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Chapter 27: Problem 50

During lab, you're given a circular wire loop of resistance \(R\) and radius \(a\) with its plane perpendicular to a uniform magnetic field. You're supposed to increase the field strength from \(B_{1}\) to \(B_{2}\) and measure the total charge that moves around the loop. Your lab partner claims that the details of how you vary the field will make a difference in the total charge; your hunch is that it won't. By integrating the loop current over time, determine who's right.

Short Answer

Expert verified
The total charge \(Q\) that moves around the loop is \(Q = - \frac {\pi a^2} {R} (B_2 - B_1)\). This does not depend on how the magnetic field is varied over time, so the student's hunch is correct.

Step by step solution

01

State Faraday's Law

Faraday's Law says that the induced emf in a circuit loop is equal to the negative of the rate of change of the magnetic flux. We write this as \(- \frac {d\Phi_{B}} {dt} = \varepsilon\). Here, \(\varepsilon\) is the induced emf and \(\Phi_{B}\) is the magnetic flux.
02

Calculate Induced Emf

The flux through the loop is \(\Phi_B = B \pi a^2\). Differentiating, \(\frac {d\Phi_{B}} {dt} = \pi a^2 \frac {dB} {dt}\). Substituting this into the earlier obtained equation from Faraday's Law, we get \(\varepsilon = - \pi a^2 \frac {dB} {dt}\)
03

Calculate Total Charge

The current \(I\) in the loop is the induced emf divided by the resistance \(R\), that is \(I = \frac {\varepsilon} {R} = - \frac {\pi a^2} {R} \frac {dB} {dt}\). The total charge \(Q\) that moves around the loop is obtained by integrating this current over time, from \(B_1\) to \(B_2\) which is \(Q = \int I \, dt = - \frac {\pi a^2} {R} \int_{B_1}^{B_2} dB = - \frac {\pi a^2} {R} (B_2 - B_1)\). We can see that the total charge \(Q\) depends only on the initial and final magnetic fields \(B_1\) and \(B_2\), the radius of the loop \(a\), and the resistance \(R\), not on how \(\frac {dB} {dt}\) varies with time. Therefore, the student's hunch is correct.

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