What is the size of the magnetic dipole moment?

[aledg97]

Consider the magnetic field B generated by a magnetic dipole. The intensity of B measured along the
axis of the dipole, at a distance of 10 cm from the dipole itself, is 1.0 10-5 T. What is the size of the
magnetic dipole moment? (μ0 =4π10-7 mkg/C2)
a) 0.050 Am2
b) 5 10 -4 Am2
c) 0,1 Am2
d) 1 10 -4 Am2 I know that the magnetic dipole moment involves the area ( in fact its unit is A* m^2). But the problem does not provide the area of the loop (?).

So I know the intensity of the field, how do I get the dipole moment?

 

[Charles Link]

See post 5 of https://www.physicsforums.com/threa...-the-suns-magnetic-field.953814/#post-6045203 Thanks to @berkeman for googling this about a week ago.

 

[aledg97]

So, which formula do I have to use? I don't understand the quantities involved in the formulas you reffered to

 

[Charles Link]

So, which formula do I have to use? I don't understand the quantities involved in the formulas you reffered to

$\vec{m}=IA \, \hat{z}$ is the magnetic dipole. Along the axis (at least for positive z) will mean that $\vec{m} \cdot \vec{r} =m r$ and $\vec{r}=r \hat{z}$. That should make it easy to evaluate the numerator of $B$.

 

[aledg97]

I still have a question. I don't know if it's correct to ask this. I think we didn't see this formula in the course. So is there another way to find the magnetic dipole moment?
Maybe with a parallelism with electric dipole moment.

 

[Charles Link]

I still have a question. I don't know if it's correct to ask this. I think we didn't see this formula in the course. So is there another way to find the magnetic dipole moment?
Maybe with a parallelism with electric dipole moment.

The answer is yes. The same result can be derived using magnetic poles and the inverse square law, directly analogous to the electric dipole moment. You assume that the dipole moment is of the same magnitude $m=IA$, but instead is the result of magnetic poles (magnetic charges), of $+q_m$ and $-q_m$ separated by a distance $d$. (In the calculation, I believe you need to make the approximation that $d$ is very small=(it's been a few years since I did this calculation, but I did get an identical result)). Edit: $\mu_o m=q_m d$. You then compute $H$ from the dipole, (analogous to $E$, but with a $\mu_o$ in place of $\epsilon_o$). Finally $B=\mu_o H$. This will give an identical formula for $B$ as in the "link" of post 2. $\\$ The magnetic pole theory isn't taught as much as it was 35-40 years ago when I was a university student. The emphasis now, in ferromagnetic materials/magnets, is on bound surface currents rather than magnetic poles. $\\$ Additional comment: The on-axis case is easier to compute than the general case. If you follow the formulas above with magnetic charges, it should get you the answer very quickly if you do a Taylor expansion in $\frac{d}{z}$. Edit: And yes, I computed it this way just now, and I got a result that agrees with the formula in the "link" in post 2 for the on-axis case. $\\$ And to write it out for you, on-axis $H=\frac{+q_m}{4 \pi \mu_o}(\frac{1}{z^2}-\frac{1}{(d+z)^2} )$. The next step is to get the expression to first order in $d$, ($d<<z$), with a series expansion of the second term, and then use $q_m d=\mu_o m$. It quickly gives the same result that you get if you evaluate the "link" of post 2. $\\$ Additional note: On-axis, you could easily begin with a small circular current loop, and compute the magnetic field $B$ using Biot-Savart. You would also get the same result.