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Unformatted text preview: Massachusetts Institute of Technology Department of Physics 8.022 Spring 2004 Solution 8: Induction & Inductors. TA: Yi Mao, ymao@mit.edu. 1. Purcell 7.11: Mutual and self inductance of coils. Refer to the figure for Problem 7.11, p.288 of Purcell. In the figure (a) that shows positive current and positive electromotive force, the signs are minus in both equations: E 1 = L 1 dI 1 dt M dI 2 dt , and E 2 = L 2 dI 2 dt M dI 1 dt . (1) For example, Coil 2s magnetic field has flux upward through Coil 1. If I 2 increases, the electromotive force in Coil 1 ( E 1 ) should be negative so as to oppose the increasement of the upward flux. If the current and electromotive force of upper coil keep their directions the same but those of lower coil reverse their directions, both signs in eq.(1) are reversed: E 1 = L 1 dI 1 dt + M dI 2 dt , and E 2 = L 2 dI 2 dt + M dI 1 dt . (2) You can see this point either by the similar argument as we had for previous case based on Lenzs law, or simply change in eq.(1) I 2  I 2 and E 2 E 2 . In the connectin of figure (b), the situation falls into the the first case described by eq.(1) with I 1 = I 2 = I ; so E = E (1) 1 + E (1) 2 = L dI dt , (3) where L = L 1 + L 2 + 2 M. (4) (5) The superscript (1) (or (2)) just indicates that quantities are given by eq.(1) (or (2)). The situation for figure (c) falls into the the second case described by eq.(2) with I 1 = I 2 = I ; so E 00 = E (2) 1 + E (2) 2 = L 00 dI dt , (6) where L 00 = L 1 + L 2 2 M. (7) (8) Since M > 0, we have L > L 00 . The selfinductance must be positive for any circuit, again due to the Lenzs law that says the electromotive force is always directed so as to oppose the change in current. Therefore, we have L 1 + L 2 > 2 M . 2. Purcell 7.17: LR circuit....
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 Spring '08
 Hughes
 Inductance, Mass

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