Skip to main content
Physics LibreTexts

17.6: Thirteen Questions

We have seen that the SI definition of magnetic moment is unequivocally defined as the maximum torque experienced in unit external field.  Nevertheless some authors prefer to think of magnetic moment as the product of the equatorial field and the cube of the distance.  Thus there are two conceptually different concepts of magnetic moment, and, when to these are added minor details as to whether the magnetic field is \(B\) or \(H\), and whether or not the permeability should include the factor \(4 \pi\), six possible definitions of magnetic moment, described in Section 17.6, all of which are to be found in current literature, arise. 

Regardless, however, how one chooses to define magnetic moment, whether the SI definition or some other unconventional definition, it should be easily possible to answer both of the following questions:

A. Given the magnitude of the equatorial field on the equator of a magnet, what is the maximum torque that that magnet would experience if it were placed in an external field?

B. Given the maximum torque that a magnet experiences when placed in an external field, what is the magnitude of the equatorial field produced by the magnet?

It must surely be conceded that a failure to be able to answer such basic questions indicates a failure to understand what is meant by magnetic moment.

I therefore now ask a series of thirteen questions.  The first six are questions of type A, in which I use the six possible definitions of magnetic moment.  The next six are similar questions of type B.  And the last is an absurdly simple question, which anyone who believes he understands the meaning of magnetic moment should easily be able to answer.

1.  The magnitude of the field in the equatorial plane of a magnet at a distance of 1 cm is 1 Oe.

     What is the maximum torque that this magnet will experience in an external magnetic field of 1 Oe, and what is its magnetic moment?  

  Note that, in this question and the following seven there must be a unique answer for the torque.  The answer you give for the magnetic moment, however, will depend on how you choose to define magnetic moment, and on whether you choose to give the answer in SI units or CGS EMU.

 

2.  The magnitude of the field in the equatorial plane of a magnet at a distance of 1 cm is 1 \(\text{Oe}\).

     What is the maximum torque that this magnet will experience in an external magnetic field of 1 \(\text{G}\), and what is its magnetic moment?

 

3.  The magnitude of the field in the equatorial plane of a magnet at a distance of 1 cm is 1 \(\text{G}\).

     What is the maximum torque that this magnet will experience in an external magnetic field of 1 \(\text{Oe}\), and what is its magnetic moment?

 

4.  The magnitude of the field in the equatorial plane of a magnet at a distance of 1 cm is 1 \(\text{G}\).

     What is the maximum torque that this magnet will experience in an external magnetic field of 1 \(\text{G}\), and what is its magnetic moment?

 

5.  The magnitude of the field in the equatorial plane of a magnet at a distance of 1 \(\text{m}\) is 1 \(\text{A m}^{-1}\).

     What is the maximum torque that this magnet will experience in an external magnetic field of 1 \(\text{A m}^{-1}\), and what is its magnetic moment?

 

6.  The magnitude of the field in the equatorial plane of a magnet at a distance of 1 \(\text{m}\) is 1 \(\text{A m}^{-1}\).

     What is the maximum torque that this magnet will experience in an external magnetic field of 1 \(\text{T}\), and what is its magnetic moment?

 

7.  The magnitude of the field in the equatorial plane of a magnet at a distance of 1 \(\text{m}\) is 1 \(\text{T}\).

     What is the maximum torque that this magnet will experience in an external magnetic field of 1 \(\text{A m}^{-1}\), and what is its magnetic moment?

 

8.  The magnitude of the field in the equatorial plane of a magnet at a distance of 1 \(\text{m}\) is 1 \(\text{T}\).

     What is the maximum torque that this magnet will experience in an external magnetic field of 1 \(\text{T}\), and what is its magnetic moment?

 

9.  A magnet experiences a maximum torque of 1 dyn cm if placed in a field of 1 \(\text{Oe}\).  What is the magnitude of the field in the equatorial plane at a distance of 1 cm, and what is the magnetic moment?

Note that, in this question and the following three there must be a unique answer for \(B\) and a unique answer for \(H\), though each can be expressed in SI or in CGS EMU, while the answer for the magnetic moment depends on which definition you adopt.

 

10.  A magnet experiences a maximum torque of 1 dyn cm if placed in a field of 1 \(\text{G}\).  What is the magnitude of the field in the equatorial plane at a distance of 1 cm, and what is the magnetic moment?

 

11.  A magnet experiences a maximum torque of 1 \(\text{N}\) m if placed in a field of 1 \(\text{A m}^{-1}\).  What is the magnitude of the field in the equatorial plane at a distance of 1 \(\text{m}\), and what is the magnetic moment?

