Jump to content

Magnetic fields


Sriman Dutta

Recommended Posts

We have two notions of the magnetic field. One is the B field, commonly called the magnetic flux density and the other one is the H field, also called the magnetic field strength. However, I want to know the difference between them. Most equations involving magnetic field uses B. However what is the use of H is still not clear to me. I have seen some relations between them in Wikipedia. However I couldn't understand that well.

So please show how they are different and in what way the H field is useful.

Link to comment
Share on other sites

https://en.m.wikipedia.org/wiki/Ampère%27s_circuital_law

 

H is useful for calculating the flux density in magnetic circuits with different magnetic materials.

 

I find it especially useful when working with permanent magnets.

http://www.electronics-tutorials.ws/electromagnetism/magnetic-hysteresis.html

 

When using Hopkinson's law for magnetic circuits, B is the "current density" and H is related to the voltage.

Edited by Bender
Link to comment
Share on other sites

Have you studied electric fields?

 

The relationship between B and H is very similar to the relationship between E and D.

 

In many situations one is a constant times the other but in non-isotropic media they may point in different directions (they are all vectors).

 

If you let me know whether you understand about E and D I will explain further.

Link to comment
Share on other sites

I know about the E field. But the concept of D field is new to me. To speak the truth I haven't seen it before.

 

But I know about the E field well enough.

It can be described as the force exerted on a test charge when it passes through the sphere of influence of another electrically charged body( considering the charge on the test particle negligible).

[math]E=\frac{F}{q}[/math]

Another popular notion is that E is the electric flux density. It is the derivative of the electric flux with respect to the surface through which the flux is considered.

[math]E=\frac{d\phi_E}{dS}[/math]

By Gauss' Law, we have yet another relation:

[math]\int_S E.dS = \frac{Q}{\epsilon_0}[/math]

Edited by Sriman Dutta
Link to comment
Share on other sites

OK thank you for that information.

 

The teaching of electricity and magnetism nearly always runs along the following path these days.

 

A force on a suitable body is noted and an electric/magnetic effect deduced to account for it.

This effect is defined in terms of the force produced and the position in space and is a vector quantity.

 

The work done as the body moves around is calculated and related to the position to define a potential field.

 

Sometimes a second vector quantity is defined as a vector derived from the first one.

Either E and B can be defined first, with D and H as the derived vectors or the other way round (which is the older method).

 

But often the significance of the second vector is not presented, hence your question.

 

So consider this statement in the light of the following.

 

You need two vectors to have a dot (scalar) product and energy is a scalar.

 

Starting with a mechanical example we have

 

Stored Mechanical Energy per unit volume = 1/2 stress x strain

 

Stored Electrical Energy per unit volume = 1/2 E x D

 

Stored Magnetic Energy per unit volume = 1/2 B x H

 

edit (note these are all dot products not cross products)

 

Are you beginning to see a pattern ?

 

There are differences in the nature of stress, E and B but scalar multiplied by another suitable vector they yield the energy stored in the system.

 

There are actually many more examples that could be listed and they all spring from a common root in continuum mechanics.

 

How are we doing?

Edited by studiot
Link to comment
Share on other sites

So electrical energy stored per unit volume is 0.5*E*D. But both units should be of N/C ( though I don't know the exact unit of D).

Moreover I knew that the formula to find the electrical potential energy is

[math] Energy= k\frac{Qq}{r}[/math]

Link to comment
Share on other sites

So electrical energy stored per unit volume is 0.5*E*D. But both units should be of N/C ( though I don't know the exact unit of D).

Moreover I knew that the formula to find the electrical potential energy is

[math] Energy= k\frac{Qq}{r}[/math]

 

I am a little unsure from this reply where you want to go from her or even if you want to carry on the discussion?

 

Can you not find out what the units of E and D and the k you mention in your version of Coulomb's law?

Link to comment
Share on other sites

  • 3 weeks later...

Sriman: you could think of H and B as roughly analogous to voltage and current in a resistive circuit. Don't read too much into the analogy, but just as V and I are proportional for a given resistor, H and B are proportional for a given material. But voltage obeys Kirchoff's Voltage Law - its integral around a closed path is zero -, whereas current obeys Kirchhoff's Current Law - the sum of currents entering or leaving any point in a circuit is zero.

 

For H and B, you have similar behavior. The integral of B over any closed surface is zero (no magnetic monopoles, divergence of B is zero), whereas the integral of H around a closed path is equal to the ampere-turns encircled by the path (curl of H = J).

 

In a medium of constant permeability both sets of relations would hold for both B and H, with appropriate proportionality factors. But that's the "physics difference" between them. H is something that satisfies a line integral / curl related law; B is something that satisfies a surface integral / divergence related law.

 

Hope this helps!

Link to comment
Share on other sites

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!

Register a new account

Sign in

Already have an account? Sign in here.

Sign In Now
×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.