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 = `\I_1` = 200 mA = 200ｘ10⁻³ A = 0.2 A

Final current = `\I_2` = 100 mA = 100ｘ10⁻³ A = 0.1 A

Change in current = Δ`\I` = 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 `\frac {△I}{Δt}`

or

L = `\frac {Ɛ}{{△I}/{Δt}}`

or

L = `\frac {Ɛ Δt}{△I}`

by putting the corresponding values

L = `\frac {40 ｘ10⁻³ V ｘ 0.005 s}{0.1 A}`

L = 2 ｘ10⁻³ H 

or

L = 2 mHV 



------------------------------------

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

ΔE = `\frac {1}{2}` L `\I₂^2` - `\frac {1}{2` L `\I₁^2`

or

ΔE = `\frac {1}{2}` L (`\I₂^2` -  `\I₁^2`)

Putting the corresponding values

ΔE = `\frac {1}{2}` 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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