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Homework Help: Expression for the time average power dissipated in a resistance R

  1. Jul 20, 2011 #1
    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. jcsd
  3. Jul 20, 2011 #2

    Philip Wood

    User Avatar
    Gold Member

    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.
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