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1. Coil 1 has an inductance of 4.50×10-2 H and 140 turns. Coil 2 has an inductance of 8.00×10^-2 H and 230 turns. The coils are rigidly positioned with respect to each other, their mutual inductance is 5.00×10^-3 H. A 7.00×10^-3 A current in coil 1 is changing at a rate of 7.0 A/s.
What flux links coil 2?
2. M = N2 * Flux Linkage / i1
3. Mutual inductance = 5*10^-3
Current in first wire = 7 * 10^-3
Turns in second wire = 230
Flux linkage through second wire = 5 * 10^-3 * 7 * 10^-3 / 230 = 1.52*10^-7 Wb
Solution doesn't work.
What flux links coil 2?
2. M = N2 * Flux Linkage / i1
3. Mutual inductance = 5*10^-3
Current in first wire = 7 * 10^-3
Turns in second wire = 230
Flux linkage through second wire = 5 * 10^-3 * 7 * 10^-3 / 230 = 1.52*10^-7 Wb
Solution doesn't work.