A system consists of two coaxial current carrying circular loops as shown in Fig. The self inductance of loop 1 is `L_(1)` and self inductance of loop 2 is `L_(2)`. Magnitude of mutual inductance is M. If at any instant current flowing through loop 1 and loop 2 are `i_(1) and i_(2)` respectively, then total magnetic energy of the system is A. `1/2 L_(1)i_(1)^(2)+1/2L_(2)i_(2)^(2)-|M|i_(1)i_(2)`B. `1/2 L_(1)i_(1)^(2)+1/2L_(2)i_(2)^(2)+|M|i_(1)i_(2)`C. `1/2 L_(1)i_(2)^(2)+1/2L_(2)i_(2)^(1)-|M|i_(1)i_(2)`D. `1/2 L_(1)i_(1)^(2)+1/2L_(2)i_(2)^(2)`

A system consists of two coaxial current carrying circular loops as sh

1.	A system consists of two coaxial current carrying circular loops as shown in Fig. The self inductance of loop 1 is `L_(1)` and self inductance of loop 2 is `L_(2)`. Magnitude of mutual inductance is M. If at any instant current flowing through loop 1 and loop 2 are `i_(1) and i_(2)` respectively, then total magnetic energy of the system is A. `1/2 L_(1)i_(1)^(2)+1/2L_(2)i_(2)^(2)-\|M\|i_(1)i_(2)`B. `1/2 L_(1)i_(1)^(2)+1/2L_(2)i_(2)^(2)+\|M\|i_(1)i_(2)`C. `1/2 L_(1)i_(2)^(2)+1/2L_(2)i_(2)^(1)-\|M\|i_(1)i_(2)`D. `1/2 L_(1)i_(1)^(2)+1/2L_(2)i_(2)^(2)`
Answer» Correct Answer - A Notice that meganetic field due to left coil is towards left. Since magnetic field due to right coil is also towards left, so it increases flux associated with left coil. To oppose it, induced flux associated with left coil. To oppose it, induced current due to mutual induction opposes the original current in left coil.

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