Electric field in a solenoid/inductor.

In summary: Strictly speaking, you should instead use the following:##\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.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?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 volt
  • #1
ElectricJosh
10
0
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/
 
Physics news on Phys.org
  • #2
ElectricJosh said:
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).
Radios and transformers use magnetic fields in a different way to electromagnets though.

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?
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.

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

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?
Look up "capacitor".
 
  • Like
Likes 1 person
  • #3
Thanks for your reply that helps a lot :)

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:
  • #4
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.
 
  • Like
Likes 1 person
  • #5
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:
  • #6
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.
 
  • Like
Likes 1 person

FAQ: Electric field in a solenoid/inductor.

1. What is an electric field in a solenoid/inductor?

An electric field in a solenoid/inductor is a type of electromagnetic field that is created when an electric current flows through a coil of wire. This field is responsible for inducing a voltage in nearby conductors and is a crucial component in the operation of many electronic devices.

2. How is the electric field in a solenoid/inductor calculated?

The electric field in a solenoid/inductor can be calculated using the formula E = -N * dΦ/dt, where E is the electric field strength, N is the number of turns in the coil, and dΦ/dt is the rate of change of the magnetic flux through the coil.

3. What factors affect the strength of the electric field in a solenoid/inductor?

The strength of the electric field in a solenoid/inductor is affected by the number of turns in the coil, the current flowing through the coil, and the permeability of the material inside the coil. The strength of the magnetic field also has an indirect effect on the electric field.

4. How does the shape of the solenoid/inductor affect the electric field?

The shape of the solenoid/inductor can affect the electric field in a few ways. A longer coil with more turns will produce a stronger electric field, while a shorter coil with fewer turns will produce a weaker electric field. Additionally, a tightly wound coil will have a stronger electric field than a loosely wound one.

5. What are some practical applications of the electric field in a solenoid/inductor?

The electric field in a solenoid/inductor is used in a wide range of electronic devices, such as transformers, motors, generators, and inductors. It is also an essential component in the operation of many household appliances, such as TVs, radios, and computers.

Back
Top