 

12.  A magnet experiences a maximum torque of 1 \(\text{N m}\) if placed in a field of 1 \(\text{T}\).  What is the magnitude of the field in the equatorial plane at a distance of 1 \(\text{m}\), and what is the magnetic moment?

I’ll pose Question Number 13 a little later.  In the meantime the answers to the first four questions are given in Table \(\text{XVII.2}\), and the answers to Questions 5 – 12 are given in Tables \(\text{XVII.3}\) and \(4\).  The sheer complexity of these answers to absurdly simple questions is a consequence of different usages by various authors of the meaning of “magnetic moment” and of departure from standard SI usage.

 

\(\text{TABLE XVII.2}\)
ANSWERS TO QUESTIONS 1 – 4 IN CGS EMU AND SI UNITS
The answers to the first four questions are identical

 

\(\text{TABLE XVII.3}\)
ANSWERS TO QUESTIONS 5 – 8 IN CGS EMU AND SI UNITS

\begin{array}{l l c \quad c \quad c \quad c \quad l}
 &  & 5 & 6 & 7 & 8 &  \\
\tau & = & (4\pi)^2 & 4\pi \times 10^7 & 4\pi \times 10^7 & 10^{14} & \text{dyn cm} \\
 & = & (4\pi)^2 \times 10^{-7} & 4\pi & 4\pi & 10^7 & \text{N m} \\
&&&&&& \\
p_1 & = & 4\pi \times 10^3 & 4\pi \times 10^3 & 10^{10} & 10^{10} & \text{dyn cm Oe}^{-1} \\
 & = & (4\pi)^2 \times 10^{-7} & (4\pi)^2 \times 10^{-7} & 4\pi & 4\pi & \text{N m (A/m)}^{-1} \\
&&&&&&\\
p_2 & = & 4\pi \times 10^3 & 4\pi \times 10^3 & 10^{10} & 10^{10} & \text{dyn cm G}^{-1} \\
& = & 4\pi & 4\pi & 10^7 & 10^7 & \text{N m T}^{-1} \\
&&&&&&\\
p_3 & = & 4\pi \times 10^3 & 4\pi \times 10^3 & 10^{10} & 10^{10} & \text{G cm}^3 \\
& = & 4\pi \times 10^{-7} & 4\pi \times 10^{-7} & 1 & 1 & \text{T m}^3 \\
&&&&&&\\
p_4 & = & 4\pi \times 10^3 & 4\pi \times 10^3 & 10^{10} & 10^{10} & \text{Oe cm}^3 \\
& = & 1 & 1 & 10^7/(4\pi) & 10^7/(4\pi) & \text{A m}^2 \\ 
&&&&&&\\
p_5 & = & (4\pi)^2 \times 10^3 & (4\pi)^2 \times 10^3 & 4\pi \times 10^{10} & 4\pi \times 10^{10} & \text{G cm}^3 \\ 
& = & (4\pi)^2 \times 10^{-7} & (4\pi)^2 \times 10^{-7} & 4\pi & 4\pi & \text{ T m}^3 \\
&&&&&&\\
p_6 & = & (4\pi)^2 \times 10^3 & (4\pi)^2 \times 10^3 & 4\pi \times 10^{10} & 4\pi \times 10^{10} & \text{Oe cm}^3 \\ 
& = & 4\pi & 4\pi & 10^7 & 10^7 & \text{A m}^2 \\
\end{array}
 

 

\(\text{TABLE XVII.4}\)
ANSWERS TO QUESTIONS 9 – 12 IN CGS EMU AND SI UNITS

\begin{array}{l l c c c c l}
& & 9 & 10 & 11 & 12 & \\
B & = & 1 & 1 & 10^4/(4\pi) & 10^{-3} & G \\
& = & 10^{-4} & 10^{-4} & 1/(4\pi) & 10^{-7} & T \\
&&&&&&\\
H & = & 1 & 1 & 10^4/(4\pi) & 10^{-3} & \text{Oe} \\
& = & 10^3/(4\pi) & 10^3/(4\pi) & 10^7/(4\pi)^2 & 1/(4\pi) & \text{A m}^{-1} \\ 
&&&&&&\\
p_1 & = & 1 & 1 & 10^{10}/(4\pi) & 10^3 & \text{dyn cm Oe}^{-1} \\
& = & 4\pi \times 10^{-10} & 4\pi \times 10^{-10} & 1 & 4\pi \times 10^{-7} & \text{N m (A/m)}^{-1} \\
&&&&&&\\
p_2 & = & 1 & 1 & 10^{10}/(4\pi) & 10^3 & \text{dyn cm G}^{-1} \\ 
& = & 10^{-3} & 10^{-3} & 10^7/(4\pi) & 1 & \text{N m T}^{-1} \\
&&&&&&\\
p_3 & = & 1 & 1 & 10^4/(4\pi) & 10^{-3} & \text{G cm}^3 \\ 
& = & 10^{-10} & 10^{-10} & 10^{-6}/(4\pi) & 10^{-13} & \text{T m}^3 \\
&&&&&&\\
p_4 & = & 1 & 1 & 10^4/(4\pi) & 10^{-3} & \text{Oe cm}^3 \\ 
& = & 10^{-3}/(4\pi) & 10^{-3}/(4\pi) & 10/(4\pi)^2 & 10^{-6}/(4\pi) & \text{A m}^2 \\ 
&&&&&&\\
p_5 & = & 4\pi & 4\pi & 10^4 & 4\pi \times 10^{-3} & \text{G cm}^3 \\ 
& = & 4\pi \times 10^{-10} & 4\pi \times 10^{-10} & 10^{-6} & 4\pi \times 10^{-13} & \text{T m}^3 \\ 
&&&&&&\\
p_6 & = & 4\pi & 4\pi & 10^4 & 4\pi \times 10^{-3} & \text{Oe cm}^3 \\
& = & 10^{-3} & 10^{-3} & 10/(4\pi) & 10^{-6} & \text{A m}^2 \\
\end{array}

