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CHAPTER 32INDUCTANCE AND MAGNETIC ENERGY ActivPhysics can help with these problems: Activity 14.1 Section 32-1:Mutual Inductance Problem 1. Two coils have a mutual inductance of 2.0 H. If current in the first coil is changing at the rate of 60 A/s, what is the emf in the second coil? Solution From Equation 32-2, E 2 1 2 60 120 = - = - = - M dI dt ( ) ( )( / ) . = H A s V (The minus sign, Lenzs law, signifies that an induced emf opposes the process which creates it.) Problem 2. A 500-V emf appears in a coil when the current in an adjacent coil changes at the rate of 3.5 A/ms. What is the mutual inductance of the coils? Solution Equation 32-2 gives M dI dt = = = E 2 1 500 3 5 143 = = = ( ) ( . / ) . V A ms mH (The sign of M depends on how the coils are coupled.) Problem 3. The current in one coil is given by I I ft p = sin , 2 where I f t p = = = 75 60 mA, Hz and time , . Find the peak emf in a second coil if the mutual inductance between the coils is 440 mH. Solution Suppose I I ft p 1 2 = sin in Equation 32-2. Then E 2 1 2 2 = - = - M dI dt fMI ft p = cos , and the peak emf (when cos ) 2 1 ft = is 2 2 60 440 75 12 fMI p = = ( )( )( ) .4 . Hz mH mA V Problem 4. Two coils have a mutual inductance of 580 mH. One coil is supplied with a current given by I t t =- + 3 2 4 2 , where I is in amperes and t in seconds. What is the induced emf in the other coil at time t = 2 5 . s? Solution Since dI dt t 1 6 2 = =- (in A/s), Equation 32-2 gives, for t = = - - = - 2 5 580 6 2 5 2 7 54 2 . , ( )( . )( / ) . s mH A s V E (see comment in solution to Problem 1). Problem 5. An alternating current given by I ft p sin 2 is supplied to one of two coils whose mutual inductance is M . (a) Find an expression for the emf in the second coil. (b) When I p = 10 . A and f = 60 Hz, the peak emf in the second coil is measured at 50 V. What is the mutual inductance? Solution (a) From Equation 32-2, E 2 1 2 2 = - = - M dI dt M f I f t p = cos( ). (b) The peak value of the cosine is 1, so M = E 2 2 50 2 60 1 133 p p fI = = = = V Hz A mH ( )( ) . (From the information given, only the magnitude of M can be determined; its sign depends on how the coils are coupled.) 2 CHAPTER 32 Problem 6. Find the mutual inductance of the two-coil system described in Problem 19 of Chapter 31. Solution The mutual inductance of this configuration of two coils is calculated in Example 32-1 (simply replace 2 by N 2 ): M = 2 1 1 1 N N A ( ) . = The data from Problem 31-19 then yields M = =- ( / )( )( ) ( ) . . 4 10 5 5000 2 30 111 7 1 4 2 H m m cm mH = Problem 7. Two long solenoids of length both have n turns per unit length. They have circular cross sections with radii R and 2 R , respectively. The smaller solenoid is mounted inside the larger one, with their axes coinciding. Find the mutual inductance of this arrangement, neglecting any nonuniformity in the magnetic field near the ends.... View Full Document

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