When the current through a cell changes from 100 mA to 200 mA in 0.005 sec, an induced emf of 40 mV is produced in the coil.
(a) What is the self-inductance of the coil?
(b) Find the increase in the energy stored in the coil. (Answer: 2 mH, 0.03 mJ)



Data Given:

Initial current = I1 = 200 mA = 200ｘ10⁻³ A = 0.2 A

Final current = I2 = 100 mA = 100ｘ10⁻³ A = 0.1 A

Change in current = ΔI = 0.2 A - 0.1 A = 0.1 A 

Change in time interval = △t = 0.005 s

Induced emf  = Ɛ = 40 mV = 40 ｘ10⁻³ V



To Find:


Self-inductance = L = ?

Increase in Energy = △E = ?

Solution:


Using the following relation:

Ɛ =  L △IΔt

or

L = Ɛ△IΔt

or

L = ƐΔt△I

by putting the corresponding values

L = 40ｘ10⁻³Vｘ0.005s0.1A

L = 2 ｘ10⁻³ H 

or

L = 2 mHV 



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Now the following elation for finding the change in energy store in an inductor 

ΔE = 12 L I₂2 - 12 L I₁2

or

ΔE = 12 L (I₂2 -  I₁2)

Putting the corresponding values

ΔE = 12 ｘ 2 ｘ10⁻³ H ((0.2 A)² -  (0.1 A)²)

ΔE = 1 ｘ10⁻³ H (0.04 A² -  0.01 A²)

ΔE = 1 ｘ10⁻³ H ｘ 0.03 A² 

ΔE = 0.03 ｘ10⁻³ J

or

ΔE = 0.03 mJ ----------------Ans.2


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