Electromagnetic induction

Suppose the field is uniform at the surface of the whorl. Its flux through the coil is :

\(\Phi = \frac{{{\mu _0}nN\pi {b^2}}}{2}(1 - \cos \theta ){i_{1,m}}\cos \omega t\)

A mutual inductance between the whorl and the solenoid can be defined :

\(M = \frac{{{\mu _0}nN\pi {b^2}}}{2}(1 - \cos \theta )\;\;\;\;\;\;\;so\;\;\;\;\;\;\;\Phi = M{i_{1,m}}\cos \omega t\)

The electric equation of the coil is then :

\(Ri + L\frac{{di}}{{dt}} + M\frac{{d{i_1}}}{{dt}} = 0\)

In a forced sinusoidal state :

\(i = {I_m}{e^{j\omega t}}{e^{j\varphi }}\)

So :

\(Ri + jL\omega i + jM\omega {i_1} = 0\)

Hence :

\({I_m}{e^{j\varphi }} = - \frac{{jM\omega }}{{R + jL\omega }}{i_{1m}} = \frac{{M\omega }}{{jR - L\omega }}{i_{1m}}\)

It can be deduced :

\({I_m} = \frac{{M\omega }}{{\sqrt {{R^2} + {L^2}{\omega ^2}} }}{i_{1m}}\)

And :

\(\tan \varphi = \frac{R}{{L\omega }}\;\;\;\;(with\;\cos \varphi = - \frac{{L\omega }}{{\sqrt {{R^2} + {L^2}{\omega ^2}} }} < 0)\)

Let's compute the radial component of the field.

Outside the \((Oz)\) axis, the magnetic field can be expressed :

\(\vec B(M)=B_r(r,z)\;\vec u_r+B_z (r,z) \;\vec u_z\)

We can notice that the plane défined by the point M and \((\vec u_r,\vec u_z)\) is a anti-symetrical plane for the current distribution along the wire, so the magnetic field is in this plane and has no coordinate along \(\vec u_{\theta}\).

Outside the axis, the magnetic field can be obtained by expressing the conservation of the magnetic flux :

\( \oint_S\vec {B}.\vec{n} \ dS=0\)

A small cylinder, centered on axis Oz, of height \(dz\) and small radius \(r\) is used.

The flux, exiting through this closed surface must be equal to zero :

\(\pi {r^2}B_z(z+dz,0)-\pi r^2B_z(z,0)+ 2\pi r dz{B_r}(r,z) = 0\)

Thus :

\(\pi {r^2}dB_z(z,0) + 2\pi r dz{B_r}(r,z) = 0\)

So :

\({B_r}(r,z) = - \frac{r}{2}\frac{{d{B_z}(z,0)}}{{dz}}\)

The Laplace's force (mainly vertical) is :

\(d\vec f = id\ell {\vec u_\theta } \wedge {B_r (r,z)}{\vec u_r} = - i{B_r(r,z)}d\ell {\vec u_z}\)

So :

\(\vec F = i\pi N{b^2}\frac{{d{B_z}(z,0)}}{{dz}}{\vec u_z}\)

However :

\(\frac{{dB}}{{dz}} = \frac{{{\mu _0}n{i_{1,m}}\cos \omega t}}{2}\sin \theta \frac{{d\theta }}{{dz}}\)

And :

\(\frac{{d\theta }}{{dz}} = - \frac{1}{a}{\sin ^2}\theta\) (Use \(tan\theta=a/z\))

Thus :

\(\frac{{dB}}{{dz}} = - \frac{{{\mu _0}n{i_{1,m}}\cos \omega t}}{{2a}}{\sin ^3}\theta\)

The force becomes :

\(\vec F = - \frac{{\pi {b^2}{\mu _0}Nni{i_{1,m}}\cos \omega t}}{{2a}}{\sin ^3}\theta \;{\vec u_z}\)

Hence :

\(\vec F = - \frac{{\pi {b^2}{\mu _0}Nn{I_m}{i_{1,m}}\cos \omega t\cos (\omega t + \varphi )}}{{2a}}{\sin ^3}\theta \;{\vec u_z}\)

The mean value : (using \(\cos \omega t\cos (\omega t + \varphi ) = \frac{1}{2}(\cos (2\omega t + \varphi ) + \cos \varphi )\))

\(\left\langle {\vec F} \right\rangle = - \frac{{\pi {b^2}{\mu _0}nN{I_m}{i_{1,m}}}}{{4a}}{\sin ^3}\theta \;\cos (\varphi ){\vec u_z}\)

With :

\({I_m} = \frac{{M\omega }}{{\sqrt {{R^2} + {L^2}{\omega ^2}} }}{i_{1m}}\;\;\;and\;\;\;\cos \varphi = - \frac{{L\omega }}{{\sqrt {{R^2} + {L^2}{\omega ^2}} }}\)

It comes :

\(\left\langle {\vec F} \right\rangle = \frac{{\pi {b^2}{\mu _0}nNML{\omega ^2}i_{1m}^2}}{{4a({R^2} + {L^2}{\omega ^2})}}{\sin ^3}\theta \;{\vec u_z}\)

It is indeed a repulsive force (oriented upwards).

Right above the solenoid, \(\theta=\pi/2\).

Levitation is possible if the repulsion force is superior to the weight.

\(\frac{{\pi {b^2}{\mu _0}nNML{\omega ^2}i_{1m}^2}}{{4a({R^2} + {L^2}{\omega ^2})}} > mg\)

It is a stable equilibrium : if the mass goes up, the repulsion force diminishes and the mass goes back down.

Same if the mass begins by going down.