Electromagnetism Part 2

by | Apr 22, 2018 | Physics | 0 comments

Energy of a System

The energy of a system will be the total worked required to assemble it. For example, if you wanted to make a system of two protons separated by a distance of 5cm (assuming you bring them in from infinity) the energy of the system will be the amount of work you do to push the two protons together against their repulsive electrostatic force.

Now, for any electrostatic system, the work is simply

    \[W=\frac{1}{2}\int V\rho d\tau = \frac{\epsilon_0}{2}\intE^2d\tau.\]

Now this is paralleled for the energy in magnetic fields. The formula we know from electrodynamics,

    \[W=\frac{1}{2}\int\vec{A}.\vec{J}d\tau=\frac{1}{2\mu_0}\int B^2d\tau,\]

Where \vec{A} is the magnetic vector potential, analogous to the potential in electric fields E=\=-\nabla V, however, it is defined as \vec{B}=\nabla\times\vec{A}.

We note that this can’t be applied to a magnetostatic field, since magnetic fields don’t do any work. But why does building up a magnetic field from scratch require work? This is because we start from no field to some field, this involves a changing magnetic field, which induces an electric field, which in turn does work, and so we need to battle against that to “assemble” our magnetic field, (by maybe increasing a current or something).

Combining the previous two results, we get that the total energy in an electromagnetic field is given by

    \[W=\frac{1}{2}\int_V\big(\epsilon_0 E^2+\frac{1}{\mu_0}B^2\big).\]

In a dielectric filled system, as you build up the free charges around your dielectric, the bound charge in the dielectric exerts a force on it, and so we have to take this into account when calculating the work, and so in a dielectric system the work becomes

    \[W=\frac{1}{2}\int\vec{D}.\vec{E}d\tau.\]

In a magnetisable material, the energy is given by

    \[W=\frac{1}{2}\int\vec{H}.\vec{B}d\tau.\]

Inductance

Now this is all well good to find energy, but, there is an easier way to calculate these quantities without performing integrals, and that is through using something called Inductance. Let’s first define what it is. Say we have two loops side by side, but not touching.

If there is a a changing current flowing in loop 1, the changing magnetic field induced will pass through loop 2, causing a change in flux in loop 2. By Faraday’s Law, the changing flux through loop 2 will create in electric field and so a current will now flow through loop 2. But we said that the two loops are not connected, they don’t even touch! How on earth does one affect the other? The energy stored in the magnetic field of loop 1 is transferred to loop 2 causing the current in loop 2. By Lenz’s law, the current will flow to oppose the changing the magnetic field that caused it. This is known as mutual inductance.

Now let’s do some math with this. What is the magnetic flux through loop two? Well we first need to figure out the magnetic field from loop 1. This is given by the Biot-Savart Law:

    \[\vec{B}=\frac{I_1\mu_0}{4\pi}\int_{loop1}\frac{d\vec{l}\times\vec{r}}{\norm{\vec{r}}^3}\]

where I_1 is the current through loop 1.

The flux through loop 2 will be given by

    \[\phi=\int_{loop 2} \vec{B}.\vec{da}\]

As you can see, the flux is going to be proportional to the current through loop 1 and so we can write

    \[\phi_2=M_{12}I_1\]

Where

    \[M_{12}=\frac{\mu_0}{4\pi}\int_{loop 2}\int_{loop 2}\frac{1}{|\vec{r}|}dl\]

Is the mutual inductance between loop 1 and loop 2. We note that since it doesn’t matter which loop we consider, M_{12}=M_{21}=M. Secodnly, M is purely geometric, it only relies on the shape of the loops.

However, more often than not, we won’t be dealing with mutual inductance between two circuits or systems, we will usually just have the one loop. In this case, we note that the same phenomenon will occur, Lenz’s states that a current will flow to oppose the change in magnetic field that caused it. But where will this current flow if we don’t have another loop? You guessed it, loop 1 itself, the source loop! This is called self inductance, and the current induced will be opposed to the changing current that caused it. You can think of this induced current as fighting the change you are trying to make to the current in loop 1, and so, it is called the back EMF.  Now as before, the flux through loop 1 will be given by

    \[\phi=LI\]

where L is the self inductance of the loop.

In this case of self inductance, we can think of the inductance L as analogous to mass, in that it is what you have to fight against in order to cause a change. The relation is clear if you think about this: if you have a larger mass, it is harder to move it, and similarly if you have a large self-inductance L, the harder it is to change the current. L is sad to measure the magnetic inertia of a system.

The energy stored in the magnetic field for a given current is then given by

    \[U=\frac{1}{2}LI^2.\]

Now this formula works for the magnetic energy in or not in a magnetisable system, all you need to do is find the inductance first!

Capacitance

You may be asked somewhere along the line of your life to find capacitance. What is capacitance? A better question could be, why doesn’t Csaba teach this if he’s going to ask questions about it? The answer to that question is probably because it’s so simple!

Capacitance tells you how the potential of a system V changes with charge, Q, specifically, free charge, Q_f as we are only interested in the charge held by the capacitor.  So capacitance is given by

    \[C=\frac{Q}{V}.\]

That’s pretty much all you need to know. This is true for self capacitance of some conductor or the mutual capacitance between two conductors where the net charge is zero, e.g. a parallel plate capacitor.

Now, the energy stored in the capacitor can be found with a new equation, namely,

    \[W=\frac{1}{2}CV^2.\]

May the force be with you!

Submit a Comment

Your email address will not be published.

%d bloggers like this: