# Power in solenoids

1. Sep 29, 2009

### physicsnoobie

so far i've been studying that P= work done (energy change) / time = I x V
and its all been good... until i came across P= I squared x R (i know how this is derived) when studying solenoids in Magnetic Resonance Imaging and this was refered to as the power loss (energy dissipation over time)

1. i found this somewhat confusing.. since it suggests that the process: electrical energy -> heat energy (losses), is the only energy change process that is going on in a solenoid

i thought about it and so far im guessing that maybe this is true in solenoids but not in normal electric circuits because electrical energy -> heat energy (losses) + light energy (bulbs)

anyway, this was in the context of why copper is not used in the solenoid of MRI.. because the large combined resistance would result in large power losses.. instead superconductors with zero/negligible resistance is used

2. then again, does that mean that power of the solenoid in MRI = zero/infinitely small because the resistance is negligible? (following P = I squared x R)

anyway, thanks to anyone who shares some light on this.. hopefully in simple terms.. Im quite a newbie in physics :(

2. Sep 29, 2009

### Gear300

The power when considering P = I^2*R is explicitly noted as the rate at which energy is supplied to a resistance. The power of the solenoid, I'm guessing, would be the rate at which energy is supplied to the magnetic field in the solenoid. If I remember correctly, the energy of the magnetic field of a solenoid may be given as U = (1/2)LI^2; so if the current is constant, the power is 0W. In the case of non-steady state circuits or AC circuits, the current does change.

Last edited: Sep 29, 2009