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Sahay Sir > Question Answers > Uncategorised (JEE Advanced Physics by BM Sharma + GMP Solutions) > A system consists of two coaxial current carrying circular loops as shown in Fig. The self inductance of loop 1 is L1 and self inductance of loop 2 is L2. Magnitude of mutual inductance is M. If at any instant current flowing through loop 1 and loop 2 are i1 and i2 respectively, then total magnetic energy of the system is

A system consists of two coaxial current carrying circular loops as shown in Fig. The self inductance of loop 1 is L1 and self inductance of loop 2 is L2. Magnitude of mutual inductance is M. If at any instant current flowing through loop 1 and loop 2 are i1 and i2 respectively, then total magnetic energy of the system is

(assume that the internal resistance of the battery and of the connecting wire are negligible) In the circuit shown in fig the time constant of the two braches are equal(=T). Then if the key S is closed at the instant t=0 the time in which the current in the circuit through the battery will rise to its final value of (E/R) will be