Energy dissipated by a loop of muscle in an MRI machine

[kostoglotov]

edit: turned out to be a calculator typo...mods feel free to delete this thread if you wish, I won't complain.

Homework Statement

[/B]
Having issues with part a)

imgur link: http://i.imgur.com/4hzLyhb.jpg

Homework Equations

Resistivity of muscle (from table in text): \rho \approx 13

Large Diameter, Small Diameter: D = 0.080m \ \ \ d = 0.010m

Energy dissipated: Q = P \Delta t
Power: P = \frac{V^2}{R}
EMF: V = \varepsilon = \left|\frac{\Delta \Phi}{\Delta t}\right|

Change in Flux: \left|\frac{\Delta \Phi}{\Delta t}\right| = \left|\frac{\Delta B}{\Delta t}\right| A_{eff}

Effective Area (assume \theta = 0): A_{eff} = A \cos{\theta} = A = \pi \left(\frac{D}{2}\right)^2
Resistance: R = \frac{\rho L}{A} = \frac{\rho \pi D}{\pi \left(\frac{d}{2}\right)^2} = \frac{\rho D}{ \left(\frac{d}{2}\right)^2}

The Attempt at a Solution

So, energy expression: Q = P \Delta t = \frac{\varepsilon^2 \Delta t}{R} = \left[\frac{\Delta B}{\Delta t}\pi \left(\frac{D}{2}\right)^2\right]^2 \Delta t \frac{d^2}{4 \rho D}

Simplifying: Q = \frac{(\Delta B)^2 \pi^2 D^3 d^2}{64 \Delta t \rho}

Plugging all the values from the problem in gives me: 6.5 \times 10^{-8} J with 2 sig figs.

The answer in the back is stated as: 5.2 \times 10^{-9} J

Where have I gone wrong?

 

[mfb]

The numbers differ by a factor of 4 pi, but I don't see wrong factors in your approach.

 

[kostoglotov]

The numbers differ by a factor of 4 pi, but I don't see wrong factors in your approach.

Yeah, that's because I have what I like to call "stupid fingers"...it was a calculator typo...