Past exam of the mathematics course of the University of Cambridge / 2009 / ib / Paper 3 / 17A / c

Table of contents
- Solution c

Solution

The Poynting vector in the annular region is

S = \frac{1}{μ _{0}} E \times B = \frac{V I}{2 π r ^{2} ln ( b / a )} e_{z} .

It points from the battery towards the resistor with the preceding current convention. Integrating through an annular cross-section gives

P = \int_{a}^{b} S_{z} 2 π r d r = \frac{V I}{l n ( b / a )} \int_{a}^{b} \frac{d r}{r} = V I = V^{2} / R .

The energy is transported by the electromagnetic field in the space between the ideal conductors; the integration recovers the power dissipated at the terminating resistor.

 Ancestors (10)

 View article source

 Discussion (0) New discussion

There are no discussions about this article yet.

 Articles by others on the same topic (0)

There are currently no matching articles.

  See all articles in the same topic Create my own version