By a change of current from 5 A to 10 A in 0.1 s, the self induced emf is 10 V. The change in the energy of the magnetic field of a coil will be

Options

A. `5J`
B. `6J`
C. `7.5J`
D. `9J`

Answer

C

Explanation

` |epsi|=L(DeltaI)/(Deltat)`
`L=(|epsi|)/(DeltaI)=(10xx0.1)/((10-5))=0.2H`
The magnetic field energies for currents `I_(1) and I_(2)` are `U_(1)=(1)/(2)LI_(1)^(2) and U_(2)=(1)/(2)LI_(2)^(2)`
Change I energy `= U_(2)-U_(1)`
`=(1)/(2)LI_(2)^(2)=-(1)/(2)LI_(2)^(2)=(L)/(2)(I_(2)^(2)-I_(1)^(2))`
`=(0.20)/(2)(10^(2)-5^(2))7.5J`