# Faraday's law in a ring

What role does the induced emf in Faraday's law play in generating current in a ring in which the magnetic flux is changing ?

-
Can someone help me ? – Carpediem Mar 27 '13 at 15:30

The induced EMF is really just like a voltage provided by a battery. Consider, for example, a situation in which an EMF of magnitude $\mathcal E$ were induced in a loop. Then this would be equivalent to a putting a battery with voltage $V=\mathcal E$ in the loop, and this could cause current to flow. If the resistance of the loop were $R$ for example, then an amount of current $I = V/R$ would flow.

-

COIL EMF AND BATTERY EMF

To understand the subtle difference between a battery emf and that produced by a coil we need to write down the basic equations for two circuits:

(i) An ideal battery(no internal resistance) with emf = $e(t)$ in series with a resistor, R:

emf=$e(t) = i(t)R$ from which $i(t)=\frac{e(t)}{R}$

(ii) An ideal coil (with negligible resistance) with variable current flowing through it

emf=$e(t)=L\frac{di(t)}{dt}$ from which $i(t)=\frac{1}{L}\int_0^t e(\tau)d\tau$

These two equations show a subtle difference between the two emfs. Note that the current that develops due to the emf of the coil is bult up through all past moments of the emf, unlike the case of the battery where the current takes the value $e(t)/R$ regardless what the value of $e(t)$ has been a short while ago.

-

If you have a stationary ring, with total resistance $R$, inside an external (changing) magnetic field, then there is flux through the ring.

First, there is flux $\Phi_1$ from the external, changing $\vec{B}$ field. Since that $\vec{B}$ field is changing, there is an emf due to that.

Second, the current from the ring itself produces it's own $\vec{B}$ field, so it's own flux, $\Phi_2$. In a quasistatic approximation, you might say that this flux is proportional to the instantaneous current $I$ and denote the proportionality by $\Phi_2=LI$. If the current is changing, then that flux is also changing, so there is an emf due to that.

So together we have a total emf: $\mathscr E = -d(\Phi_1+\Phi_2)/dt$. Based on the resistance, we have $$RI=\mathscr E=-d(\Phi_1+\Phi_2)/dt=-d\Phi_1/dt-LdI/dt.$$

This is a differential equation, and the solution depends on how the external $\vec{B}$ field is changing (to get $-d\Phi_1/dt$).

-