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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Infinity Learn · Accepted Answer

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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