Eddy Current Pendulum

 

 

 

 

 

 

 

Demo: The solid pendulum’s bob stops between the magnet after being released. The bob made with slits oscillates freely.Three bobs, made from copper, with the same surface area are available for this demonstration. One has deep narrow slits to diminish the Eddy currents’ circulation. It has the same mass as one of the solid bobs, while the third solid bob has half the mass.

Materials: Eddy current pendulum setup including three bobs (kept with pendulum setup). Notes: The pendulum is very heavy, so a classroom that can be accessed by lab cart is preferred. Additionally, the magnet on the stand is very strong, so be careful when other magnets or iron-based materials are around it.

Setup: The bobs are attached with a small screw at the bottom of the pendulum arm.

Explanation:

Eddy currents are circular loops of current created in response to a changing magnetic field such that the resulting magnetic field opposes the original change. These currents correspond to Lenz’s law, which states that loops of current flow in a direction such that the induced magnetic field that they create opposes the changing external magnetic field.

While it may be tempting to assume that this demonstration can be explained by the simple Lorentz force equation, \vec{F} = q\vec{v} \times \vec{B}, that only works for point charges that cannot be polarized. It turns out that if you have a conductor moving through a static magnetic field, there will be no force applied to it. This is because the material will polarize itself, and, if the field is constant, the forces will cancel (seen below).

Figure 1: Polarizable object in a static magnetic field with velocity V. The negative charges will be attracted to the positive side of the magnetic field and the positive charges will be attracted to the negative side of the field. This, in turn, causes a force felt by the negative half of the object to be pointing out of the page and a force felt by the positive half of the object pointing into the page. Together, these forces cancel out because they are of equal magnitude in opposite directions.

 

 

 

 

 

 

 

 

 

 

To explain this demonstration fully, we first will explain a simple case, and then apply that simple case to the case at hand.

First, we start with one of Maxwell’s equations: \vec{\nabla} \times \vec{E} = -\frac{\partial \vec{B}}{\partial t}

While it may not seem initially that an electric field would be present, when in the reference frame of the pendulum, the moving magnetic field creates a transverse electric field in the y+ direction. The coordinate axis we are working with and the respective directions of dL, dA, and B can be seen in the figure below.

Figure 2: Coordinate axis, and directions of dA, dL, and B. Both B and dA point in the +z direction (out of the page) and dL points in the anti-clockwise direction.

 

 

 

 

 

 

For the general case, we take a piece of area, A, enclosed by a curve of length L. The differential area and length segment of that surface are dA and dL, respectively. If we take the surface integral of both sides of Maxwell’s equation over area A, we get

\int (\vec{\nabla} \times \vec{E}) \cdot d\vec{A} = -\int \frac{\partial \vec{B}}{\partial t} \cdot d \vec{A}

Using Stokes theorem, we can transform the left side of the equation as follows:

\int (\vec{\nabla} \times \vec{E}) \cdot d\vec{A} = \oint E \cdot d \vec{L}

Note: We are converting dA to dL, so signs are important here. The sign of dA and dL follow the right hand rule, so if you curl your fingers around in the direction that dL is travelling, your thumb will be pointing in the direction of a positive dA. We are defining dA out of the page to be the positive direction in this instance, so the corresponding dL would be a length of the loop going in the counter clockwise direction. This will come up later when discussing the induced current in the circuit. See Figure 2.

Plugging that into our integrated Maxwell equation, we get

\oint E \cdot d\vec{L} = -\int \frac{\partial \vec{B}}{\partial t} \cdot d \vec{A}

We know \vec{B} is static, so it does not depend on time t. Therefore, we can take the \frac{\partial}{\partial t} out of the integrand, giving us:

\oint E \cdot d\vec{L} = -\frac{\partial}{\partial t} \int \vec{B} \cdot d \vec{A}

Here, we define \Phi \equiv \int \vec{B} \cdot d \vec{A} where \Phi is the electromotive force, EMF. This gives us the equation for the EMF:

\oint E \cdot d \vec{L} = - \frac{\partial}{\partial t} \Phi

At this point, we are done with the background information and will transfer our explanation to a very simple version of the pendulum and magnet: a simple loop of wire being pushed through a magnetic field, as seen below. The loop of wire is passing through a static magnetic field of magnitude B_o coming out of the page. The loop of wire has side length L, resistivity R, and is moving through the magnetic field with a velocity v.

