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Question
An inductor of inductance L carries an electric current i. How much energy is stored while establishing the current in it?
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Solution
When current is established in the circuit, inductance opposes the growth of current. So, work is done against this opposition by external agency. This work done is stored as magnetic potential energy.
Induced emf at any instant is
ε = - L`"di"/"dt"`
Work done dW = -ε dq = -εidt
∴ dq = idt
Substituting for ε from first equation
ε = - L`"di"/"dt"`
dw = `- (- "L" "di"/"dt")"i" "dt"`
dw = Li di
Total work done
W = ∫ dw = `int_0^"i" "Li di" = "L" ["i"^2/2]`
W = `1/2 "Li"^2`
This work done is stored as magnetic potential energy
∴ UB = `1/2` Li2
Energy density is the energy stored per unit volume of space
UB = `"U"_"B"/"Al"`
∴ Volume of solenoid = Al
`"u"_"B" = "Li"^2/(2"Al") = ((mu_0"n"^2 "Al")"i"^2)/(2"Al")`
∵ L = `mu_0"n"^2"Al"`
`= (mu_0 "n"^2"i"^2)/2`
`"u"_"B" = "B"^2/(2mu_0)` ∵ B = `mu_0"ni"`
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