# $ε_0$ affects electric field intensity, but $μ_0$ doesn't affect magnetic field intensity?

I'll be honest: this question is actually a homework problem. I've spent the past hour going through Google and several textbooks trying to answer the question "Why does $ϵ_0$ affect electric field intensity but $μ_0$ does not affect the magnetic field intensity?" I don't understand much about electromagnetism, but I haven't found an explanation for this. From what I've seen, $μ_0$ seems to be inherently related to the magnetic field strength in the same way that $ϵ_0$ is related to the electric field strength.

Can anyone help me by providing a very general answer or by pointing me to a good resource? I don't want or expect anyone to do my homework for me, but a nudge in the right direction would be much appreciated!

• Is there any chance that the question is referring to what Griffiths calls the Auxiliary Field $\mathbf{H}$ rather than $\mathbf{B}$? Commented Sep 30, 2014 at 22:29
• Is there no hint on what the professor might be thinking in your most recent lecture notes, or in the book you have been reading? Commented Oct 1, 2014 at 0:21
• He explained it again today. He's looking to know why the constant for E (1/(4*pir^2*ϵ0)) includes ϵ0, but the constant for H is simply 1/(4*pir^2) and does not include ϵ0's corresponding term, μ0. Commented Oct 2, 2014 at 1:56

What's probably happening here is the following: The fundamental or microscopic fields $\mathbf{E}$ and $\mathbf{B}$ are technically called the electric field strength and the magnetic induction, while $\mathbf{D}$ and $\mathbf{H}$, their macroscopic counterparts, are called the electric displacement and the magnetic field, a quite weird nomenclature, since you would think $\mathbf{E}$ and $\mathbf{B}$ would be simply called the fields, but that's history for you.

In this context, saying that $\mu_0$ doesn't affect the magnetic field intensity would mean that it doesn't affect $\mathbf{H}$, in much the same way that $\epsilon_0$ doesn't affect the electric displacement, that is, $\mathbf{D}$. What $\mu_0$ does affect is the magnetic induction $\mathbf{B}$, which is often simply called the magnetic field.

• This seems like a plausible answer - but it doesn't answer the "why" part of the question, just the "what". But then again, what does "affect" really mean. Is it more than a scaling term? Commented Oct 1, 2014 at 0:20
• @Floris: That's a good point about answering the "why"; I think I'll wait until OP confirms if this is indeed the crux of the question before thinking up a more indepth explanation. Commented Oct 1, 2014 at 1:39
• No, unfortunately, I don't believe that the main idea behind the question :/ The professor explained it again yesterday. He's looking to know why the constant for E (1/(4*pir^2*ϵ0)) includes ϵ0, but the constant for H is simply 1/(4*pir^2) and does not include ϵ0's corresponding term, μ0. Commented Oct 2, 2014 at 13:22
• I'm not sure how you're getting a $1/r^2$ magnetic field, but let's put that aside. The issue is related to what I wrote, namely, the fact that $H$ is not the "true" magnetic field, which is $B$. If you take a look at Biot-Savart's law, you'll see a $\mu_0/4\pi$ right there. Commented Oct 2, 2014 at 16:50

The only thing I can see them going for is the fact that only two of $\epsilon_0$, $\mu_0$ and $c$ are independent, and typically, a modern view will fold $\epsilon_{0}$ into the definition of charge, and declare $c$ to be the fundamental constant used to transform space into time in special relativity, making $\mu_{0}$ a prediction of the theory.

I couldn't tell you whether this is what your book is going for, though.

• Thanks! Unfortunately, it's a problem written by the professor. At this point, I'm really hoping I misunderstood the question when it was given verbally, though I made sure to confirm with the professor what the question was. I guess I'll go back to him tomorrow and ask for further clarification. Thank you for your help! Commented Sep 30, 2014 at 22:43