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When a coil connected to an AC generator creates an EMF in another nearby coil (mutual inductance), is self inductance simultaneously occurring in both coils?

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  • $\begingroup$ I believe it is practically true. Self induction is more common and prevalent as compared to mutual induction. If the magnetic flux of one coil/circuit is linked with the other coil/circuit, in all likelihood, it is also linked with itself. $\endgroup$ – Satwik Pasani Feb 27 '14 at 5:14
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When a current passes through a closed loop, a magnetic flux through the cross-sectional area of that loop arises due to that current. This magnetic flux is related to the current by the relation:

$$\phi=Li$$

$L$ is called the self-inductance of the loop and depends upon the configuration of that loop. $L$ remains constant as long as the configuration of the loop is not changed.

If the current in the loop changes with time, the magnetic flux through the loop changes in proportion to the current. This change in magnetic flux causes an electromotive force (EMF) to be induced across the terminals of the loop. This phenomena is called self-induction. The EMF generated is given by Faraday's law as:

$$e=-\dfrac{d\phi}{dt}=-L\dfrac{di}{dt}$$

A common misunderstanding in beginners is:

Let an AC voltage source be connected to a circular loop. As the current through the circuit changes, it causes the magnetic flux to change. This change in magnetic flux acts to change the current in the circuit which then changes the magnetic flux and so on i.e. we are revolving in a circle.

In other words, change in current $\rightarrow$ Change in magnetic flux $\rightarrow$ again change in current ... the circle never terminates. This is a misconception.

To clarify how a balance is achieved, consider an AC voltage applied to an ideal (zero resistance) loop. We will solve the circuit by applying KVL. The voltage applied is:

$$V(t)=V_0 \sin(\omega t)$$

A current $I(t)$ is established in this loop. The misconception is that a change in $V$ causes a change in $I$. Since $I$ is changed, the magnetic flux also changes, which would cause self-induction, changing the net voltage across the loop. Hence the existing electric field changes so the current will again change and so on. But this misconception has a flaw. By KVL, the sum of voltages around the loop must be zero i.e.:

$$V(t)+ \text{induced emf} =0$$

The astonishing fact is that there is $0$ Electric field inside the wire of the loop because the resistance of the loop is $0$. For a current to flow in an ideal inductor there is no need of $E$. The varying current only exists to counter the applied voltage.

We know that induced EMF is related to the current in the loop by:

$$e=-\dfrac{d\phi}{dt}=-L\dfrac{di}{dt}$$

This relation tells us that the current is a function of induced EMF. Since induced EMF is also a function of applied voltage, the current is a function of applied voltage and can be calculated. This is also the expression for an ideal inductor and the solution is, as we know:

$$I(t)=\dfrac{V_0}{\omega L}\cos(\omega t)$$

This explains why, despite the circle of reasoning, the current can be found as a function of applied voltage. This is because the induced EMF has to be equal to the applied voltage. In fact, the changes in quantities are related by a differential equation whose solution will give us the value of unknown current.

Consider two loops as shown:

image 2
(source: talkingelectronics.com)

There are two loops, blue and black. When the black loop is not present, the current in the blue loop is $i_1$. This current is constant and there is no self-induction; the battery has zero voltage since the loop is perfectly conducting. For the time being, consider this battery as a short circuit.

What happens when the black loop is placed in the vicinity of the blue loop? As the blue loop is moved to approach the black loop, the magnetic flux through both the blue and black loops change with time. This causes an induced EMF to appear in both loops. Both loops are ideal conductors, so we do not need any $E$ inside their wires to cause a current.

Suppose the initial current flowing through the black loop was $i_2$. Both $i_1$ and $i_2$ will change when the loops approach each other, but how?

Let the blue loop be loop$1$ and the black loop be loop$2$. The magnetic flux crossing the black loop due to the current in the blue loop is $\phi_{12}(x,t)$. This flux is related to $i_1(t)$ by the equation:

$$\phi_{12}(x,t)= M_{12}(x)i_1(t)$$

$M_{12}(x)$ is a function of the distance $x$ between the two loops and their configuration. This parameter $M(x)$ is called mutual inductance. If $x$ is not changed, the mutual inductance will remain constant.

Similarly, the magnetic flux crossing loop$1$ due to current in loop$2$ is:

$$\phi_{21}(x,t)=M_{21}(x)i_2(t)$$

If both loops have the same configuration, they create the same pattern of magnetic field around them when passing the same current. So $M_{12}(x)=M_{21}(x)$ for similar loops and may be denoted simply as $M(x)$ in that case.

