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Energy dissipated by a loop of muscle in an MRI machine

  1. Dec 22, 2015 #1
    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)


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

    2. Relevant equations

    Resistivity of muscle (from table in text): [tex]\rho \approx 13[/tex]

    Large Diameter, Small Diameter: [tex]D = 0.080m \ \ \ d = 0.010m[/tex]

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

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

    Effective Area (assume [itex]\theta = 0[/itex]): [tex]A_{eff} = A \cos{\theta} = A = \pi \left(\frac{D}{2}\right)^2[/tex]
    Resistance: [tex]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}[/tex]

    3. The attempt at a solution

    So, energy expression: [tex]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}[/tex]

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

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

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

    Where have I gone wrong?
    Last edited: Dec 22, 2015
  2. jcsd
  3. Dec 22, 2015 #2


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    Staff: Mentor

    The numbers differ by a factor of 4 pi, but I don't see wrong factors in your approach.
  4. Dec 22, 2015 #3
    Yeah, that's because I have what I like to call "stupid fingers"...it was a calculator typo...
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