# Flux linkage in a wire coil

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1. Jul 15, 2015

### JSmith85

Flux linkage in a wire coil is given by BANcos(theta) with B being flux density, N being the number of turns in a coil and A being the cross sectional area of the coil and theta being the angle between the normal to the plane of the coil and the magnetic field.

Is anybody able to explain why it is the cross sectional area of the coil?

I understand it in terms of a length of wire as the cross section of that is cutting field lines but in terms of a coil I'm confused as you have the whole central area of the coil which is empty and therefore not cutting field lines so why is the cross sectional area of the coil used and not just the area of the coil cutting the field lines?

2. Jul 15, 2015

### Hesch

It's simply because voltage induced in a coil is not depending on "wires cutting field lines". It depends on the (various) flux, that the wire surrounds:

Say you have some square shaped winding ( 1m x 1m ) in a magnetic field. Now you somehow expand the winding to 2m x 2m. Of course the wires must cross the magnetic field due to this expansion, but that doesn't matter. It's the expansion of the cross section area ( 1m2 → 4m2 ) that matters. The winding surrounds more flux by the expansion.

Emf = dΨn/dt , Ψn = Ψ*N , Ψ = B * A (flux)

3. Jul 15, 2015

### JSmith85

Ok I think you have lost me a little bit now. So how does this fit with the idea of flux through a single wire.

4. Jul 15, 2015

### Hesch

I'm not sure, what you mean by the question.
Say you have a straight piece of wire, you hold it horizontal while you are running over the North-pole, thereby inducing emf between the ends of the wire.

Now you want to measure this voltage, so you have a volt-meter and connect its probes to the ends of the wire. Maybe you could tape the leads of the volt-meter to the wire, so they are fixed. Then you start running again, and you will measure nothing, because a counter-emf is induced in the leads of your voltmeter as you are running.

You cannot measure/sense an emf, without forming a closed loop including your volt-meter. Currents can only flow through closed circuits.

That's why areas surrounded by a wire ( not the wire itself ) are regarded by calculations of an emf in a coil.

Having a varying magnetic field through a wire will only induce a circulation emf (closed loop) inside the wire, resulting in Eddy-currents in the wire.

5. Jul 15, 2015

### JSmith85

Sorry I think you have misunderstood I'm not talking about the emf just the magnetic flux.

Let me try again if I have a single wire in a magnetic field of a known cross sectional area flux is given by BA with A being the cross sectional area of the wire however if i have a coil ( for arguments sake let us pretend it is square with each side being made up by identical wires to the first example) then A becomes the cross sectional area of the coil rather than the sum of the cross sectional areas of the wires making it up, why is this?

6. Jul 16, 2015

### Hesch

B*A*N*cos(θ) is an expression that equals Ψn ( the flux linkage ). A is the area surrounded by the wire: the area within the winding.
The emf in the winding = dΨn/dt.
This emf is used to calculate transformers ( E = 4.44 * f * N * A * B , sinusoidal voltage ), motors, mutual induction, self-induction, etc.
You could define A being the area of the wire itself, but then you are calculating something else ( Eddy voltage ).
I have understood that you find it strangely that flux in "thin air" induces voltage in a wire, but so it is. That's a phenomenon made by the nature.
Using A = Awire, you can calculate Eddy-currents within the wire, because now you are regarding a closed loop here.

So you must consider what you want to calculate by these flux linkages: Eddy-losses or mutual/self inductions?
Regarding only flux linkages as a "stand alone" ( no emf, no current ) doesn't make much sense: Who cares about a flux linkage if it has connection with nothing?