# Electric field in a solenoid/inductor.

1. Mar 2, 2014

### ElectricJosh

Hello,

I am trying to create a decent magnetic field in an inductor, I come from an electronics background so I am not a master physicist (I might use this knowledge later if I build an radio system or transformer).

I have been studying Maxwell's equations, more specifically Amperes law.

My present understanding of Amperes law is, that a magnetic H field can be created by a constant current density 'J' and/or a varying electric flux density 'D' (partial derivative of 'D').

I am interested in the changing electric flux density as I do not want to use a high current DC source but rather use a high voltage AC source and utilize the inductive reactance to stop high currents flowing and conserve battery life.

Firstly is my understanding correct?

Secondly can anyone help me link the changing potential difference across the inductor to electric flux density?

I have also come to understand by Gauss' law that Electric flux density is created by charge density 'Pv', in which case, does anyone know how to find the charge across two electric potentials?

I realize that this question is a bit ambiguous but that is why I am asking, any help will be appreciated.

Thank you,

Josh.

Website I have been using: http://www.maxwells-equations.com/

2. Mar 2, 2014

### Simon Bridge

Radios and transformers use magnetic fields in a different way to electromagnets though.

Kinda - if you alternate the flux density, you will get an alternating magnetic field in the coil.
$\qquad\qquad\vec\nabla\times\vec H = \partial_t\vec D + \vec J$
... is the relation you are thinking of.

It is the curl (∇x) of H that means there is a magnetic field around a wire carrying a current.

The electric field along a length of wire is E=V/L - where V is the applied potential difference and the L is the length of wire thus SI units "volts per meter". D=εE

In a solenoid - look up "inductor".

Look up "capacitor".

3. Mar 4, 2014

### ElectricJosh

Just one question, If E = V/L is E the electric field in a vacuum, hence do I have to covert it to D using the permittivity of free space constant?

Last edited: Mar 4, 2014
4. Mar 5, 2014

### Simon Bridge

In that context, E is the electric field in the wire - given a static DC voltage.
The relative permittivity of an ideal conductor is 1 in any case, but, in this example, the permitivity is already included when you measured V.

Strictly: $\vec E = -\frac{d}{dx}V(x)\hat{\imath}$ for the 1D case in the example.
Things get a bit more interesting when we consider dielectrics, as your reading about capacitors should show you.

5. Mar 5, 2014

### ElectricJosh

Ah ok thanks, so I could just divide E by the permittivity of free space to find the value of D to put into amperes law?

Last edited: Mar 5, 2014
6. Mar 5, 2014

### Simon Bridge

In the DC case, all metals have relative permitivity 1.
In general - all metals have a purely imaginary dielectric constant that depends on frequency.

Note - when you measure the voltages (DC case), you have taken into account the relative permittivity already.

The relationships are:

$\vec E = -\vec\nabla V$

$\epsilon \nabla^2 V = \rho$ where $\rho$ is the free charge density.