The magnetic flux existing in loop$1$ due to its own current $i_1$ is:

$$\phi_{11}(t)=L_1i_i(t)$$

The magnetic flux in loop$2$ due to $i_2$ is:

$$\phi_{22}(t)=L_2i_{2}(t)$$

The EMF generated in loop$1$ and loop$2$ are:

$$e_{11}=-\dfrac{d\phi_{11}}{dt}$$ $$e_{22}=-\dfrac{d\phi_{22}}{dt}$$ $$e_{12}=-\dfrac{d\phi_{12}}{dt}$$ $$e_{21}=-\dfrac{d\phi_{21}}{dt}$$

For KVL to hold, the condition is:

$$e_{11}+e_{21}=0=e_{22}+e_{12}$$

But $e_{11}$ and $e_{22}$ represent self-induction in loop$1$ and loop$2$ respectivly. So $i_1$ and $i_2$ both vary with time. Both $\phi_{12}$ and $\phi_{21}$ will also change due to changes in $i_1$ and $i_2$ and any change in $x$. So we can simplify $e_{12}$ as:

$$e_{12}=-\dfrac{d\phi_{12}}{dt} =\dfrac{d(M(x)i_1(t))}{dt} =\dfrac{dM_{12}}{dt}i_1+\dfrac{di_1}{dt}M_{12}$$

$e_{12}$, the induced EMF in loop$2$, is produced by a change in x and/or a change in loop$1$ current, $i_1$. The phenomena of inducing an emf in a loop by changing the current in some other loop is called mutual-induction. If $L_1$ is small, we can visualize loop$1$ as a bar magnet and self induction will only be observed in loop$2$. Further, if we assume loop$2$ has resistance $R_2$ very large compared to $L_2$, we will see no self induction but simply Faraday's law i.e. $e_{12}=i_2R_2$.

Now let's deal your specific question:

"When a coil connected to an AC generator creates an EMF in another nearby coil (mutual inductance), is self inductance simultaneously occurring in both coils?"

To analyze this, consider again the previous loop$1$ and loop$2$, replacing the DC battery with an AC voltage source. The distance $x$ between the loops is not changed. Both loops are perfectly conducting and have comparable self-inductances and equal mutual-inductances. Since both the loops are ideal conductors, $E$ inside the wires of both will be $0$.

$$-\dfrac{d(\phi_{11}+\phi_{21})}{dt}+V_0 \sin(\omega t) =0 \implies V_0\sin(\omega t)=L_1\dfrac{di_1}{dt}+M\dfrac{di_2}{dt} \tag{1}$$

Also:

$$\dfrac{\phi_{22}+\phi_{12}}{dt}=0 \implies L_2i_2=-Mi_1+k \implies i_2=-M/L_2i_1+k$$

Using this value of $i_2$ in eq 1 we get

$$V_0\sin(\omega t)=(L_1-\dfrac{M^2}{L_2})\dfrac{di_1}{dt}$$

Let ${L_1-\dfrac{M^2}{L_2}}=\alpha$. Then we get an expression for $i_1$ as:

$$i_1=-\dfrac{V_0}{\omega \alpha}cos(\omega t)$$

We see $i_1$ varies with time and hence $i_2$ also so both the loops undergo self-induction and mutual-induction. This was a very conditioned analysis. Resistance was ignored but the thing is even including the resistance both self-inductance and mutual-inductance will occur.

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Imagine two coils, coil 1 and coil 2 with (self) inductance $L_1$ and $L_2$. If the two coils are not coupled, we have:

$$v_1 = L_1 \frac{di_1}{dt}$$

$$v_2 = L_2 \frac{di_2}{dt}$$

Now, if the coils are coupled, we have:

$$v_1 = L_1 \frac{di_1}{dt} + M\frac{di_2}{dt}$$

$$v_2 = M\frac{di_1}{dt} + L_2 \frac{di_2}{dt}$$

where $M$, the mutual inductance is given by

$$M = k\sqrt{L_1L_2}, \, 0 \lt k \le 1 $$

When a coil connected to an AC generator creates an EMF in another nearby coil (mutual inductance), is self inductance simultaneously occurring in both coils?

First of all, your question is oddly phrased. Obviously, if either (self) inductance is zero, the mutual inductance is zero in the above formula.

But, perhaps you're asking something else. If for example, coil 2 is not connected to a circuit, then $i_2$ is zero so there can be no self induced voltage.

However, there can be a mutually induced voltage across coil 2 due to a changing current in coil 1.

Regardless, the (self) inductances $L_1$ and $L_2$ must be non-zero for there to be non-zero mutual inductance $M$.

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Yes, sometimes it is negligible, like in the case of two coupled loop. But it would be important for two coils where each has large number of turns.

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