Why is the magnitude of the magnetic field proportional to the permeability of free s

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Since the permeability of free space is a measure of how free space resists a magnetic field, shouldn't it be inversely proportional like the permittivity of free space with respect to the electric field?

B = μqrsin / 4πr^2
E = q / 4πεr^2
 

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  • #2
Meir Achuz
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The permeability of free space is NOT a measure of how free space resists a magnetic field.
 
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The permeability determines how much space itself is magnetized in the presence of a magnetic field. Since it is less than unity I think of it of like a resistance where the permeability equals zero in ideal space
 
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In Albert Shadowitz's The Electromagnetic Field, page 319-320, (you can find it at google.books http://books.google.com/books?id=31...resnum=2&ved=0CB4Q6AEwAQ#v=onepage&q&f=false") it's explained that

"When the theory of electricity and magnetism was first being developed it was believed that the vectors E and H were analogous to each other."

That made the constitutive equations D = epsilon E and B = mu H analogous.

But, "Today it is generally accepted that E and B are the analogous quantities, rather than E and H. For E is produced by any kind of charge, free or bound, just as B is caused by any kind of current, conventional or bound. D and H, on the other hand, are only produced by free charge and conventional current, respectively. From this point of view it is unfortunate that epsilon0 and mu0 were placed in opposite positions in the two defining equations."
 
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In Albert Shadowitz's The Electromagnetic Field, page 319-320, (you can find it at google.books http://books.google.com/books?id=31...resnum=2&ved=0CB4Q6AEwAQ#v=onepage&q&f=false") it's explained that

"When the theory of electricity and magnetism was first being developed it was believed that the vectors E and H were analogous to each other."

That made the constitutive equations D = epsilon E and B = mu H analogous.

But, "Today it is generally accepted that E and B are the analogous quantities, rather than E and H. For E is produced by any kind of charge, free or bound, just as B is caused by any kind of current, conventional or bound. D and H, on the other hand, are only produced by free charge and conventional current, respectively. From this point of view it is unfortunate that epsilon0 and mu0 were placed in opposite positions in the two defining equations."
E & H are analogous in 2 ways. First, E is in units of V/m, while H is in A/m. The ratio E/H is V/A or ohms. The E/H ratio of an e/m wave is the impedance of the medium. Also, E X H, the cross product, has units W/m^2. This is the radiated power per area, aka "Poynting Vector".

If we examine electric & magnetic fields in the boundary region between differing media, we get the following. The normal components of the B fields in the 2 media, Bn1 & Bn2, are equal. The normal components of the D fields in each medium are either equal or they differ by a mere constant, i.e. Dn1 - Dn2 = rho_s. The "rho_s" is the surface charge density. The B & D quantities behave in a similar fashion.

But the tangential fields at a boundary displays a different property. The tangential E field components are equal for materials 1 & 2, i.e. Et1 = Et2. Also, Ht1 = Ht2. So E & H behave analogously.

Based on the above, there is compelling reason to regard E & H as a pair, likewise w/ B & D. But hold on. The Lorentz force equation is as follows:

F = q*(E + (u X B)). Here, the Lorentz force is determined by E for the electric part, & B for magnetic. Hence a case can be made tying E to B.

So the conclusion is that there is no conclusion. Is E the counterpart of B or H? I don't think we have an answer.

Claude
 
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  • #6


I need some information of magnetic permeability of aluminium material,

its numerical value.
while my other velocities are in m/s, and density is kg/m^3
thanks
 

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