simply put, the first EMF is the dominant one as it is the result of the greater change in flux (or the greater force applied on the charges).

As for the math, the previous answer seems to have the hang of it and avoid repeating their work. Cheers