Scratch that. I'll reply momentarily... (I was suggesting you factor ## b^2-a^2=(b-a)(b+a) ##, but from what I see what you have is incorrect)...[Editing note: The OP's solution is correct. The book has an error in their answer. It was difficult to see the OP's posting, so that in the following post, I misinterpreted the diagram. That later gets corrected in the discussion in the subsequent postings]. ## \\ ## In this problem, the EMF ## \mathcal{E} ## is larger around the outside part of the loop at radius ## b ## , than it is at radius ## r=a ##. The EMF will be a function of radius. For the outer part, the area in the ## \mathcal{E}=- \frac{d \Phi}{dt} ## equation is ## A=\pi b^2 ##. I think you may need to convert the resistance per unit length ## \lambda ## to a resisitivity ## \rho ##, with the assumption that the loop has thickness ## w ## which will ultimately drop out of the calculation, along with recognizing the other dimension is ## l=b-a ##, in order to compute resistivity ## \rho ## from ## \lambda ##.## \\ ## Meanwhile, current density ## J=\sigma E ## with ## \sigma=\frac{1}{\rho} ##, where ## E=E(r) ## is the induced electric field, and current density ## J=J(r) ##. The induced electric field at a radius ## r ## is calculated as ## E(r)= \frac{ \mathcal{E}(r)}{2 \pi r } ##. The current in the loop is then the current density integrated over a cross section of area: ## I=\int J(r) \, dA ##, where ## dA= w \, dr ##, and ## r ## goes from ## a ## to ## b ##. ## \\ ## Perhaps there is a simpler approach, but this is how I would approach the solution. From what I can tell, this problem is somewhat unique, and is not a simple one. ## \\ ## And note: The EMF on a loop of radius ## r## is given by ## \mathcal{E}(r)=-\frac{d \Phi}{dt}=- k \, \pi r^2 ##. You applied the EMF equation incorrectly in the OP. ## \\ ## And if my interpretation is correct, this is not an introductory E&M problem, but rather a somewhat advanced one. I provided more of the solution than is normally recommended for Homework Helpers to provide, because this one is not an introductory exercise. At least that is my interpretation. Again, perhaps someone else will see a simpler way of solving it. :) ## \\ ## And one other comment: The loop needs to be closed or you will get minimal current flowing. I question at this point the validity of the problem. If this is in an introductory course, the answer should presumably be zero. ## \\ ## Scratch my whole solution above: I thought the material covered the range between radius ## r=a ## and ## r=b ##. My mistake. :-)
