Energy stored in a magnetic field
A coil has an inductance of 53 mH and a resistance of 0.35 `Omega`.
(a) If a 12 V emf is applied across the coil, how much energy is stored in the magnetic field after the current has built up to its equilibrium value?
(b) After how many time constants will half this equilibrium energy be stored in the magnetic field?

Explanation

Thus, to find the energy `U_(Boo)` stored at equilibrium, we must first find the equilibrium current. From Eq. 30-43, the equilibrium current is
`i_(oo)=E/R=(12V)/(0.35Omega)=34.3A` (30-53)
Then substitution yields
`U_(Boo)=1/2Li_(oo)^(2)=(1/2)(53xx10^(-3)H)(34.3A)^(2)`
=31J
Now we are being asked: At what time t will the relation
`U_(B)=1/2U_(Boo)`
be satisfied? Using Eq. 30-51 twice allows us to rewrite this energy condition as
`1/2Li^(2)=(1/2)1/2Li_(oo)^(2)`
or `i=(1/(sqrt2))i_(oo)`. (30-54)
This equation tells us that, as the current increases from its initial value of 0 to its final value of `i_(oo)` the magnetic field will have half its final stored energy when the current has increased to this value. In general, we know that `i_(oo)` is given by Eq. 30-43, and here ( (see Eq. 30-53) is E/R, so Eq. 30-54 becomes
`E/R(1-e^(-t//tau_(L)))=E/(sqrt2R)`
By canceling E/R and rearranging, we can write this as
`e^(-t//tau_(L))=1-1/(sqrt2)=0.293`
which yields
`t/(tau_(L))=-"ln"0.293=1.23`
or `t~~1.2tau_(L)`
Thus, the energy stored in the magnetic field of the coil by the current will reach half its equilibrium value 1.2 time constants after the emf is applied.