Falling circuit in a magnetic field

[henryc09]

Homework Statement

1) a circuit is composed of a square wire of side length l, mass m,
resistance R and thermical capacity c. It is placed on the xz plane,
with corners at the points (0,0), (0,l), (l,0), (l,l). The magnetic field
is 0 for positive values of z, aligned along the y-axis and with absolute
value B for negative values of z. The circuit is left falling freely in
the gravitational field. Describe its temperature as a function of
time (not forgetting that in the beginning only one of the horizontal
wires with current feels the magnetic field!)

Homework Equations

\DeltaQ = mc\DeltaT

The Attempt at a Solution

I'm not really clear on why the temperature would be changing at all. Gravitational energy would be changed to kinetic energy as it falls. I suppose that as it falls the magnetic flux through the circuit changes and induces a voltage but I'm confused to what exactly this would do, and how it would affect temperature.

Any help would be appreciated!

 

[henryc09]

OK been thinking about this a bit and have come up with this:

if Z is the position of the bottom of the circuit,

Z(t) = 1/2*g*t²

the change in magnetic flux through the surface, d\phi/dt = d/dt \intB.ds = -V (potential difference).

I think then that \intB.ds = BlZ = 0.5 Blgt².

Differentiating this gives V = Blgt

Can then consider the power dissipated by the resistance as V²/R = B²l²g²t²/R

The work done on the resistance is equal to \DeltaQ and so integrating the power between 0 and t gives \DeltaQ and hence the change in temperature, as long as the flux is still changing, i.e. the cicuit has not all entered the magnetic field yet. The time this stops can be worked out from standard constant acceleration formulae.

Can anyone confirm if I have done this correctly or not?