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Prove that energy is conserved during e...

Prove that energy is conserved during electromagnetic induction.

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i. During LC osillation in LC circuits, the energy of the system oscillates between the electric field of the capacitor and the magnetic field of the inductor. Although, these two forms energy remains constant. It means that LC oscillations take place in accordance with the law of conservation of energy.
Total energy, `U=U_(E)+U_(B)+(q^(2))/(2C)+(1)/(2)Li^(2)`
Let us consider 3 different stages of LC oscillations and calculate the total energy of the system.
Case i. When the charge in the capacitor, `q=Q_(m)`, and the current through the inductor, i = 0, the total energy is given by
`U=(Q_(m)^(2))/(2C)+0=(Q^(2))/(2C)`
The total energy is wholly electrical.
Case ii. When charge = 0 , current = `I_(m)`, the total energy is
`U=0+(1)/(2)LI_(m)^(2)=(1)/(2)LI_(m)^(2)`
`(1)/(2)xx((Q_(m)^(2))/(LC))" since "I_(m)=Q_(m)omega=(Q_(m))/(sqrt(LC))`
`=((Q_(m)^(2))/(LC))`
Case iii. When charge =q, current = i, the total energy is
`U=(q^(2))/(2C)+(1)/(2)Li^(2)`
Since `q=Q_(m)=cosomegat,`
`i=(dq)/(dt)=Q_(m)omegasinomegat`
The negative sing in current indicates that the charge in the capacitor decreases with time.
`U=(Q_(m)^(2)cos^(2)omegat)/(2C)+(Lomega^(2)Q_(m)^(2)sin^(2)omegat)/(2)`
`=(Q_(m)^(2))/(2C)(cos^(2)omegat+sin^(2)omegat)`
`U=(Q_(m)^(2))/(2C)`
From above three cases, it is clear that the total energy of the system remains constant.
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