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MRI and impedance

  1. Feb 23, 2015 #1
    1. The problem statement, all variables and given/known data
    In magnetic resonance imaging (MRI), we use radiofrequency (RF) coils to transmit oscillating magnetic fields into the patient. The coils are essentially series RLC antennas and must be tuned to resonate at the characteristic frequency of the experiment (called the Larmor frequency) for efficient transmission. This frequency changes as a function of the field strength of the MRI machine. At 1.5 Tesla field strength, it is 64 MHz and at 3.0 Tesla, it is 128 MHz. A simple RF coil is a loop of wire (which provides the inductance) broken by capacitors in series.

    a)If the value of each of the four capacitors is 20 pF for a coil tuned to operate in a 1.5 Tesla MRI, what is the inductance of the coil?

    b) If you wanted to use the same coil at 3.0 Tesla, what new value would you need to use for the capacitors?

    the image can be found at http://www.chegg.com/homework-help/...oils-transmit-oscillating-magnetic-f-q3608466
    2. Relevant equations

    Z_c = 1/(jwC)

    3. The attempt at a solution

    a) as impedances add in series:

    Z = 4 * 1/[(64MHz)(20pF)] = 3125 ohms

    b) setting Z = 3125 and w = 128 MHz and solving for C:

    C = 1/[(3125 ohms)(128 Mhz) = 2.5*10^-12 C

    --> I just wanted to confirm if my answer was correct - it seems a little too simple. Am I missing something?
     
  2. jcsd
  3. Feb 23, 2015 #2
    Wait for b:

    would it be C = 4/[(3125 ohms)(128 Mhz) = 1*10^-11 C
     
  4. Feb 23, 2015 #3

    gneill

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    Staff: Mentor

    In parts (a) and (b) you need to use either the angular frequency ω or multiply the frequency f by the factor ##2\pi## when calculating the reactance of the capacitors. Also for part (a) you are asked to find a value for the inductance, which would be in Henries.

    In part (b) note that the reactance of the inductor will change with the frequency. The reactance you found in part (a) will not be the same here.
     
  5. Feb 23, 2015 #4
    Okay, so:

    a) Z = 497 H

    b) I'm a little confused as to how to solve for the capacitance if the impedance is different?
     
  6. Feb 23, 2015 #5

    gneill

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    Staff: Mentor

    Reactance is not inductance. Inductance is measured in Henries. For part (b) the same inductor is used but it will have a different reactance.
     
  7. Feb 23, 2015 #6
    Oh, okay. So then for b:

    Z = 497 = 4/[C(128MHz)(2pi)] ---> C = 4/[(497)(2pi)(128MHz)] = 10 pF
     
  8. Feb 24, 2015 #7

    gneill

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    I still haven't seen a value for the inductance L for the coil. You are presenting the reactance as though it is an inductance. But reactance has units of Ohms whereas inductance has units of Henries.

    When the coil from part (a) is used at the higher frequency it will have a different value of reactance for the same value of Henries.
     
  9. Feb 24, 2015 #8
    how am i to find that?
     
  10. Feb 24, 2015 #9

    gneill

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    What formula would you use to find the reactance of an inductor?
     
  11. Feb 24, 2015 #10
    Z_c = 1/(jwC)

    But we're asked to find the capacitance, so it would be 2 unknowns?
     
  12. Feb 24, 2015 #11

    gneill

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    No, that's the formula for the impedance of a capacitor. You want the similar one for an inductor.
    In part (a) you are given the capacitance and are asked to find the matching inductance. Your answer should be in Henries.

    In part (b) you take the inductance from part (a) (in Henries) and are asked to find the matching capacitor values for the new operating frequency.
     
  13. Feb 24, 2015 #12
    So then:

    for a) Z = 497 ohms and finding the inductance:

    Z = jwL so L = 1.24 H

    b) L = 1.24 H so Z = jwL = 2Pi(128 MHz)(1.24) = 1/(jwC) --> C = 1.25*10^-16 F
     
  14. Feb 24, 2015 #13

    gneill

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    Yes!

    (A quick aside regarding reactance vs impedance: you might have noticed the "j" in the impedance formulas. This is the same as i, the square root of negative one. Impedance values are complex numbers. Reactance is the magnitude of the imaginary part of impedance. For simple components such as capacitors and inductors the impedance is purely imaginary.

    The reason I mention this is because some markers may take exception if they see an impedance calculation where the "j" is left off the resulting value. For reactance values you use the same formulas but leave out the 'j' in them. The results are always real values in Ohms.)

    Okay, now remember that there are four capacitors. What you've calculated so far is their net (or equivalent) capacitance.
     
  15. Feb 24, 2015 #14
    Alright: so for b then I would have to divide by 4 to get their individual capacitance so 3.125*10^-17 F
     
  16. Feb 24, 2015 #15
    Wait in part a : L = 1.24 * 10 ^-6 H

    So that changes the capacitance to 4.99*10^-10 F
     
  17. Feb 24, 2015 #16

    gneill

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    Ah, so close. Capacitors in parallel add. They combine differently in series; the "sum" is always less than any individual capacitor value. For identical capacitors as we have here you'll want to multiply by 4 rather than divide by 4.
     
  18. Feb 24, 2015 #17

    gneill

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    Which capacitance are you referring to? The value was a given in part (a). For part (b) you had to find a new capacitance to match the same coil operating at the new higher frequency.
     
  19. Feb 24, 2015 #18
    Sorry, I meant that the new capacitance that I was solving for is 4.99*10^-10 F
     
  20. Feb 24, 2015 #19

    gneill

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    That seems too large. It's larger than the original capacitors (20 pF) which were operating at a lower frequency. You should expect smaller values at the higher frequency. pF ---> 10-12 F.
     
  21. Feb 24, 2015 #20

    gneill

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    That seems too large. It's larger than the original capacitors (20 pF) which were operating at a lower frequency. You should expect smaller values at the higher frequency. pF ---> 10-12 F.
     
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