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Chapter 27: Problem 40

A stent is a cylindrical tube, often made of metal mesh, that's inserted into a blood vessel to overcome a constriction. It's sometimes necessary to heat the stent after insertion to prevent cell growth that could cause the constriction to recur. One method is to place the patient in a changing magnetic field, so that induced currents heat the stent. Consider a stainless- steel stent 12 mm long by 4.5 mm diameter, with total resistance \(41 \mathrm{m} \Omega\). Treating the stent as a wire loop in the optimum orientation, find the rate of change of magnetic field needed for a heating power of \(250 \mathrm{mW}.\)

Short Answer

Expert verified
After the calculations, the rate of change of magnetic flux, and hence magnetic field, is equal to -E (substitute the value of E calculated in step 4), is needed to heat the stent with a power of 250 mW.

Step by step solution

01

Understand the nature of power in the loop

Power \(P\) in an electric circuit is given by \(\frac{V^2}{R}\), where \(V\) is the voltage and \(R\) is the resistance. Since we are dealing with induced electromotive force (emf) rather than traditional voltage, we can use the similar power equation \(P = \frac{E^2}{R}\), where \(E\) is the emf and \(R\) is the resistance.
02

Solve the power equation for the squared emf

We rearrange the above equation to solve for emf: \(E^2 = P \cdot R\).
03

Substitute the known values

We know the heating power \(P\) is \(250 \mathrm{mW}\) (or \(0.25 \mathrm{W}\)) and the resistance \(R\) is \(41 \mathrm{m}\Omega\) (or \(0.041 \Omega\)). Substituting these values gives us \(E^2 = 0.25 \mathrm{W} \times 0.041 \Omega\). Now, calculate the value of \(E^2\).
04

Find the emf

To solve for \(E\), (the voltage equivalent in this context), we need to take the square root of the previous answer: \(E = \sqrt{0.25 \mathrm{W} \times 0.041 \Omega}\). Calculate the value of \(E\).
05

Apply Faraday’s law

The emf (E) induced in a loop by a changing magnetic field is given by \(E = -N \frac{d\Phi}{dt}\), where \(N\) is the number of turns in the coil (for our single loop, \(N = 1\)), \(\Phi\) is the magnetic flux, \(t\) is time, and \(\frac{d\Phi}{dt}\) represents the rate of change of magnetic flux. As we want to find the rate of change of the magnetic field \(\frac{d\Phi}{dt}\), we can rearrange the formula to \(\frac{d\Phi}{dt} = \frac{E}{-N}\). And since \(N = 1\), this simplifies to \(\frac{d\Phi}{dt} = -E\).
06

Substitute the emf value

With \(E\) found, it can be substituted back into the formula of the rate of change of magnetic flux, giving us \(\frac{d\Phi}{dt} = -E\). This will give us the answer with respect to electromagnetic induction.

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Most popular questions from this chapter

A \(1-\) H inductor carries \(10 \mathrm{A},\) and a \(10-\mathrm{H}\) inductor carries \(1 \mathrm{A}\). Which contains more stored energy?

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