MRI & Impedance: Tune RF Coil for 1.5/3.0 Tesla

[Lucille]

Homework Statement

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/questions-and-answers/magnetic-resonance-imaging-mri-use-radiofrequency-rf-coils-transmit-oscillating-magnetic-f-q3608466

Homework Equations

Z_c = 1/(jwC)

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?

 

[Lucille]

Wait for b:

would it be C = 4/[(3125 ohms)(128 Mhz) = 1*10^-11 C

 

[gneill]

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.

 

[Lucille]

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?

 

[gneill]

Reactance is not inductance. Inductance is measured in Henries. For part (b) the same inductor is used but it will have a different reactance.

 

[Lucille]

Oh, okay. So then for b:

Z = 497 = 4/[C(128MHz)(2pi)] ---> C = 4/[(497)(2pi)(128MHz)] = 10 pF

 

[gneill]

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.

 

[Lucille]

how am i to find that?

 

[gneill]

how am i to find that?

What formula would you use to find the reactance of an inductor?

 

[Lucille]

Z_c = 1/(jwC)

But we're asked to find the capacitance, so it would be 2 unknowns?

 

[gneill]

Z_c = 1/(jwC)

No, that's the formula for the impedance of a capacitor. You want the similar one for an inductor.

But we're asked to find the capacitance, so it would be 2 unknowns?

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.

 

[Lucille]

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

 

[gneill]

So then:

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

Z = jwL so L = 1.24 H

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.)

b) L = 1.24 H so Z = jwL = 2Pi(128 MHz)(1.24) = 1/(jwC) --> C = 1.25*10^-16 F

Okay, now remember that there are four capacitors. What you've calculated so far is their net (or equivalent) capacitance.

 

[Lucille]

Alright: so for b then I would have to divide by 4 to get their individual capacitance so 3.125*10^-17 F

 

[Lucille]

Wait in part a : L = 1.24 * 10 ^-6 H

So that changes the capacitance to 4.99*10^-10 F

 

[gneill]

Alright: so for b then I would have to divide by 4 to get their individual capacitance so 3.125*10^-17 F

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.

 

[gneill]

Wait in part a : L = 1.24 * 10 ^-6 H

So that changes the capacitance to 4.99*10^-10 F

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.

 

[Lucille]

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.

Sorry, I meant that the new capacitance that I was solving for is 4.99*10^-10 F

 

[gneill]

Sorry, I meant that the new capacitance that I was solving for is 4.99*10^-10 F

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.

 

[gneill]

Sorry, I meant that the new capacitance that I was solving for is 4.99*10^-10 F

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.

 

[Lucille]

** i meant 5 E-12

 

[Lucille]

if you don't mind could you help me with this question? https://www.physicsforums.com/threads/bandpass-filter.799647/