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Energy Stored in Inductor
When is applied to an inductor of inductance L the current in it grows from zero to maximum steady value I0. If I did the current at any instant t then the magnitude of induced developed in the inductor at that instant is
E = L (dI/dt)
The self induced is also called the back as it opposes any change in the current in the circuit.
Physically, the self inductance plays the role of inertia. It is the electromagnetic analogue of mass in mechanics. Therefore, work needs to be done against the back E in establishing the current. This woke done is stored as magnetic potential energy.
For the current I at an instant t, the rate of doing work is dW/dt = EI
If we ignore the resistive losses, and consider only inductive effect then
Using the eqn. we have,
dW/dt = EI = L (dI/dt) × I or dW = LI dI
Total amount of work done in establishing the current I is
W = ∫ dW = 0∫1dI = ½ LI2
Thus energy required to build up current in an inductor is
UB = W = ½ LI2
(a) Obtain an expression for magnetic energy stored in a solenoid in terms of magnetic field B, and A and length I of the solenoid.
(b) How hoes this magnetic energy compares with the electrostatic energy stored in a capacitor.
(a) From the above equation the magnetic energy is
UB = ½ LI2
From the above equation B = μ0 (NI/I), therefore, (BI/μ0) N
∴ UB = ½ L (BI/μ0 N)2 = ½ (μ0 N2 A/ I) (BI/μ0 N)
UB = (B2/2μ0) AI
(b) The volume that contains flux V = A X I
∴ Magnetic energy per unit volume,
uB = (UB/V) = B2/2μ0
it is known that electrostatic energy stored per unit volume in a parallel plate capacitor
u E = ½ e0 E2
In both the cases energy is directly proportional to eh square of the field strength.
Note Eq. has been derived for special cases- a solenoid and a parallel plate capacitor respectively. But they are valid for any region of space in which a magnetic field or/ and an electric field exist.
A 1.5 mH inductor in LC circuit stores a maximum energy of 17 μJ. What is the peak current?
Here, L = 1.5 (m/h) = 1.5 × 10-3 H,
E0 = 17 μJ = 17 X 10-6 J, I0 =?
From Eo = ½ LI20
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