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Consider I 1 and I 2 are the currents flowing simultaneously in two nearby coils 1 & 2, respectively. If L 1 = self inductance of coil 1, M 12 = mutual inductance of coil 1 with respect to coil 2, then the value of induced emf in coil 1 will be :

1. ε 1 = –L 1 dt dI 1 + M 12 dt dI 2
2. ε 1 = –L 1 dt dI 1 –M 12 dt dI 1
3. ε 1 = –L 1 dt dI 1 – M 12 dt dI 2
4. ε 1 = –L 1 dt dI 2 –M 12 dt dI 1

Correct answer: ε 1 = –L 1 dt dI 1 – M 12 dt dI 2

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