 

The thirteenth and last of these questions is as follows:  Assume that Earth is a sphere of radius \(6.4 \times 10^6 \text{m} = 6.4 \times 10^8 \text{cm}\), and that the surface field at the magnetic equator is \(B = 3\times 10^{-5}  ,\ \text{T}=0.3 \ \text{G}, \text{or} \ H = 75/\pi \ \text{A m}^{-1} = 0.3  \ \text{Oe}\), what is the magnetic moment of Earth?  It is hard to imagine a more straightforward question, yet it would be hard to find two people who would give the same answer.

The SI answer (which, to me, is the only answer) is

\[B = \frac{\mu_0}{4\pi}\frac{p}{r^3}, \ \therefore \ p=\frac{4\pi r^3 B}{\mu_0}=\frac{4\pi \times (6.4 \times 10^6)^3 \times 3 \times 10^{-5}}{4\pi \times 10^{-7}}=7.86 \times 10^{22} \ \text{N m T}^{-1}.\]

This result correctly predicts that, if Earth were placed in an external field of \(1 \ \text{T}\), it would experience a maximum torque of \(7.86 \times 10^{22} \ \text{N m}\), and this is the normal meaning of what is meant by magnetic moment.

A calculation in GCS might proceed thus:

\[B = \frac{p}{r^3}, \ \therefore \ p = r^3 B = (6.4 \times 10^8)^3 \times 0.3 = 7.86 \times 10^{25} \ \text{G cm}^3.\]

Is this the same result as was obtained from the SI calculation?  We can use the conversions \(1 \ \text{G} = 10^{-4} \text{T}\) and \(1 \ \text{cm}^3 =  10^{-6} \text{m}^3\), and we obtain

\[p = 7.86 \times 10^{15} \text{T m}^3.\]

We arrive at a number that not only differs from the SI calculation by \(10^7\), but is expressed in quite different, dimensionally dissimilar, units.

Perhaps the CGS calculation should be

\[H = \frac{p}{r^3}, \ \therefore \ p = r^3H = (6.4 \times 10^8)^3 \times 0.3 = 7.86 \times 10^{25} \ \text{Oe cm}^3.\]

Now \(1 \ \text{Oe} = 1000/(4\pi) \ \text{A m}^{-1}\) and \(1 \ \text{cm}^3 = 10^{-6} \text{m}^3\), and we obtain

\[p=6.26 \times 10^{21} \text{A m}^2\]

This time we arrive at SI units that are dimensionally similar to \(\text{N m T}^{-1}\), and which are perfectly correct SI units, but the magnetic moment is smaller than correctly predicted by the SI calculation by a factor of 12.6.

Yet again, we might do what appears to be frequently done by planetary scientists, and we can multiply the surface field in \(\text{T}\) by the cube of the radius in \(\text{m}\) to obtain

\[p=7.86 \times 10^{15} \ \text{T m}^3.\]

This arrives at the same result as one of the CGS calculations, but, whatever it is, it is not the magnetic moment in the sense of the greatest torque in a unit field.  The quantity so obtained appears to be nothing more that the product of the surface equatorial field and the cube of the radius, and as such would appear to be a purposeless and meaningless calculation.

It would be a good deal more meaningful merely to multiply the surface value of \(H\) by 3.  This in fact would give (correctly) the dipole moment divided by the volume of Earth, and hence it would be the average magnetization of Earth – a very meaningful quantity, which would be useful in comparing the magnetic properties of Earth with those of the other planets.