Hover over the menu bar to pick a physics animation.

For the tube enviromnment the direct force at the leading edge of the magnet seems
to be exactly canceled by the direct force at the trailing edge. However, we
do know that oppositely rotating eddy currents are induced at both edges. Therefore
at least some gravitational energy is being lost in the tube. To understand this we need to
consider that the conduction electrons are rotating in circles relative to the magnet.
But these sideways moving electrons run into the atoms of a falling tube and
this reduces the tube's speed. This concept is like the ficticious
Coriolis Force
associated with the
rotation of the earth. In the northern hemisphere, air moving north
appears to veer to the west relative to the terrain underneath
because it had more circumferential speed earlier when it was farther south.
If the magnet is moving and the tube is stationary, then it is the magnet that gets the
retarding acceleration.
Here is an article on Eddy Currents.
Magnetic braking inside a conducting tube is a high symmetry way
to demonstrate the effects of magnet-induced eddy
currents on the motion of a permanent magnet. The eddy currents considered here will be circumferential in
a conducting tube. Consider the case of a long permanent magnet whose field axis is parallel to the
axis of the tube as shown in Canvas 0. Note that the `bbB_E` field of the eddy currents due to the north pole of the magnet
induce `bbB_E` field force that would tend to advance the magnet
while the eddy currents due to the south pole of the magnet would retard the magnet by ** exactly the same
force **. Therefore, for at least one geometry,
the tube would seem to have no effect on the magnet motion.
However, since the conductivity of copper is finite (resistivity, `rho`, is not zero), the
currents can absorb some kinetic or gravitational energy if a magnet falls through
the conducting tube and this will slow (or brake) the magnet.

It is very important to understand that, without the energy loss due to resistance in the conductor,
the magnetic field generated by the currents in the conductor do not **directly** brake
the speed of a magnet.

By one of Maxwell's equations we have

`gradxxbbE=-(delbbB)/(delt)`

where `bbB` is the magnetic field and `bbE` is the electric field that will induce the eddy currents. To make use of this equation we will apply Stokes Theorem`intintgradxxbbE*dbbA=ointbbE.dbbl`

where the surface integral on left is over the area of a closed loop and the line integral on the right is over the boundary of that loop. The magnetic vector field, `bbB(rho,z)`, due to the permanent magnet is computed in Appendix 1 below.Applying this equation to calculating the circumferential electric field in the tube we have

`intintgradbbE*dbbA=ointbbE.dbbl=-(del)/(delt)intint(bbB*dbbA)`

which is a statement of Faraday's law of induction. which, if our tube has radius `a_t`, states that the induced voltage in the tube is`V=-1/(2*pi*a_t)(del)/(delt)intint(bbB(a_t)*dbbA)`

where we need only the `z` component of the area averaged magnetic field to compute the circumferetial eddy currents. In order to get the time derivative of the quantity on the right, we need to consider that the magnet is moving in the `z` direction at velocity `v_z`:`V=-v_z/(2*pi*a_t)intint((del)/(delz)B_z(a_t)dA)`

The eddy current in the tube will be proportional to `V` as computed here. Since the eddy currents from the top of the magnet are in opposite directions from those from the bottom of the magnet, the sum of their magnetic fields, `bbB_E`, at the center of the permanent magnet is zero and they have no effect on the motion of the permanent magnet. So they will not be calculated here but they are computed in Appendix 2.The eddy currents move in the circumferential direction in the conducting tube walls. The magnetic field due to the PM at the ends of the moving magnet where the eddy currents are largest is mostly radial and therefore perpendicular to the eddy current direction. So the eddy current charges, `q_E`, in the conductor feel an axial force called the Lorentz Force

`bbF=q bbv_e xx bbB`

where `bbv_e` is the speed of the charge of the eddy current and `bbB` is the magnetic field due to the PM. In terms of the eddy current density, the charge speed can be written `v_e=J_e/(nq)bbhatphi` where `n` is the moving charge density in the conductor. Since `bbv_e` is perpendicular to `bbB` the axial force per charge works out to just:`F_z=q_Ev_eB_r`.

By examining Canvas 1 you can see that the largest values of both `B_r` and `(delB_z)/(delz)` occur at the top and bottom edges of the PM. So the largest contributions to braking forces derive from these edges also.

Since the conductor has a force `bbF_C` opposite to the direction of motion of the permanent magnet, if we hold the conductor stationary, we can invoke Newton's law of motion, reaction=action. and clearly say that the force, `bbF_M`, on the magnet is the opposite of force, `bbF_C`, that we have calculated for the moving conductor i.e.

`bbF_M=-bbF_C`

which is also opposite to the direction of motion of the magnet.To move the PM in the 'z' direction, I have chosen to use a sinusoid

`z_M=Asin(omegat)`

`v_z=omegaAcos(omegat)`

Shows the axial component, `B_z`, of the field, `bbB_(PM)`, as color coded contours.

It also shows the directions of the field as color coded arrows.

The black curve is the value of the axial field, `B_z`, at the cylinder walls.

The colored curves that are seen while the PM is in motion show at the cylinder wall `dB_z/dz` (blue) and the retarding force (red).

Note that the sign of red curve is the same at both ends of the magnet so that the force add, while the blue curves have opposite sign at bottom from that at top.

