Derivation: 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} $.

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Rectangular current loop in a magnetic field.

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:

Magnetic dipole moment

$$ \begin{align} \class{red}{\boldsymbol{\mu}} ~=~ \class{red}{I} \, \boldsymbol{A} \end{align} $$

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:

$$ \begin{align} \boldsymbol{M} ~=~ \boldsymbol{r} \times \boldsymbol{F} \end{align} $$

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:

$$ \begin{align} \boldsymbol{F}_1 ~:=~ F ~=~ I \, \boldsymbol{a} ~\times~ \class{violet}{ \boldsymbol{B} } \end{align} $$

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:

$$ \begin{align} \boldsymbol{F}_2 ~:=~ -F \end{align} $$

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:

$$ \begin{align} \boldsymbol{M} & ~=~ \boldsymbol{r}_1 \times \boldsymbol{F}_1 ~+~ \boldsymbol{r}_2 \times \boldsymbol{F}_2 \\\\
& ~=~ \frac{\boldsymbol{b}}{2} \times \boldsymbol{F} ~+~ \left(-\frac{\boldsymbol{b}}{2}\right) \times \left(-\boldsymbol{F}\right) \\\\
& ~=~ \boldsymbol{b} \times \boldsymbol{F} \end{align} $$

Now insert the Lorentz force 3 into Eq. 5:

$$ \begin{align} \boldsymbol{M} & ~=~ \boldsymbol{b} \times \left( I \, \boldsymbol{a} ~\times~ \class{violet}{\boldsymbol{B}}\right) \\\\
& ~=~ I \, \boldsymbol{b} \times \boldsymbol{a} ~\times~ \class{violet}{\boldsymbol{B}} \end{align} $$

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:

$$ \begin{align} \boldsymbol{M} ~=~ I \, \boldsymbol{A} ~\times~ \class{violet}{\boldsymbol{B}} \end{align} $$

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Magnetic dipole in magnetic field experiences torque

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:

Torque of a dipole in the external magnetic field

$$ \begin{align} \boldsymbol{M} ~=~ \class{red}{\boldsymbol{\mu}} \times \class{violet}{\boldsymbol{B}} \end{align} $$

The magnitude of the torque 8 is thus:

$$ \begin{align} M ~=~ \mu \, \class{violet}{ B } \, \sin(\varphi) \end{align} $$

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:

$$ \begin{align} W ~=~ \int F \, \text{d}s \end{align} $$

The magnitude $M$ of the torque from Eq. 2 is given by:

$$ \begin{align} M ~=~ r \, F \, \sin(\alpha) \end{align} $$

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:

$$ \begin{align} M ~=~ r \, F_{\perp} \end{align} $$

With the help of Eq. 12 the force in integral 10 can be expressed with the torque $M$ ($F:=F_{\perp}$):

$$ \begin{align} W ~=~ \int \frac{M}{r} \, \text{d}s \end{align} $$

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:

$$ \begin{align} W ~=~ -\int M \, \text{d}\varphi \end{align} $$

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):

$$ \begin{align} W & ~=~ -\mu \, \class{violet}{B} \int_{\pi/2}^{\varphi} \sin(\varphi) \, \text{d}\varphi \\\\
& ~=~ -\mu \, \class{violet}{B} \, \bigl[ - \cos(\varphi) \bigr]^{\varphi}_{\pi/2} \\\\
& ~=~ \mu \, \class{violet}{B} \, \bigl[ \cos(\varphi) - \cos(\pi/2) \bigr] \\\\
& ~=~ \mu \, \class{violet}{B} \, \cos(\varphi) \end{align} $$

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:

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

Or expressed compactly with the help of the scalar product:

Potential energy of a dipole in the magnetic field

$$ \begin{align} W_{\mu} = -\class{red}{\boldsymbol{\mu}} \cdot \class{violet}{\boldsymbol{B}} \end{align} $$

Force

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

$$ \begin{align} \boldsymbol{F} ~=~ - \nabla W_{\text{pot}} \end{align} $$

Here, $\nabla$ is the nabla operator. Substituting the potential energy 17 of the magnetic dipole into Eq. 18 yields:

$$ \begin{align} \boldsymbol{F} ~=~ - \nabla \left( -\boldsymbol{\mu} \cdot \class{violet}{\boldsymbol{B}} \right) \end{align} $$

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

Force on a magnetic dipole

$$ \begin{align} \boldsymbol{F} ~=~ \nabla \left( \class{red}{\boldsymbol{\mu}} \cdot \class{violet}{\boldsymbol{B}} \right) \end{align} $$

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Magnetic dipole in an inhomogeneous magnetic field.

If $\boldsymbol{\mu}$ is position-independent, Eq. 20 can alternatively be written as follows:

$$ \begin{align} \boldsymbol{F} ~=~ (\nabla \class{violet}{\boldsymbol{B}}) \, \boldsymbol{\mu} \end{align} $$

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.