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

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 2.0 T  to 0 T in 0.4 s , as it might in an MRI solenoid.(Figure 1)

Part A

What is the induced emf in the loop?

Express your answer using three significant figures using the proper unit.

Part B

What is the power dissipated by the loop while the magnetic field is changing?  Hint: Given the resistivity of muscle tissue, the loop would have a resistance of 41.6k.

Part C

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.

Part D

By how much will the temperature of the tissue increase?  Assume that muscle tissue has specific heat 3600J/(kgC) .  Hint: Given the density of muscle tissue, the loop would have a mass of 22g.

Express your answer using three significant figures.

Part E

This question will be shown after you complete previous question(s).

Explanation / Answer

A) induced emf e = rate of change of magnetic flux = A*(dB/dt) = (3.142*(0.01/2)^2)*(2/0.4) = 3.92*10^-4 V

B) power dissipated is P = e^2/R = (3.92*10^-4)^2/(41.6*1000) = 3.69*10^-12 W

C) heat created is 3.69*10^-12 J/sec

so heat created in 0.4 sec is (3.69*10^-12*0.4) = 1.476*10^-12 J

D) dQ = 1.476*10^-12 = m*S*dT

change in tepmeprature is dT = (1.476*10^-12)/(22*10^-3*3600) = 1.86*10^-14 deg C

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