How do we maximize core and winding losses in an electromagnet (and calculate them) for variable DC input?

Question

Bigger Picture:

Imagine a magnet approaching the solenoid with some initial constant velocity on \$v_0\$ on x axis. The approaching magnet will induce an EMF and induced current in solenoid. We want to maximize the power loss so that the solenoid acts as a brake. We intend to do so by maximizing the power losses in the solenoid (both winding and core).

Technical Setup:

Most of the time we deal with minimizing copper and core losses in an electromagnet. Let us now consider an electromagnet with a core: the shape, size, material etc. and all the other variables are left free for design.

Given an electromagnet in which the winding is supplied with some variable DC input current and voltage, and their values can be determined by two given functions over time \$t: V(t)\$ and \$I(t)\$. The maximum input voltage and current are 10s of volts and at most 10 A. We can thus simply determine the value of input power:

\$P(t)= V(t).I(t)\$

Technical Objective:

We need to maximize the power losses in the electromagnet (in form of eddy currents etc.) in terms of percentage w.r.t. to total input power to the winding. How do we achieve this?

Question 1: To maximize the core losses what shape, size, material, etc. for the core do we need? Wikipedia (https://en.wikipedia.org/wiki/Magnetic_core) mentions: > If the core is electrically conductive, the changing magnetic field induces circulating loops of current in it, called eddy currents, due to electromagnetic induction. The loops flow perpendicular to the magnetic field axis. The energy of the currents is dissipated as heat in the resistance of the core material. The power loss is proportional to the area of the loops and inversely proportional to the resistivity of the core material.

So should we as a rule of. thumb choose a highly conductive solid core (soft iron with maximum diameter in case of a cylindrical shape)?

How do we deal with this rigorously from an engineering point of view (both design and calculating the power losses)? Can we achieve losses of over 75% (of input power)?

Question 2: What kind of winding parameters do we need ((1). gauge of wire, (2) number of turns (3) shape of solenoid) to maximize the winding losses?

Answer 1

It seems based on your comments that you just want to stop the magnet. Whether there is a core or not is irrelevant.

Let's consider the case of a magnet entering into a coil of wire wound around a plastic bobbin. A cross section is shown below.

In order to stop the magnet, you need to make the magnet transfer its kinetic energy into the coil.

The coil is an inductor and therefore it can store energy.

$$E(t) = \frac12i(t)^2L(x(t))$$

Where...

• E(t) is the stored energy.
• i(t) is the coil current.
• L(x(t)) is the coil inductance. Note that L varies which varies with magnet position x. Usually, it will vary with the square of the depth to which the magnet is inserted into the coil.

We can also say that...

$$V(x(t)) = L(x) * \frac{di}{dt}$$

Where...

• di/dt is the rate of change of current in the coil.
• V(x(t)) is the inductor voltage (induced by motion of the magnet).

The inductance will be proportional to the square of the number of turns. So we may write...

$$ L(x(t)) = \beta(x(t)) * N^2 $$

Where...

• Beta(x(t)) is a function that depends on the position of the magnet relative to the coil, the geometry of the magnet and coil, their relative position. Usually, Beta varies proportional to x^2 and proportionally to the permeability of the magnet material.
• N is the number of turns of wire.

As the magnet enters the bobbin the field lines move through the wires with a perpendicular component. As they move it will induce a voltage in the wires.

$$V(x(t)) = \alpha(x(t)) * N * \frac{dx}{dt}$$

Where...

• Alpha(x(t)) is a "coupling factor" that depends on the position of the magnet relative to the coil, the geometry of the magnet and coil, their relative position, and the strength of the magnet.

• V(x(t)) is the induced voltage.

• dx/dt is the velocity of the magnet as it enters.

• x(t) is the position of the magnet relative to the solenoid.

An important thing to note is that all other things being equal, if you use thinner wire but pack more turns into the same volume, then alpha(x(t)) should be largely unaffected, but V will increase proportional to the number of turns.

Now to maximizing braking...

• Note that current must flow in a loop. So, the ends of the solenoid wires must be part of a circuit. If the wires were open, you would not transfer energy to the coil.

• Secondly, any resistance in series with the coil will reduce the amount of current that builds up, and therefore reduce the energy. Ideally, we want the resistance as low as possible.

In the case of shorting the coil ends together, assuming no coil resistance, a fixed motion of the magnet, and combining the above equation we get...

$$E = \frac12\int \frac{\alpha(x(t))}{\beta(x(t))N}\frac{di}{dt}$$

The alpha and beta terms just depend on the geometry. We don't even need to find them to know that if all other things are equal, the maximum braking occurs with a low number of turns and low resistance. Both of those factors are minimized by just using a metal tube (thicker tube walls are better here).

Also note that alpha and beta are probably going to be pretty hard to calculate by hand. Normally one would build a model in some EM solver tool and calculate them that way.

Answer 2

If you do want to maximize eddy current losses...

• Laminated cores reduce losses, so don't use them.
• Cores made of magnetic powder with an insulating binder material reduce losses.
• Solid conductive metal cores maximize losses. For example, a solid iron core.

But note that...

• For a DC current the core losses will be zero.
• For a very slowly changing current the core losses will still be very close to zero.

The main thing that will be dissipating heat in a solenoid with a DC current through it is the resistance of the wires that are wound around the core.

In fact, if the solenoid is not moving and you are putting a DC current into it, then all of the input power is turning into heat via the winding resistance.

Also note that the current doesn't need to change direction to have an AC component. For example, you could have:

$$I = 5~\mathrm{A}+4~\mathrm{A}\cdot\sin(2\cdot\pi\cdot f\cdot t)$$

In this case the current would vary between 1 A and 9 A at frequency f. The current never completely changes direction but has an AC component of 4 A.