# EMF Induced by Changing Magnetic Field

• BrandonInFlorida
In summary, the conversation discusses a possible misunderstanding of the premise regarding the expansion of a circular region and the increase of field strength, B. The two cases of the radius, r, being less than or greater than the radius, R, are considered and the resulting equations for E are given. The issue with the book's diagram and labels is discovered and clarified, leading to a resolution of the misunderstanding. Instructions on how to post a photo are also provided.f

#### BrandonInFlorida

Homework Statement
This is a section of my book I don't understand, not literally homework. They (Halliday and Resnick) show a circular region containing a homogeneous magnetic field, which is increasing in strength. They then consider concentric circles within the region.

Faraday's Law states that the induced EMF around each loop is the negative of the time rate of change of the flux. They show how larger and larger circles within the circular region have larger and larger fluxes, more importantly larger and larger time rates of change of flux and, therefore, larger induced EMFs.

I'm with them so far. What I don't understand is when they say that concentric circles outside the region have zero induced EMF. It seems to me that once the circle considered becomes larger than the circular region containing the field, the flux and its time derivative remain constant for bigger and bigger circles. It doesn't fall to zero, so the induced EMF wouldn't fall to zero, according to Faraday's Law.
Relevant Equations
EMF = derivative of flux with respect to time
Can anyone show me where I'm wrong?

You must be misunderstanding the premise. Are you certain H and R is not expanding the size of the circle at a constant dr/dt ?? Then their statement would be true (for constant B).

Let's say the circular region in which the field is non-zero has radius ##R##. Consider concentric cicles of radius ##r##. There are two cases.

Case I: ##r<R##
From Faraday's law in integral form, the result of the integration on both sides is $$E~2\pi~r=-\pi r^2\frac{dB}{dt}\implies E=-\frac{r}{2}\frac{dB}{dt}.$$Case II: ##r>R##
From Faraday's law in integral form, the result of the integration on both sides is $$E~2\pi~r=-\pi R^2\frac{dB}{dt}\implies E=-\frac{R^2}{2r}\frac{dB}{dt}.$$Clearly the emf is continuous at ##r=R.## It's hard to believe that H&R, which has been around more than 60 years, has an error like that. Can you post a legible photo of the relevant section?

• hutchphd
You must be misunderstanding the premise. Are you certain H and R is not expanding the size of the circle at a constant dr/dt ?? Then their statement would be true (for constant B).
Absolutely sure. The size of the circle is unchanging. The field strength, B, is increasing.

Yeah then I need to see the section in Halliday...

Ctrl+P or use 12th icon

Let's say the circular region in which the field is non-zero has radius ##R##. Consider concentric cicles of radius ##r##. There are two cases.

Case I: ##r<R##
From Faraday's law in integral form, the result of the integration on both sides is $$E~2\pi~r=-\pi r^2\frac{dB}{dt}\implies E=-\frac{r}{2}\frac{dB}{dt}.$$Case II: ##r>R##
From Faraday's law in integral form, the result of the integration on both sides is $$E~2\pi~r=-\pi R^2\frac{dB}{dt}\implies E=-\frac{R^2}{2r}\frac{dB}{dt}.$$Clearly the emf is continuous at ##r=R.## It's hard to believe that H&R, which has been around more than 60 years, has an error like that. Can you post a legible photo of the relevant section?
Yes. How does one post a photo? None of these icons seems to correspond to upload photo. Maybe I can just attach it as a file.

Ctrl+P or use 12th icon
In going to take the photo, I found the issue. Although the book's diagram shows concentric circles both within and outside of the circular field, and although the labels #1, #2, #3, and #4 seem to correspond to those circles, they actually dont. They actually correspond to very small Farraday loops, which suddenly makes the whole thing make sense.

Thanks, guys. You never fail me.

• hutchphd
Yes. How does one post a photo? None of these icons seems to correspond to upload photo. Maybe I can just attach it as a file.
To post a photo click the link "Attach files", lower left.

• hutchphd