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There are two basic definitions in magnetism for circuit theory:

1-) Magnetic flux density = Magnetic flux / Area

B = 𝜱 / A

2-) Magnetic field strength = Magnetomotive force / Length of the flux path

H = mmf / l (mmf = N * I) so:

H = N * I / l


But there is also a relation between H and B such as:

B = μ * H

I don't understand how the equations 1 and 2 yield B = μ * H.

1 and 2 are defined independently but still B and H has a relation which is also defined separately. I'm confused at this point. 1 and 2 are definitions but I cannot link B and H using these two definitions.

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    \$\begingroup\$ Voltage and current are related by resistance; MMF and flux are related by reluctance; magnetic field strength and flux density are related by permeability. They are all constants defining the linear relationships between related variables. \$\endgroup\$
    – Chu
    Jul 3, 2017 at 0:59
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    \$\begingroup\$ @user134429: Why do you expect equations 1 and 2 to yield B=μH? \$\endgroup\$
    – Curd
    Aug 2, 2017 at 20:04

4 Answers 4

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These equations don't yield \$ B = \mu \cdot H\$.
\$ \mu \$ is a measure of of the permeability of a magnetic material and is defined as \$ \mu = \frac{B}{H} \$. This is not always the most useful figure as if we look at a typical B-H curve

enter image description here

You will notice it is not always a simple relationship. The incremental permeability is often more useful as \$ \mu_i \$ is proportional to the inductance. For some materials such as air the B-H curve is a straight line so air gaps are often introduced into magnetic materials if you want a defined inductance.

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According to one of my textbooks from college,

\$ H = \frac{\mu I \textbf{a}_\phi}{2 \pi r} = \frac{B}{\mu} \$, where \$ \textbf{a}_\phi \$ is just a unit vector that goes into the direction of the magnetic field when a wire is exposed to electric current. This is how \$ B \$ is related to \$ H \$.

The relationship is just simple theory of a wire with some type of current creates a magnetic field around the wire. This is known as the Right hand rule for a magnetic field in a wire. See the picture below.

enter image description here

\$ B \$ , \$ H \$ , and \$ \phi \$ all "wrap" around a wire as the current travels in one direction. \$ H \$ is the magnetic field density (Teslas per unit area) and \$ B \$ is just the magnetic field (Teslas).

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  • \$\begingroup\$ There must be a way to derive eq. 2 by using eq. 1 for a solenoid. Thats the only way to reveal the link. But I dont know how. What you've written as an answer is not what Im asking. \$\endgroup\$
    – user1245
    Jul 2, 2017 at 20:33
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The equations can never hope to uncover the physics that lies in the value of mu, the permeability of free space. Mu is a real physical property of the universe and is related to the speed of light along with epsilon, the permittivity of free space. Both mu and epsilon define the speed of light in the material it passes.

Your first equation doesn't help uncover mu, it just defines basically what density is and, although magnetic flux (as a value) is a quantity related to mu it isn't explicitly revealed in equation 1.

Your second equation just defines the driving force behind magnetism i.e. amps, turns and distance. Nothing about mu contained here.

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  • \$\begingroup\$ Here how my thinking works: There is some link is missing. One person defines B = phi /A and another person independently defines a quantity H = N * I / l. But when they check B and H they notice they are related with mu. There must be a link. \$\endgroup\$
    – user1245
    Jul 2, 2017 at 20:09
  • \$\begingroup\$ Mu is the link. B and H are related by mu. Phi is total flux and, when divided by area becomes flux density, B. I'm not sure what you think is missing. Mu is the tricky one related to the laws of physics, the definition of the henry and metre. \$\endgroup\$
    – Andy aka
    Jul 2, 2017 at 20:56
  • \$\begingroup\$ There must be a way to derive eq. 2 by using eq. 1 for a solenoid. Thats the only way to reveal the link. But I dont know how. \$\endgroup\$
    – user1245
    Jul 2, 2017 at 20:59
  • \$\begingroup\$ You can't derive voltage with knowledge only of current. You need resistance, as a concept, to convert amps to volts. It sounds like you are trying to relate volts to the amps flowing. Flux is a by product of electric field strength (H) and mu. \$\endgroup\$
    – Andy aka
    Jul 2, 2017 at 21:01
  • \$\begingroup\$ I feel like there must be a relation between 𝜱 I and N for a solenoid. So embedding one to another should reveal B = mu * H. Probably it is simple but I couldn't figure out. \$\endgroup\$
    – user1245
    Jul 2, 2017 at 21:04
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Equations 1 and 2 are not the reason of eq 3 (B=uH). All these equations are consequenses of the basic laws of electromagnetism. Those basic laws are partial differential equations of vector fields and are best known as Maxwell's equations. Only your eq 3 can be considered to be one f those basic laws if we accept the factor mu to be something more complex vector related than a number that is used to multiply another number.

Your equations 1 and 2 are valid in some simple, but practical situations that are well presented in the elementary textbooks of electricity. They cannot be used elsewhere than in those simple cases. You had thought that they could be used to derive one more general law of electromagnetism, but didn't understand how. Nobody can't because that derivation is impossible.

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  • \$\begingroup\$ here is the answer youtube.com/watch?v=1dCChkEGi_c&feature=youtu.be&t=988 \$\endgroup\$
    – user1245
    Jul 3, 2017 at 0:31
  • \$\begingroup\$ @user134429 the lecturer used B=uH as the starting point, not proved B=uH. \$\endgroup\$
    – user136077
    Jul 3, 2017 at 0:45
  • \$\begingroup\$ No I wasn't asking the proof of B=uH. I couldnt make a link between eq 1 and 2. \$\endgroup\$
    – user1245
    Jul 3, 2017 at 0:50
  • \$\begingroup\$ I disagree that Maxwell's laws are second order differential equations. All 4 equations contain the del operator: div B = 0, div D = rho, curl H = J + dD/dt, curl E = -dB/dt and the last 2 equations have a first order time derivative. Only combining these equations yields a second order differential (wave) equation, which contains the laplacian, and a second order time derivative. Excuses for not using vector notation, and partial differential symbols. I am not sure if this works in comments. \$\endgroup\$
    – Bart
    Sep 6, 2017 at 6:58

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