Magnetic energy stored in a coil
Take 20 minutes to prepare this exercise.
Then, if you lack ideas to begin, look at the given clue and start searching for the solution.
A detailed solution is then proposed to you.
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A coil of length \(\ell\), radius \(a\) and axis \((Oz)\), is made of the winding of \(n\) joined circular whorls by length unit.
Let us assimilate the coil to an infinite solenoid, and suppose the steady-state.
Question
The current \(I\) travels through the coil.
Determine the magnetic field created by the coil.
Indice
The point of this exercice is to review electromagnetic energy. Follow the guidelines...
Can you write the local and then integral electromagnetic energy ?
Question
What is the magnetic energy of the coil ? Deduce the value of the inductance L of the coil.
Solution
Magnetic energy can be expressed in two ways :
\(\frac{{B^2 }}{{2\mu _0 }}(\pi a^2 \ell ) = \frac{1}{2}LI^2\)
Hence :
\(L = \mu _0 n^2 \ell \pi a^2 = 100\;mH\)
Question
The coil is put in a series circuit.
There is a resistor \(R\) and an electromotive force generator of constant value \(U_0\).
Determine the expression \(I(t)\) of the current in the coil in function of time.
Solution
We use Kirchoff' laws :
\(U_0=Ri+L\frac{di}{dt}\)
Classically :
\(I(t) = \frac{U_0}{R}(1 - \exp ( - t/\tau )),\;\tau = \frac{L}{R}\)
Question
Compute the magnetic and electric fields created by the coil in any point at the moment \(t\).
Solution
Let's write :
\(B_0 = \frac{{\mu _0 nU_0}}{R}\)
Inside :
\(\vec B = B_0 (1 - e^{ - t/\tau } )\vec u_z\)
Outside the coil, the field is equal to zero.
The electric field is orthogonal to the radius (study the symmetries).
It depends on \(r\) and time.
Let us use Stoke's theorem on a circle :
If \(r<a\) : \(\vec E(r,t) = - \mu _0 n\frac{r}{2}\frac{{dI(t)}}{{dt}}\vec u_\theta\)
If \(r>a\) : \(\vec E(r,t) = - \mu _0 n\frac{{a^2 }}{{2r}}\frac{{dI(t)}}{{dt}}\vec u_\theta\)
Question
Determine the density of magnetic and electric energy.
What can be noticed about the ratio of these two energies ? Conclude
Solution
The density of magnetic energy is :
\(e_B = \frac{{B^2 }}{{2\mu _0 }} = \frac{{\mu _0 n^2 I^2 }}{2}\)
The density of electric energy is, for example in \(r=a\) where it is maximum : (using \(\varepsilon _0 \mu _0 c^2 = 1\))
\(e_E = \frac{1}{2}\varepsilon _0 E(a)^2 = \frac{{a^2 \mu _0 n^2 }}{{8c^2 }}\left( {\frac{{dI}}{{dt}}} \right)^2\)
Let's calculate the ratio :
\(\frac{{e_E (r = a)}}{{e_B }} = \frac{{a^2 }}{{4c^2 }}\left( {\frac{{(dI/dt)}}{I}} \right)^2 = \frac{{a^2 }}{{4c^2 }}\frac{1}{{\tau ^2 }} \approx 1,7.10^{ - 5}\), with \(R = 10\;k\Omega\).
The electric energy is negligible.
In a steady-state approximation, a coil is almost only magnetic.
Question
What is the expression of Poynting's vector through a surface delimiting the volume of the coil ? Comment on that.
Solution
Let's compute Poynting's vector in \(r=a\) :
\(\Pi (a,t) = \frac{1}{{\mu _0 }}\vec E(a,t) \wedge \vec B(a,t) = \frac{1}{{\mu _0 }}\left( { - \mu _0 n\frac{a}{2}\frac{{dI(t)}}{{dt}}\vec u_\theta } \right) \wedge \left( {\mu _0 nI\;\vec u_z } \right) = - \frac{1}{2}\mu _0 n^2 aI(t)\frac{{dI(t)}}{{dt}}\vec u_r\)
The flux entering the coil is then :
\(\Phi = \left( {\frac{1}{2}\mu _0 n^2 aI(t)\frac{{dI(t)}}{{dt}}} \right)2\pi a\ell = LI(t)\frac{{dI(t)}}{{dt}} = \frac{d}{{dt}}\left( {\frac{1}{2}LI^2 } \right)\)
This flux indeed corresponds to the variation of the magnetic energy stored by the coil by time unit.