Magnetic Dipole Moment Calculator

Our magnetic dipole moment calculator allows you to obtain the magnetic dipole moment of a current loop as well as a solenoid's magnetic moment. Within a short number of paragraphs, you will learn what the magnetic dipole moment formula is, the unit of magnetic dipole moment, and understand the force on a dipole in a magnetic field. Let's jump right into it!

The magnetic moment $\overline{\mu}$ of a magnet is a measure of the strength and direction of the magnetic field it produces. Therefore it is a vector quantity. 'But wait, isn't that the definition of the magnetic dipole moment as well?' Well, yes, and no.

The magnetic moment is derived from the magnetic potential, which comprehends a series of terms that describe a magnetic field. The first term is associated with a magnetic monopole (magnet with either a north or south pole), the second term with dipoles (a regular magnet), the third with a quadrupole, etc.

Now, a magnetic monopole hasn't been discovered in nature yet. However, most magnets or bodies that produce a magnetic field can be easily represented as dipoles, so that's why we only keep the dipole term from the magnetic potential and call it 'magnetic dipole moment' instead.

In short, the magnetic dipole moment will tell us how strong a magnet is and how it will behave in the presence of an external magnetic field.

Before we introduce the magnetic dipole moment formula and let you experiment with the magnetic dipole moment calculator, let's make an analogy first, so it's easier to understand it. You should already be familiarized with the magnetic field of a current-carrying wire. If not, check our magnetic field of straight current-carrying wire calculator.

When said wire is wrapped around a loop, each part of the circumference adds up to a one-directional magnetic field in the center. This field is illustrated in the picture below:

Notice how the magnetic field looks similar to that of a regular magnet? If we take the area of the loop as if it was tiny, or if we are far away from it, we can consider them both almost identical (of course, in physics, we think 'almost' as good enough for our practical purposes 😉). Therefore, this wire will behave as if it was a magnet, and so, we can calculate the magnetic dipole moment of a current loop.

The loop's magnetic field magnitude depends on the current circulating through it, and since we can control the current, we can control its magnetic field's strength. Now, how do we express its orientation? We use the right-hand rule, similarly as in the cross product: We wrap the loop with our right hand in the direction of the current; our thumb will then point in the same direction as the field lines in the picture above:

Now that we have a way to measure the loop's magnetic field's strength and direction, how do we calculate the magnetic moment? With the magnetic dipole moment equation:

This equation translates as the intensity of the current ($I$), multiplied by the loop's area ($A$), therefore, the unit of the magnetic dipole moment is the ampere-square meter ($\text{A}\cdot \text{m}^{2}$).

It is a vector quantity, so the area should include its orientation according to the direction of the current, as we expressed above. We use this equation in the magnetic moment calculator to calculate the magnetic moment of an electron too.

The magnetic dipole moment equation is almost the same for a solenoid, which is a wire wrapped around in N turns. We multiply the previous equation by N:

Let's see how to solve some simple problems related to the magnetic dipole moment.

Example 1:

What is the magnitude of the magnetic dipole moment for a 2 m wire wrapped around a loop through which a 2 A current circulates?

Here we list the steps to solve this problem:

1. To understand how to calculate the magnetic moment, first, we must translate the length of the wire to the area of the loop. We do this using $A=\frac{L^{2}}{4\pi}$, where $L$ is the wire's length.
2. Now, we just multiply the result by 2 A, which is the intensity to get $\overline{\mu} \approx 0.6366 \ \text{A}\cdot \text{m}^{2}$
3. If we wrapped the wire around in N loops, we would get a solenoid, and its magnetic dipole moment would be the result we have obtained multiplied by N.
4. Check the result with the magnetic dipole moment calculator!

Example 2:

How strong should the current circulating through a 50 cm radius, 150-turn solenoid be to produce a magnetic dipole moment with a magnitude of 15 A⋅m²?

Lets's list the data we need to use the magnetic dipole moment formula:

• Characteristics of the body (solenoid, $N=150$) ✔
• Area ($A=\pi \cdot radius^{2}$) ✔
• Magnetic moment magnitude ✔
• Current intensity ❌

Now, if we go back and plug those values in the solenoid's magnetic dipole moment equation (remember to use meters instead of cm if you're not using our magnetic dipole moment calculator), we get:

There will exist a force on a dipole in a magnetic field $\overline{B}$ that causes it to align itself with the field's direction. It can be explained by the torque $\overline{\tau}$ the magnetic field exerts on each point in the magnet produced by the Lorentz Force. Since calculating that for every point is not practical for most cases, we can use the magnetic dipole moment instead. The following equation summarizes the total torque on the magnet:

This can be related to work, and thus, energy. The system, then, will try to align the magnet with the field since it is in that position that the system will require the minimum amount of energy to keep it there.