# Energy dissipated by a loop of muscle in an MRI machine

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1. Dec 22, 2015

### kostoglotov

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

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

Having issues with part a)

2. Relevant 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: $$R = \frac{\rho L}{A} = \frac{\rho \pi D}{\pi \left(\frac{d}{2}\right)^2} = \frac{\rho D}{ \left(\frac{d}{2}\right)^2}$$

3. 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?

Last edited: Dec 22, 2015
2. Dec 22, 2015

### Staff: Mentor

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

3. Dec 22, 2015

### kostoglotov

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