We must find an expression for `bbB(rho,z)` when the magetic dipole moment densities, `bbM=+-1`, are constant on the poles of the magnet.

We have a cylindrical magnet centered at z=0 and of length `L` and radius `a` oriented along the `z` axis and having unit magnetic charge density on the opposite signs on its poles. To compute the `bbB` field we will us the following coordinates.

`deltaz_1=z-L/2`

`deltaz_2=z+L/2`

`deltarho=(rho-rcosphi)`

`bbhatrho_1=deltarhobbhatrho/sqrt(deltarho^2+deltaz_1^2)`

`bbhatrho_2=deltarhobbhatrho/sqrt(deltarho^2+deltaz_2^2)`

`bbhatz_1=deltaz_1bbhatz/sqrt(deltarho^2+deltaz_1^2)`

`bbhatz_2=deltaz_2bbhatz/sqrt(deltarho^2+deltaz_2^2)`

`bbB_(rho_1)=deltarho_1bbhatrho/(deltarho^2+deltaz_1^2)^(3/2)`

`bbB_(rho_2)=deltarho_2bbhatrho/(deltarho^2+deltaz_2^2)^(3/2)`

`bbB_(z_1)=deltaz_1bbhatz/(deltarho^2+deltaz_1^2)^(3/2)`

`bbB_(z_2)=deltaz_2bbhatz/(deltarho^2+deltaz_2^2)^(3/2)`

These results are for just one (r,phi) source element. For each pole face (1,2) we must sum over the entire ranges `(r=0->a)` and `(phi=0->2pi)`. In order to get the total `B_rho,B_z` values here we sum the results for all of the surface elements on the two pole faces.

`B_z(rho,z)=int_(phi=0)^(phi=2*pi)int_(r=0)^(r=a)[[(z-L/2))/(den_1)-((z+L/2))/(den_2)]drdphi`

`B_r(rho,z)=int_(phi=0)^(phi=2*pi)int_(r=0)^(r=a)[(rho-rcosphi)(1/(den_1)-1/(den2))]drdphi`

`den_1=[(z-L/2)^2+rho^2+(rcosphi)^2-2 r rho cosphi]^(3/2)`

`den_2=[(z+L/2)^2+rho^2+(rcosphi)^2-2 r rhocosphi]^(3/2)`

As an example of showing that the divergence of `bbB` is zero, here is a simple 2 dimensional case where we have the expression

`bbB=x/(x^2+y^2)bbhatx+y/(x^2+y^2)bbhaty`

which could be a field divided by its distance from a charge point at (0,0) multiplying the unit vectors to points (x,y).
`B_x=x/(x^2+y^2)`

`B_y=y/(x^2+y^2)`

`bbgrad.bbB=1/(x^2+y^2)-2x^2/(x^2+y^2)^2+1/(x^2+y^2)-2y^2/(x^2+y^2)^2`

`=1/(x^2+y^2)^2[(x^2+y^2)-2x^2+(x^2+y^2)-2y^2]=0/(x^2+y^2)^2`

The permanent magnet induces eddy currents which are similar to a currents in a wire solenoid. We will compute the internal `bbB` field due to a solenoid of a given number of turns using the Biot-Savart (BS) law.

`bbB(bbr)=mu_0int_C(Idbblxxbbr')/|bbr'|^3`

where `C` is the full path of the current in the length increments, `bbl`, and `bbr'=bbr-bbl` and `I` is the current in the given eddy current loop. We'll assume that radius of the 'wire' in a single eddy current loop is zero.`l(phi)=a_E(cosphibbhatx+sinphibbhaty)dphi`

A description for the length increment `dl` is found by taking the derivative with respect to `phi`.`dl(phi)=a_E(-sinphibbhatx+cosphibbhaty)dphi`

The expression for the field point will be`bbr=b bbhatx+zbbhatz`

where the loop field is symmetrical about the loop axis so we don't need to specify a `y` component for the field point. The value of `bbr-bbl` is then`bbr-bbl=(b-a_Ecosphi)bbhatx-a_Esinphibbhaty+zbbhatz`

A cross product operation, `xx`, is defined in `(x,y,z)` coordinates as`bbuxxbbv=|[bbhatx,bbhaty,bbhatz], [u_x,u_y,u_z], [v_x,v_y,v_z]|=|(u_yv_z-u_zv_y)bbhatx-(u_xv_z-u_zv_x)bbhaty+(u_xv_y-u_yv_x)bbhatz|`

We could multiply these out algebraicallyly it's easier to let the computer do so after specifying the components `bbr-bbl` and `bbdl`. We now have everything we need for the BS integral`bbB(bbr)=mu_0int_C(Idbblxxbbr')/|bbr'|^3`

We will set `mu_0=1` and, to start, let `I=1" ampere"` for the current loop of largest current. The value of `I` in other eddy current loops will depend on the value of `B_z` of the permanent magnet at their `z'` location. Of course, by Lenz's Law the value of `I` depends on the direction and speed, `v_z`, of the permanent magnet in the copper tube. More explicitly we can write:`I(bbB,v_z)=A(delB_z)/(delz)v_z`

where `v_z` is the axial speed of the magnet `A` is the cross sectional are of the tube and `(delB_z)/(delz)` is the variation of the z component of the magnet's field at the radius `a_E` and axial position, `z'`, of the eddy current loop.