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Braking of a Magnet in a Conducting Tube

Introduction

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.

Computing the Eddy Currents

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.

Getting the force on the magnet due to the eddy current .

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.

Forces on the PM when the conducting tube is held stationary

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.

Description of Animation

To move the PM in the 'z' direction, I have chosen to use a sinusoid

`z_M=Asin(omegat)`
`v_z=omegaAcos(omegat)`

where `A` is the amplitude of the motion and `omega` is the radian frequency of the motion. As a result, I'm able to show the changes in the direction of eddy current in the cylinder walls. It also shows the effect of changes of magnitude of `v_z` of the PM. The largest magnitude of `v_z`, `|v_z|`, occurs when the magnet is the center of the cylinder and this also results in the largest braking force since the eddy currents are largest when `v_z` is largest.

Description of Graphics

Canvas 1

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.

Canvas 2:

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.

Appendix 1:Computing the Magnetic Field `bbB` of the Permanent Magnet

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)`

where (rho,z) are the observation (field) coordinates and (r,L/2,phi) are the source coordinates. Unit vectors pointing from source toward the field points are

`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)`

The `bbB` field is the unit vector divided by the source distance squared which results in

`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`

where `den_1` and `den_2` are:

`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)`

where z=0 at the `xy` plane.

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`

Note that the result is just 0 except when (x,y)=(0,0). This is a classical solution of Laplace's equation `grad.B=0`. This form of the `bbB` expression satisfies the Maxwell equation requirement `grad.bbB`=0. This form is what we use for `bbB` here

Appendix 2: Computing Magnetic Field `bbB_E` due to the Eddy Currents in the Copper Tube

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.

Field for a Single Eddy Current Loop

We need to compute the BS integral for a circular loop for any radius, `rho` as well as axial locations, `z`. Let the current loop lie at `z=0` and its radius be `a_E`. Then a description for any point `l` on the loop would be

`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.