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Unformatted text preview: CHAPTER 31—ELECTROMAGNETIC INDUCTION ActivPhysics can help with these problems: Activities 13.9, 13.10 Sections 31- 2 and 31-3:—Faraday’s Law and Induction and the Conservation of Energy Problem 1. Show that the volt is the correct SI unit for the rate of change of magnetic flux, making Faraday’s law dimensionally correct. Solution The units of d dt B φ = are T m s (N/A m)(m /s) (N m/A s) J/C V 2 2 ⋅ = ⋅ = ⋅ ⋅ = = / . Problem 2. A bar magnet is moved steadily through conducting rings, as shown in Fig. 31-44. Sketch qualitatively the current and power dissipation in the ring as functions of time. Take as positive a current flowing out of the plane of the page at the top of the ring, and indicate the position of the magnet on your time axis. FIGURE 31-44 Problem 2. Solution Let t = 0 when the magnet is midway through the ring (center picture in Fig. 31-44). The flux through the ring increases as the magnet approaches ( ) t < 0 and decreases as it recedes ( ). t The current is proportiooal to the negative time derivative of the flux, hence reverses direction at t = 0 as shown. (The details of the reversal depend on the dimensions and other properties of the magnet. The sketches are for a small magnet, which spends a time ∆ t going through the ring.) The power dissipated is proportional to the square of the current. Problem 2 Solution. 2 CHAPTER 31 Problem 3. Find the magnetic flux through a circular loop 5 0 . cm in diameter oriented with the loop normal at 30 ° to a uniform 80-mT magnetic field. Solution For a stationary plane loop in a uniform magnetic field, the integral for the flux in Equation 31-1 is just φ B = ⋅ = B A BA cos ( ) ( . ) cos . . θ π = ° = ×- 80 2 5 30 136 10 2 4 mT cm Wb (The SI unit of flux, T m 2 ⋅ , is also called a weber, Wb.) Problem 4. A circular wire loop 40 cm in diameter has 100- Ω resistance and lies in a horizontal plane. A uniform magnetic field points vertically downward, and in 25 ms it increases linearly from 5 0 . mT to 55 mT. Find the magnetic flux through the loop at (a) the beginning and (b) the end of the 25 ms period. (c) What is the loop current during this time? (d) Which way does this current flow? Solution (a) As in the previous solution, φ π π B d B t = ⋅ = = = × =- B A 1 4 2 1 4 2 4 1 40 5 6 28 10 ( ) ( ) . , cm mT Wb at and (b) 6 91 10 25 3 2 . × =- Wb at ms. t (c) Since the field increases linearly, d dt t B B φ φ = = = =- × ∆ ∆ ( . . ) 6 91 0 628 10 25 3- = Wb ms 0.251 V. = From Faraday’s law, this is equal to the magnitude of the induced emf, which causes a current I R = = = E= = 0 251 100 2 51 . . V mA Ω in the loop. (d) The direction must oppose the increase of the external field downward, hence the induced field is upward and I is CCW when viewed from above the loop....
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- Spring '08
- Physics, Magnetic Field, Faraday