Test your understanding of the lesson
Question
Give the volume expression of the Laplace's force applied to a conductor in a magnetic field \(\vec B\).
What is the Laplace's force which applies to an element of conductor of length \(d\vec\ell \), travelled by intensity \(i(t)\) when immersed in a magnetic field \(\vec B\) ?
Solution
Volume expression of the Laplace's forces :
\(d\vec f=\vec j d\tau \wedge \vec B\)
Laplace's Force :
\(d\vec f = id\vec\ell \wedge \vec B\)
Question
Write Lenz's and Faraday's laws.
Solution
Lenz's law :
The induced current has such a sign that the induced flux created opposes the variations of inductor flux.
Or :
The electromotive force tends because of its consequences to oppose the cause which made it appear.
Faraday's law :
The electromotive force \(e\) induced in a closed, steady circuit in the Galilean laboratory is opposed to the time derivative of the magnetic field \(\Phi_{\vec B}\) through the circuit :
\(e=-\frac {d\Phi_{\vec B}}{dt}\)
Question
Prove Faraday's law from one of Maxwell's equations.
Solution
Maxwell-Faraday equation :
\(\vec {rot}\vec E=-\frac {\partial\vec B}{\partial t}\)
Let \((C)\) be a closed wire circuit.
\(\iint_{(S)} \vec {rot}\vec E.\vec n\;dS=\oint_{(C)} \vec E.d\vec \ell=-\frac{d}{dt}(\iint_{(S)} \vec B.\vec n\ dS)\)
Hence :
\(e=-\frac {d\Phi_{\vec B}}{dt}\)
Question
Define self-inductance \(L\) and mutual inductance \(M\).
Solution
Self inductance :
A closed thread-like circuit \((C)\) is travelled by an intensity \(I\).
The flux of the circuit's own magnetic field through an outline oriented by the positive sense of the chosen current, or, proper flux, is proportional to \(I\) :
\(\Phi_P=LI\)
\(L\) is the self inductance of the circuit \((C)\), also known as induction coefficient.
Mutual inductance :
Two thread-like circuits \((C_1)\) and \((C_2)\) are travelled by two intensities \(I_1\) and \(I_2\).
The flux of the magnetic field \(\vec B_2\) through the closed outline \((C_1)\), oriented with the positive sign of the current \(I_1\) is proportional to \(I_2\) :
\(\Phi_{2->1}=MI_2\)
Likewise, the flux of the magnetic field \(\vec B_1\) through the closed outline \((C_2)\), oriented with the positive sign of the current \(I_2\) is proportional to \(I_1\).
\(\Phi_{1->2}=MI_1\)
\(M\) is the mutual inductance of the two circuits.
Question
What is the relation between the Laplace's force and the power of the electromotive force of induction ?
Solution
The power \(P_L\) of the electromotive force of induction \(e\) is compensated by that of the Laplace's force applied to the circuit :
\(P_L+ei=0\)
Question
What is the self-inductance of a solenoid ?
Solution
The magnetic field inside an “infinite” solenoid is :
\(\vec B = \mu_0 n I\ \vec u_z\)
The proper flux through the \(N\) whorls over the length \(\ell\) is :
\(\Phi_P=(\mu_0\frac{N}{\ell}I)NS=LI\)
Hence the self-inductance :
\(L=\frac{\mu_0 N^2S}{\ell}\)
Order of magnitude :
For a coil of length \(10\;cm\) made of a \(100\) whorls of diameter \(1\;cm\), the self-inductance is around \(0,01\;mH\) : Henry is a big unit.
Question
What is the magnetic energy of two coupled circuits ?
Let \(L_1\), \(L_2\) and \(M\) be the coefficients of self and mutual inductance.
Solution
The magnetic energy of a system of two circuits of self-inductance \(L_1\) and \(L_2\) and mutual inductance \(M\), which are travelled by the two respective currents \(I_1\) and \(I_2\) is :
\(E_m=\frac{1}{2}L_1I_1^2+\frac{1}{2}L_2I_2^2+MI_1I_2\)
Question
What is the relation between the voltage of the primary winding and the voltage of the secondary for a perfect transformer ?
What is an isolation transformer ?
Solution
Let \(N_1\) and \(N_2\) be the number of whorls in the primary and secondary windings :
\(\frac{V_{secondary}}{V_{primary}}=\frac{N_2}{N_1}\)
In the case \(N_1=N_2\), \(V_{secondary}=V_{primary}\).
The secondary voltage difference is the same as in the primary, yet it is isolated from the primary.
Mass problems are avoided.
Question
Define Foucault's currents (or Eddy's currents).
Solution
The currents induced in massive mobile conductors placed in permanent magnetic fields are called Foucault's currents.
They cause Laplace's forces which tend to oppose the movement which made them appear.
This is the principle of electromagnetic braking, used in particular for heavy weights, haulers and high speed trains.
Foucault's currents are also used in induction cookers.
Question
A coil made of \(1000\) whorls is shortcut and placed in a magnetic field oriented along the axis of the coil.
The value of the magnetic field changes at the speed of \(10^{-3}\;T.s^{-1}\).
What is the thermal power released in the coil, considering its section to be \(40\;cm^2\) and its internal resistance to be \(R=160\;\Omega\) ?
Solution
The electromotive force is given by Faraday's law of induction :
\(e=-NS\frac{dB}{dt}\)
The intensity is :
\(i=\frac{e}{R}\)
And the dissipated power is :
\(P=Ri^2=\frac{N^2S^2}{R}(\frac{dB}{dt})^2=10^{-7}W\)