Magnetostatiscs: dipole moment and magnetic field

[Angela G]

Homework Statement

A closed current loop consists of two semicircles with the radius and center at the origin. Den
one is in the yz plane and the other in the xz plane as the figures on the left above show. On a
very large distance R from the origin is a short solenoid coil with N turns, radius b, length
c, center of the point $( \frac{R}{ \sqrt (2) }, \frac{R}{\sqrt 2}$, 0), and the axis along the diagonal of the xy plane, as the figures shows. We assume that R >> a, R >> b, R >> c

a) Determine the magnetic dipole moments for the loop at the origin and for the solenoid.

b) Determine the mutual inductance of the two loops

Relevant Equations

$\vec m = I \int \vec da$

$\vec B = \frac{\mu_0 I }{4 \pi} \int \frac{\vec dl \times \vec r}{r^2}$

$\vec A(r) = \frac{\mu_0}{4\pi} \frac{ \vec m \times \vec r}{r^2}$

$\vec B = \nabla \times \vec A$

$\Phi = M_{12} I_{1 or 2}$
$\Phi = \vec B \cdot \vec da$

Hello!
I tried to solve a) see figure below, is it correct?
b) so what I think I can do is to solve $M_{12}$ from the equation of the magnetic flux then I will get $\frac{\Phi}{I} = M_{12}$ Then I can even use the equation får the magnetic flux and the magnetic field $$ \Phi = \int \vec B \cdot \vec da \Rightarrow \frac{1}{I} \int \vec B \cdot \vec da = M_{12} $$ Now the problem is to determine the magnetic field, I think I can get it either from the Biot-savarts law or the rotation of the vector potential A, since we have the magnetic dipole moment we can determine the vector potential A. I'm stuck here, because I have difficult to dealing with crossproducts and the coordinates. How can I continue? Is there anothe way to get the magnetic field? or the mutual inductance?

 

[TSny]

I tried to solve a) see figure below, is it correct?

Looks good to me except for a small error in the calculation of the magnetic moment of the solenoid. You wrote $\vec{da}$ as $da (\hat x +\hat y)$. But $\vec {da}$ should equal $da$ times a unit vector in the direction of $\vec {da}$.

b) ... Now the problem is to determine the magnetic field

There's a well-known formula for the magnetic field produced by a magnetic dipole at points far away on the axis of the dipole. Are you sure you haven't already covered that?

 

[Angela G]

Looks good to me except for a small error in the calculation of the magnetic moment of the solenoid. You wrote da→ as da(x^+y^).

exactly when I do the rigth- hand rule I got that the unit vecto i pointing at the direction of $\hat x + \hat y$, or is it not right?

 

[ergospherical]

What @TSny was pointing out is that if $d\mathbf{a}$ is an element of vector area, then it is related to the element $da$ of scalar area by $d\mathbf{a} = \mathbf{n} da$, where $\mathbf{n}$ is a vector of magnitude $|\mathbf{n}| = 1$. This is to say that $d\mathbf{a} = \dfrac{1}{\sqrt{2}}(\hat{\mathbf{x}} + \hat{\mathbf{y}}) da$.

 

[Angela G]

What @TSny was pointing out is that if $d\mathbf{a}$ is an element of vector area, then it is related to the element $da$ of scalar area by $d\mathbf{a} = \mathbf{n} da$, where $\mathbf{n}$ is a vector of magnitude $|\mathbf{n}| = 1$. This is to say that $d\mathbf{a} = \dfrac{1}{\sqrt{2}}(\hat{\mathbf{x}} + \hat{\mathbf{y}}) da$.

Ohhh, I see thank you

 

[Angela G]

Looks good to me except for a small error in the calculation of the magnetic moment of the solenoid. You wrote $\vec{da}$ as $da (\hat x +\hat y)$. But $\vec {da}$ should equal $da$ times a unit vector in the direction of $\vec {da}$.There's a well-known formula for the magnetic field produced by a magnetic dipole at points far away on the axis of the dipole. Are you sure you haven't already covered that?

Oh, I found those equations $$ \vec B_{dip} = \vec \nabla \times \vec A = \frac{\mu_0 m }{4 \pi r^3} \left ( 2 \cos \theta \hat r + \sin \theta \hat \theta \right ) $$ and $$ \vec B_{dip} = \frac{\mu_0 m }{4 \pi r^3} \left [3 (\vec m \cdot \hat r)\hat r - \vec m \right ] $$
But the first one is derived from having the magnetic dipole at the origin and with the z- direction and as I said I have diffucult with coordinates. I think I can use the second one but I'm not sure. or which do you mean?

