CourseNotes.44 - 11.6 Problems Problem 30.15 In figure 33,...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 11.6 Problems Problem 30.15 In figure 33, a stiff wire bent into a semicircle of radius a = 2, cm is rotated at constant angular speed f = 40 rev/s in a uniform 20 mT magnetic field. What are the (a) frequency and (b) amplitude of the emf induced in the loop? Figure 33: Figure for problem 30.15 The induced emf is generated by the changing magnetic flux through the circuit. In this case, the changing magnetic flux is not due to the magnetic field changing but rather the amount of area though which the field passes is changing. Since the area oscillates back and forth at the frequency f , the frequency of the emf must be the same. To find the amplitude of the emf we must do some calculations. We want to write down the flux through the loop as a function of time. Lets return to the definition of flux. ΦB = B · dA = BA1 + B dA2 cos θ where θ is the angle between the magnetic field and the normal to the changing surface area (in or out will both work just fine). The first term is the magnetic flux through the square part of the loop which does not change. Since it does not change, it will not contribute to the emf and can therefore be neglected. Integrating over dA2 gives 1 ΦB = B ( πa2 ) cos θ 2 Since the loop is turning, θ is a function of time. If the loop turns at a rate of f , then theta changes at a rate of ω = 2πf Hence, the magnetic flux through the circuit as a function of time is ΦB = 1 Bπa2 cos(2πf t) 2 Finally, from Faraday’s law the induced emf will be E (t) = −Bπ 2 a2 f sin(2πf t) from which we can read off the amplitude as |E| = Bπ 2 a2 f 44 ...
View Full Document

Ask a homework question - tutors are online