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**Unformatted text preview: **CHAPTER 32—INDUCTANCE 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, Lenz’s 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 π 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 π in Equation 32-2. Then E 2 1 2 2 = - = - M dI dt fMI ft p = π π cos , and the peak emf (when cos ) 2 1 π ft = ± is 2 2 60 440 75 12 π π 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 π 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 = π π cos( ). (b) The peak value of the cosine is 1, so M = E 2 2 50 2 60 1 133 p p fI = = π π = × = 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 = μ 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 π π H m m cm mH = Problem 7. Two long solenoids of length λ 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....

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