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In Figs. (a) and (b), two air-cored solenoids P and Q have been shown. They are placed near each other. In Fig. (a), when Ip the current in P, changes at the rate of 5 A/s, an emf of 2 mV is induced in Q.The current in P is then switched off, and a current changing at 2 A/s is fed through Q as shown in diagram. What emf will be induced in P?

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a

8×10-4V

b

2×10-3V

c

5×10-3V

d

8×10-2V

answer is A.

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Detailed Solution

The mutual inductance M between solenoid P and Q is given by emf induced in Q due to changing current inP = M × (rate of change of current Ip in P)2×10-3=M×5∴M=4.0×10-4HSimilarly, emf induced in P due to changing current inQ = M × (rate of change of current IQ in Q)⇒ induced emf in P = 4.0×10-4×2=8.0×10-4V

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In Figs. (a) and (b), two air-cored solenoids P and Q have been shown. They are placed near each other. In Fig. (a), when Ip the current in P, changes at the rate of 5 A/s, an emf of 2 mV is induced in Q.The current in P is then switched off, and a current changing at 2 A/s is fed through Q as shown in diagram. What emf will be induced in P?