3
$\begingroup$

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!

Improve this question
$\endgroup$
3
  • $\begingroup$ Is there any chance that the question is referring to what Griffiths calls the Auxiliary Field $\mathbf{H}$ rather than $\mathbf{B}$? $\endgroup$
    – Michael
    Commented Sep 30, 2014 at 22:29
  • $\begingroup$ 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? $\endgroup$
    – Floris
    Commented Oct 1, 2014 at 0:21
  • $\begingroup$ 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. $\endgroup$
    – wisner
    Commented Oct 2, 2014 at 1:56

2 Answers 2

Reset to default
3
$\begingroup$

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.

Improve this answer
$\endgroup$
4
  • 1
    $\begingroup$ 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? $\endgroup$
    – Floris
    Commented Oct 1, 2014 at 0:20
  • $\begingroup$ @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. $\endgroup$
    – Javier
    Commented Oct 1, 2014 at 1:39
  • $\begingroup$ 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. $\endgroup$
    – wisner
    Commented Oct 2, 2014 at 13:22
  • $\begingroup$ 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. $\endgroup$
    – Javier
    Commented Oct 2, 2014 at 16:50
0
$\begingroup$

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.

Improve this answer
$\endgroup$
1
  • $\begingroup$ 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! $\endgroup$
    – wisner
    Commented Sep 30, 2014 at 22:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.