## Energy Stored In Magnetic Field

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# Energy Stored in Magnetic Field

Consider an electric circuit containing inductance. When the circuit is closed, a back e.m.f. is induced in the circuit which opposes the growth of current in it. Therefore a certain amount a work has to be done by the current in increasing back e.m.f. The work done by the current is stored in the magnetic field of the coil as potential energy. When the circuit is switched off, and e.m.f. is induced in the opposite direction which opposes the decay of current. Thus the work done by the current during the growth is recovered. Let I be the current in the inductor L at any instant t. Let the rate of growth of current be dI/dt.The back e.m.f induced in the inductor ε = - L dI/dt

The work done in moving charged dq against this e.m.f is

dW = - ε dq = LdI/dt dq = L dq/dtdI = LI dI

Total work done in increasing the current from zero to maximum value I0 is

W = ∫ LI dI = 1/2LI

_{0}

^{2}

Thus the energy stored in the inductor is

U = ½ LI

_{0}

^{2}

Now we can derive an expression for the energy stored up per unit volume (i.e., energy density) in the magnetic field. For this we consider a very long air cored solenoid of length l, cross-sectional area A, and total number of turns N.

Let I be the current in the solenoid.

Magnetic Induction along axis inside the solenoid B = μ

_{0}NI/l

Or I = BI/μ

_{0}N

The self- inductance of a solenoid L = μ

_{0}N

^{2}A/l

. : Energy stored in the magnetic field is

U = 1/2 LI

^{2}= 1/2μ

_{0}N

^{2}A/l = ½ μ

^{0}N

^{2}A/l (BI/μ

_{0}N)

^{2}= 1/2μ

_{0}B

^{2}Al

Volume of the solenoid = Al

. : Energy stored per unit volume or the energy density u in the magnetic field is

u = U/Al = ½ B

^{2}/μ

_{0}

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