Currents induced by rapid field changes in an MRI solenoid can in some cases, he

Question

Currents induced by rapid field changes in an MRI solenoid can in some cases, heat tissues in the body, but under normal circumstances the heating is small. We can do a quick estimate to show this. Consider the "loop" of muscle tissue shown in the figure. This might be muscle circling the bone of your arm or leg. Muscle tissue is not a great conductor, but current will pass through muscle and so we can consider this a conducting loop with a rather high resistance. Suppose the magnetic field along the axis of the loop drops from 1.3 T to 0 Tin 0.2 s as it might in an MRI solenoid. How much heat is created by the loop while the magnetic field is changing? (Assume that all of the power dissipated is changed into heat.) Express your answer using three significant figures and include the appropriate unit. By how much will the temperature of the tissue increase? Assume that muscle tissue has specific heat 3600 J/(kg middot C degree). Express your answer using three significant figures.

Explanation / Answer

I will be using notations, (l) for loop and (w) for wire or muscle in this case, (d) is for delta and ne(x) where n is a number e(x) is 10^x, so n*10^x. And I will not be using units, I put them in SI.
First, we have to write down what we are given in this problem, those are as follows,

B(i)=1.3
B(f)=0
dB= -1.3
dt=0.2
p= 13 (this is resistivity)
rho=1.1e(3) (this is density)
c=3600 (specific heat)
d(w)=0.01 (this we can clearly see is diameter of the muscle)
d(l)=0.08 (this is not so obvious, especially on your diagram, but it is in fact a diameter of the loop)


Second step would be to actually know what it is that we need to find. Those would be W as in energy dissipated in the loop and dT as a change in temperature of the tissue.

The general formula, that unites W and dT is, c=W/(m*dT), but we can't use it right now, we don't know W, dT and m (mass).

- Lets start by finding W. We know that power P=W/dt, we also know that in electric circuit P=(Emf^2)/R.
So, we have W/dt=(Emf^2)/R. Now, we can derive equation for W, W=(dt*Emf^2)/R.
There are two more things we need to know in order to find W, those are induced Emf and R.

Let us find R first. As you know R=pL(w)/A(w). L(w)=2pi*r(l)=pi*d(l)= 0.25132. It's a circle, remember the formula for length of a circle. A(w)=pi*r(w)^2= 7.85e(-5). Now, we can calculate R, which should be R=41619.87. This is high resistance indeed.

Now we go to Emf. Induced Emf=| (change if magnetic flux)/dt |. This formula is in module, so the answer is always positive. dflux=A(l)*dB= -6.53e(-3) and dt=0.2. Emf= 0.0268

And finally, we can find W=(dt*Emf^2)/R= 3.451e(-9).

It will show as 3.45e(-9) on calculator, multiply by e(-9) for more significant figures.

- The final step is to find dT. We derive a formula for it from general formula, which is dT=W / (c*m). But what is m?
As you know, rho=m(w) / V(w), so m=rho*V.

Volume V(w)=A(w)*L(w)= 1.97e(-5), so m=0.0217. Then we get dT= 6.65e(-11)

final answer is

Part c)

3.451e(-9) J

Part D)

6.65e(-11) K.

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.