# Magnetic Dipole - Torque, Energy and Force

## Table of contents

Consider a rectangular conducting loop with **side lengths** \(a\) and \(b\). Let the loop be rotatable around the axis parallel to the side \(a\). Let us call this axis \(\text{C}\). Let the loop be non-rotatable around the axis parallel to \(b\). In addition, the loop is traversed by an **electric current** \( \class{red}{I} \).

Let \(\boldsymbol{A}\) be the **orthogonal surface vector** representing the **area** \(A\) enclosed by the loop and orientation of this surface area (it is orthogonal to the surface area). With the right hand rule and the given current direction, the direction of \(\boldsymbol{A}\) is uniquely determined. The **magnetic dipole moment** is thus given by definition by the following formula:

If the loop is now placed in a homogeneous external **magnetic field** \( \class{violet}{\boldsymbol{B}} \), then the loop will rotate until \( \boldsymbol{A} \) and \( \class{violet}{\boldsymbol{B}} \) point in the same direction. During the rotation around the axis \(\text{C}\) the loop has a **torque** \(\boldsymbol{M}\) along \(\text{C}\). A formula for the torque is derived below.

## Torque

The torque \(\boldsymbol{M}\) about an axis of rotation \(\text{C}\) is defined as the cross product of the **distance** \( \boldsymbol{r} \) of the conductor piece from the axis of rotation and the **force** \( \boldsymbol{F} \) acting on the conductor piece:

The first parallel piece of conductor of **length** \(a\), which is in a homogeneous **magnetic field** \( \class{violet}{\boldsymbol{B}} \), experiences the **Lorentz force** (magnetic force) acts:

Here the vector \(\boldsymbol{a}\) runs along the piece of conductor. Analogously, the force can also be written with the current vector: \(a \, \boldsymbol{I} ~\times~ \class{violet}{ \boldsymbol{B} }\). Then the current is a vector indicating the direction of the current along the conductor and \(a\) is a scalar.

A magnetic force also acts on the second conductor piece, in which the current points in the opposite direction. However, this force points in the opposite direction:

The vector \(\boldsymbol{r}\) in Eq. 2

is the distance of the conductor piece of length \(a\) from the axis \(\text{C}\). That is \(\boldsymbol{r}\) is parallel to \(\boldsymbol{b}\). This distance is for the first conductor piece: \( \boldsymbol{r}_1 = \frac{\boldsymbol{b}}{2} \). The opposite second conductor piece, on the other hand, is at the same distance but in the opposite direction: \( \boldsymbol{r}_2 = -\frac{\boldsymbol{b}}{2} \). Both parts with the corresponding forces 3

and 4

contribute to the torque \(\boldsymbol{M}\). Thus, Eq. 2

becomes:

& ~=~ \frac{\boldsymbol{b}}{2} \times \boldsymbol{F} ~+~ \left(-\frac{\boldsymbol{b}}{2}\right) \times \left(-\boldsymbol{F}\right) \\\\

& ~=~ \boldsymbol{b} \times \boldsymbol{F} \end{align} $$

Now insert the Lorentz force 3

into Eq. 5

:

& ~=~ I \, \boldsymbol{b} \times \boldsymbol{a} ~\times~ \class{violet}{\boldsymbol{B}} \end{align} $$

The double cross product can be simplified. The cross product \(\boldsymbol{b} \times \boldsymbol{a} \) is *orthogonal* to \(\boldsymbol{a}\) and \(\boldsymbol{b}\) and corresponds exactly to the **surface orthogonal vector** \(\boldsymbol{A}\). Thus, Eq. 6

becomes:

If the magnetic dipole moment 1

is inserted into Eq. 7

, the relationship between the torque, magnetic dipole moment and external magnetic field is obtained:

The magnitude of the torque 8

is thus:

Here \(\varphi\) is the angle between the vectors \( \boldsymbol{\mu} \) and \( \class{violet}{ \boldsymbol{B} } \).

## Potential energy

With the help of the derived torque the **potential energy** \(W_{\mu}\) of the magnetic dipole can be derived.

The **work** \(W\) done by the external magnetic field to rotate the dipole is given by:

The magnitude \(M\) of the torque from Eq. 2

is given by:

The angle \(\alpha\) between the position vector \(\boldsymbol{r}\) and the force \(\boldsymbol{F}\) is 90 degrees here, because the force can be split into a part \(\boldsymbol{r}\) parallel to \(\boldsymbol{F}_{|}\) and into a part \(\boldsymbol{r}\) orthogonal to \(\boldsymbol{F}_{\perp}\). The parallel part does not contribute to the work done. Therefore, Eq. 11

can also be written as follows:

With the help of Eq. 12

the force in integral 10

can be expressed with the torque \(M\) (\(F:=F_{\perp}\)):

The \(\text{d}s\) element is expressed in polar coordinates as \(\text{d}s = -r \,\text{d}\varphi \) (minus sign because \(\text{d}s\) points opposite to \(\text{d}\varphi\)). When substituting into Eq. 11

, \(r\) cancels out. In this way, the work done by the magnetic field is expressed with the help of the given torque:

Now put the torque magnitude 10

into Eq. 14

. Integrate over the angle from \(\pi/2\) to a variable angle \(\varphi\). (The lower limit \(\pi/2\) was chosen so that the zero point of the potential energy is set to zero):

& ~=~ -\mu \, \class{violet}{B} \, \bigl[ - \cos(\varphi) \bigr]^{\varphi}_{\pi/2} \\\\

& ~=~ \mu \, \class{violet}{B} \, \bigl[ \cos(\varphi) - \cos(\pi/2) \bigr] \\\\

& ~=~ \mu \, \class{violet}{B} \, \cos(\varphi) \end{align} $$

The potential energy \(W_{\mu}\) of the magnetic dipole corresponds to the *negative* work \(W\) done by the external magnetic field: \( W_{\mu} = - W\). Thus the potential energy of the magnetic dipole is given by:

Or expressed compactly with the help of the scalar product:

## Force

A conservative **force** \(\boldsymbol{F}\) can be written as a gradient of the potential energy \(W_{\text{pot}}\):

Here, \(\nabla\) is the nabla operator. Substituting the potential energy 17

of the magnetic dipole into Eq. 18

yields:

The minus signs cancel out and we get the formula for the force:

If \(\boldsymbol{\mu}\) is position-independent, Eq. 20

can alternatively be written as follows:

Eq. 21

is to be understood in such a way: A matrix \(\nabla \class{violet}{\boldsymbol{B}}\) (in the three-dimensional case a 3x3 matrix) is applied to the vector \(\boldsymbol{\mu}\). The result is again a vector, namely the force \(\boldsymbol{F}\).

**Validity of the derived equations**: Even if the torque (and thus also energy and force) was derived with the help of a *rectangular* loop, the derived equations are valid for arbitrarily shaped loops, because the equations do not contain any geometrical quantities.