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# Energy Stored In Inductor, Electrostatics, Physics Assignment Help

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Electrostatics - Energy Stored In Inductor, Electrostatics, Physics

**Physics Assignment Help >> Electrostatics >> Energy Stored in Inductor**

**Energy Stored in Inductor**

When is applied to an inductor of inductance L the current in it grows from zero to maximum steady value I_{0}. 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}∫^{1}dI = ½ LI^{2}

Thus energy required to build up current in an inductor is

UB = W = ½ LI^{2}

(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 LI^{2}

From the above equation B = μ0 (NI/I), therefore, (BI/μ0) N

∴ UB = ½ L (BI/μ_{0} N)^{2} = ½ (μ_{0} N^{2} A/ I) (BI/μ_{0} N)

UB = (B^{2}/2μ_{0}) AI

(b) The volume that contains flux V = A X I

∴ Magnetic energy per unit volume,

uB = (UB/V) = B^{2}/2μ_{0}

it is known that electrostatic energy stored per unit volume in a parallel plate capacitor

**u E = ½ e**_{0} E^{2}

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,

E_{0} = 17 μJ = 17 X 10^{-6} J, I_{0} =?

**From Eo = ½ LI**^{2}_{0}

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**Popular Tags:** Energy Stored In Inductor Assignment Help, Energy Stored In Inductor Homework Help, Energy Stored In Inductor Tutors, Energy Stored In Inductor Solutions, Energy Stored In Inductor Tutors, Electrostatics Help, Physics Tutors, Energy Stored In Inductor Questions Answers

**Physics Assignment Help >> Electrostatics >> Energy Stored in Inductor**

**Energy Stored in Inductor**

_{0}. If I did the current at any instant t then the magnitude of induced developed in the inductor at that instant is

_{0}∫

^{1}dI = ½ LI

^{2}

^{2}

^{2}

_{0}N)

^{2}= ½ (μ

_{0}N

^{2}A/ I) (BI/μ

_{0}N)

^{2}/2μ

_{0}) AI

^{2}/2μ

_{0}

**u E = ½ e**

_{0}E^{2}

**L = 1.5 (m/h) = 1.5 × 10**,

^{-3}H

E

E

_{0}= 17 μJ = 17 X 10^{-6}J, I_{0}=?**From Eo = ½ LI**

^{2}_{0}

**Electrostatics Assignment Help - Live Physics Tutors 24x7 Hrs**

**Energy Stored In Inductor Assignment Help, Energy Stored In Inductor Homework Help, Energy Stored In Inductor Tutors, Energy Stored In Inductor Solutions, Energy Stored In Inductor Tutors, Electrostatics Help, Physics Tutors, Energy Stored In Inductor Questions Answers**

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