 

[TSny]

Oh, I found those equations $$ \vec B_{dip} = \vec \nabla \times \vec A = \frac{\mu_0 m }{4 \pi r^3} \left ( 2 \cos \theta \hat r + \sin \theta \hat \theta \right ) $$ and $$ \vec B_{dip} = \frac{\mu_0 m }{4 \pi r^3} \left [3 (\vec m \cdot \hat r)\hat r - \vec m \right ] $$
But the first one is derived from having the magnetic dipole at the origin and with the z- direction and as I said I have diffucult with coordinates. I think I can use the second one but I'm not sure. or which do you mean?

You could use either formula. However, there is a typo in the second formula where you wrote $$ \vec B_{dip} = \frac{\mu_0 m }{4 \pi r^3} \left [3 (\vec m \cdot \hat r)\hat r - \vec m \right ] $$The factor of $m$ outside the bracket should not be there. That is, it should be
$$ \vec B_{dip} = \frac{\mu_0 }{4 \pi r^3} \left [3 (\vec m \cdot \hat r)\hat r - \vec m \right ] $$
Suppose you want to use this formula to find the magnetic field produced by the current loop at the location of the solenoid. The unit vector $\hat r$ then points in the direction from the current loop toward the solenoid.

How does the direction of $\hat r$ compare to the direction of $\vec m$ for the current loop?

How does the expression $(\vec m \cdot \hat r)\hat r$ simplify?

 

[Angela G]

the direction of $\vec m$ is the same as the direction of $\hat r$ so will have that $( \vec m \cdot \hat r ) \hat r = m\hat r$ isn't it? But this will be the dipole moment again

 

[TSny]

the direction of $\vec m$ is the same as the direction of $\hat r$ so will have that $( \vec m \cdot \hat r ) \hat r = m\hat r$ isn't it? But this will be the dipole moment again

Yes.

 

[Angela G]

ok, I understand so I got this, my question know is which area in the equation 1 in the pic do I have to use? I know it has to be the cross sectional area of the solenoid but I'm not sure if I have to multiply the area of a circular loop times the number of turns in the solenoid or how can I do it? Arre my calulations right?

 

[TSny]

Yes, when calculating the "flux linkage" $\Phi$ through the solenoid, you should include the number of turns of the solenoid. Increasing the number of turns will increase the mutual inductance.

When obtaining the magnitude of $\vec B_{\rm dip}$, you neglected the magnitude of $(\hat x + \hat y)$.

Otherwise, your work looks good.

 

[Angela G]

ohh, so the magnetic field should be $$
\vec B_{dip} = \frac{\mu_0 }{4 \pi r^3} \left [3 (\vec m \cdot \hat r)\hat r - \vec m \right ] = \frac{\mu_0 }{4 \pi r^3}\left [3 ( m)\hat r - \vec m \right ] = \frac{\mu_0 }{4 \pi r^3}\left [3 \frac{1}{2}I_1 \pi a^2\frac{1}{\sqrt 2} (\hat x + \hat y ) - \frac{1}{2}I_1 \pi a^2 (\hat x + \hat y ) \right ] \\= \frac{\mu_0 }{4 \pi r^3}\frac{3\sqrt 2-2}{4}I_1 \pi a^2 (\hat x + \hat y )
$$

And then we will get that the mutual inductance is
$$ M_{12} = \frac{ \mu_0 \pi(3 \sqrt 2 -2)}{16 R^3 }N b^2a^2$$ Is it correct?

 

[TSny]

ohh, so the magnetic field should be $$
\vec B_{dip} = \frac{\mu_0 }{4 \pi r^3} \left [3 (\vec m \cdot \hat r)\hat r - \vec m \right ] = \frac{\mu_0 }{4 \pi r^3}\left [3 ( m)\hat r - \vec m \right ] = \frac{\mu_0 }{4 \pi r^3}\left [3 \frac{1}{2}I_1 \pi a^2\frac{1}{\sqrt 2} (\hat x + \hat y ) - \frac{1}{2}I_1 \pi a^2 (\hat x + \hat y ) \right ] \\= \frac{\mu_0 }{4 \pi r^3}\frac{3\sqrt 2-2}{4}I_1 \pi a^2 (\hat x + \hat y )
$$

You wrote correctly that $$\vec B_{dip} = \frac{\mu_0 }{4 \pi r^3} \left [3 (m)\hat r - \vec m \right ]$$ but what you wrote after that is not correct. In particular, the factor of $\frac{1} {\sqrt{2}}$ after the third equality should not be there. (The magnitude of $\vec m$ is not $\frac{1}{2}I_1 \pi a^2$. $\, \,m$ contains a factor of $\sqrt 2$ which cancels the $\frac{1}{\sqrt 2}$ in $\hat r$.)

