# Expression for the time average power dissipated in a resistance R

1. Jul 20, 2011

### paulb42

1. The problem statement, all variables and given/known data

A square loop with side-length a is positioned at the centre of a long thin solenoid, which has
radius r (with r > a), length l and N turns. The plane of the loop is perpendicular to the
5axis of the solenoid. A current I = I0 sin ωt ﬂows through the loop. Derive an expression for
the time-averaged power dissipated in a resistance R connected between the terminals of the
solenoid. You may assume that R is much greater than the resistance of the solenoid and
that the self-inductance of the solenoid is negligible.

2. Relevant equations

P = J E d (this is meant to be the integral of the total volume of the shape in question)
P = I^2 * R

3. The attempt at a solution

I understand how the power dissipated is calculated over a macroscopic object. But how does this change when it is the referring to a solenoid and the terminals?

2. Jul 20, 2011

### Philip Wood

The power is dissipated in R. Neglect any dissipated in the solenoid, because its resistance is negligible compared to that of R. [Not that there would be any problem with calculating the power in the solenoid if we had to - it's only a piece of wire, but wound into a coil !]

But first you need to find the emf induced in the solenoid. Look up 'mutual inductance', or work from Faraday's Law.