# Mutual Inductance: Coil 1 Excited, Coil 2 Open Circuited

1. ### ezcheez

5
I have a very basic question regarding mutual inductance.

Consider a basic square loop with two coils wound on opposite sides from each other, with coil one excited while coil two si left open circuited. The total flux linking coil 1 is

Φ11 = Φl121

where

Φ11 is the total flux linking coil 1
Φl1 is the leakage flux and
Φ21 is the mutual flux linking both coils

My question is why there is mutual flux if the second coil is open-circuited, in other words, why is Φ21 non-zero? I would expect flux through the iron core due to coil one but how/why does coil two contribute to this flux?

2. ### n.karthick

244
Here $$\phi_{l1}$$ is the leakage flux i.e., the flux which is linking only with first winding and not with second winding and is due to imperfect coupling. The flux $$\phi_{l21}$$ is the one which interlinks with both windings. Since $$\phi_{l21}$$ has nothing to do with the second coil's open or short circuit conditions it wont be zero. You may refer the Wikipedia's page below
http://en.wikipedia.org/wiki/Leakage_inductance

3. ### ezcheez

5
Let me take step back and ask this question:

What would happen to the flux in the magnetic core if the second coil is removed? Does only the leakage flux remain, or would some flux still flow through the whole core?

4. ### n.karthick

244
In fact when the second coil is open circuited, the first coil behaves as if it is the only inductor present in the core. So even if the second coil is removed, the flux going through the core $$\phi_{l21}$$ will be the same.

5. ### ezcheez

5
That helps a great deal. I am going to ponder that for a little while...

6. ### geetika

3
Hey Karthick,
This is different you have been discussing,Should I use N1+N2 for the no.of turns in the B=U0NI/L where both the coils are one over the other , but one is NOT Connected?
Or should I take only N1 or N2 which ever coil is driven by DRIVE CIRCUIT.

Regards
Geetika