Earlier in post #10 you were correct in obtaining $$\vec B_{dip} = \frac{\mu_0 }{4 \pi R^3} \left [2 \vec m \right ] = \frac{\mu_0 }{4 R^3} I_1 a^2 (\hat x + \hat y ) $$

You know that the flux linkage through the solenoid is $N_{sol}\, \pi b^2 |\vec B_{dip}|$.

Note that $|\vec B_{dip}| = \frac{\mu_0 }{4 R^3} I_1 a^2 | (\hat x + \hat y )|$.

What is the value of $| (\hat x + \hat y )|$?

 

[Angela G]

I think I understand now. In a) when I determined the magnetic dipole moment of the current loop I got that the vector of the magnetic dipole is $\vec m = \frac{1}{2}I_1 \pi a^2 (\hat x+ \hat y)$ but the magnitude of the magnetic dipole moment is $$ m = \left (\left[\frac{1}{2}I_1 \pi a^2 \right ]^2 + \left [ \frac{1}{2}I_2 \pi a^2 \right ]^2 \right) ^\frac{1}{2} = \left ( 2\left[\frac{1}{2}I_1 \pi a^2 \right ]^2 \right) ^\frac{1}{2} = \frac{\sqrt 2}{2}I_1 \pi a^2
$$

then the vector $\hat r$ is $\hat r = \frac{1}{\sqrt 2} (\hat x + \hat y)$ so we have that the magnetic dipole moment actually is $\vec m = \frac{1}{2} I_1 \pi a^2 ( \hat x +\hat y)$ and it is what we got in a)

So putting it in the magnetic field equation we will have $$
\vec B_{dip} = \frac{\mu_0 }{4 \pi R^3} \left [2 \vec m \right ] = \frac{\mu_0 }{4 R^3} I_1 a^2 (\hat x + \hat y)$$

The magnitude of the magnetic field is $$
|\vec B_{dip}| = \frac{\mu_0 }{4 R^3} I_1 a^2 | (\hat x + \hat y )| = \frac{\mu_0 }{4 R^3} I_1 a^2 | (\hat x + \hat y )| = \frac{\mu_0 }{4 R^3} I_1 a^2 \sqrt 2 $$

So, since we have that the magnetic flux on the solenoid from the current loop is $$ \Phi = \int \vec B \cdot da $$ ( the magnetic field and da is parallel) and then we have that$$ \Phi = \int \vec B \cdot da = \int | \vec B | da = | \vec B | N \pi b^2 $$ as you said
so we have that $$ \Phi = \int \vec B \cdot da = N \pi b^2 \frac{\mu_0 }{4 R^3} I_1 a^2 \sqrt 2 $$
so the mutual inductance is $$ M_{12} = \sqrt 2 N \pi\frac{\mu_0 }{4 R^3} b^2 a^2
$$

 

[TSny]

That all looks good and I agree with your final answer.

I would have written $\Phi$ as $\Phi = N \int \vec B \cdot \vec {da}$, where the area corresponds to the cross-section of the solenoid, $\pi b^2$. Thus, $$ \Phi = N\int \vec B \cdot \vec {da} = N\int | \vec B | da = | \vec B | N \pi b^2 $$ which is the same as what you got. You included $N$ in the integration over the area as if you are integrating over the areas of all of the turns. I think that's OK. It's a matter of choice.

[Edit: Often, one sees $M$ defined as the constant in $N_2 \Phi_2 = M I_1$ where $N_2$ is written explicitly. Then, $\Phi_2$ is the flux through one turn of coil #2. These different ways of writing it are all OK as long as the meaning of the symbols in each case is clear.]

 

[Angela G]

thank you very much for helping me

 

[TSny]

thank you very much for helping me

You are welcome. Nice work.