To solve for \Phi, we need to first find B and dA. Because B is not time-dependent, the magnitude is just B_o. The value for dA is slightly more tricky to find. The portion of the loop in the magnetic field is the vertical length of the wire, L, multiplied by the amount of the loop in the x direction that has passed through the field. To calculate this, we multiply the velocity, v with the differential time that has passed, dt.

\vec{B} \cdot d \vec{A} = B_o L v dt, so \Phi = \int B_o L v dt = B_o Lvt

Plugging this into the Electromotive force equation above gives us

\oint E \cdot d\vec{L} = - \frac{\partial}{\partial t} (B_o L v t) = EMF

Therefore, EMF = -B_oLv

The electromotive force is a difference in potential, (a voltage) which gives rise to a current. Using Ohm’s Law (V = IR), the EMF we found above, and the general resistance R of the loop of wire, we can find the current that flows through the circuit.

I = -\frac{B_oLv}{R}

The negative sign here indicates that the current is flowing in the opposite direction of dL. dL is counterclockwise, so the current will be flowing clockwise. Due to this current, there will be a force exerted in the wire. Because the “loop” we are using is really a square, it is easy to separate the total force exerted on the loop into its x and y components. There is no z component because none of the wire is coming out from or going into the page.

The red lines in the above picture show the parts of the wire that are affected by the magnetic field, and therefore, the only parts that will have forces exerted on them. The forces acting on the top and bottom portions of the wire will cancel out because the current is flowing in opposite directions, so the only part of the wire that will cause a force to be exerted on the system is the rightmost piece.

There is a force exerted by current-carrying wires due to a Lorentz force acting on each individual electron as they flow. This force is \vec{F} = L \vec{I} \times \vec{B}, which gives us \vec{F} = L (-\frac{B_o L v}{R}) (B_o) = -\frac{{B_o}^2 L^2 v}{R}. In this instance, \vec{I} is pointing in the -y direction and \vec{B} is in the +z direction.

We can find the direction of the force using the right hand rule. I is going in the -y direction, B is going in the +z direction, so the force points along the -x axis, antiparallel to the velocity. If the loop of wire is being pushed in the +x direction, it will experience a force in the -x direction. It turns out that the direction of the force will always be exactly opposite to the direction of the velocity. It does not matter if the velocity is going on a diagonal with both x and y components, the force that will be exerted on the pendulum will always be exactly opposite to the components of the velocity vector.

So how does this apply to the eddy current pendulum itself? We can approximate the pendulum in the demo as a series of concentric loops of wire, as shown below, where each black line represents a loop of wire of negligible thickness.

The physics behind this remains the same as for the single square loop of wire, the parts of the wire that would exert a y force cancel each other out, so as it is falling into the magnet, the only force generated on it acts directly opposite to the velocity vector.

So why do slits in the pendulum negate this effect in a solid pendulum? The EMF is created due to the gradient in magnetic field. Only half of the wire is being exposed to a magnetic field, which is what drives the current through the loop of wire. If you cut slits in the pendulum, it makes possible current loops much smaller, as seen below

Due to the slots in the pendulum, the size of the eddy currents that can be created are much smaller than the size of the eddy currents that can be created in the un-slotted pendulum. Additionally, the eddy currents are created due to the gradient in the magnetic field, from part of the pendulum being inside of the magnetic field and part of the pendulum being outside of the magnetic field. When a slotted pendulum is used, there are only small pieces of metal that are both inside and outside of the magnetic field at once. Because of this, the force exerted on that part of the pendulum is not strong enough to stop the motion of the pendulum in its entirety.

Written by: Sophia Sholtz