How can I calculate the z-component of the magnetic field using the dot product?

In summary: I still don't get why only the first ##\hat r## is replaced with ##\hat z## and same for m for the second one.Because the dot product is a scalar, it depends on the coordinates (x,y,z) (or ##|r|,\theta,\phi)## but its a scalar, it is a quantity that "propagates" as a multiplicative constant to all three components ##B_x,B_y,B_z## of the magnetic field.
  • #1
gabriellelee
21
1
Homework Statement
Find B in the z-drection
Relevant Equations
B = (u/4pi)((3z(z.m)-m)/(|z|^3))
Screen Shot 2019-09-21 at 8.55.33 PM.png

I thought I could replace r^ with z^ and |r|=z since it's only in the z-direction. Is it okay for me to do that?
Also, do I need to consider the magnetic dipole moment, m, only in the z-direction as well? Or can I just keep it as m?

So assuming that I can replace r^ with z^ and keep m as just m. My answer is
Bz(r) = (mu * m) / (2 * pi * (z^3)).
Is this correct? Can I say ^r*^r=1?
 
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  • #2
Hi
Keep ##\mathbf{r} \cdot \mathbf{m}## and ##|r|^3## as they are because they are scalar.
Take z component of vector ##3\mathbf{r}##.
 
  • #3
mitochan said:
Hi
Keep ##\mathbf{r} \cdot \mathbf{m}## and ##|r|^3## as they are because they are scalar.
Take z component of vector ##3\mathbf{r}##.

But then how would I get rid of the unit vector if I replace ##3\mathbf{r}## with ##3\mathbf{z}##? And it's actually Bz(r). Does that change anything?
 
  • #4
In addition to what is said in #2, you should also take the z-component of the magnetic dipole moment and put it in place of ##m## at the end (after the minus). (leave the m as it is in the dot product ##\hat r\cdot m##.
 
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  • #5
Delta2 said:
In addition to what is said in #2, you should also take the z-component of the magnetic dipole moment and put it in place of ##m## at the end (after the minus). (leave the m as it is in the dot product ##\hat r\cdot m##.

I think I can consider m as a constant because that's what I will be calculating after I measure the B field of a magnet. So if it's a constant, ##\mathbf{\hat{r}} \cdot \mathbf{m}## becomes ##\mathbf{\hat{r}} \mathbf{m}## and if I replace ##3\mathbf{\hat{r}}## with ##3\mathbf{\hat{z}}## like it says in #2, how would i get rid of the unit vector? Because this is a simple multiplication, not a dot or a cross product. Can I say ##\mathbf{\hat{r}} (\mathbf{\hat{z}}) = 0##? So in the end does it become, ## B_z= \mu_0m/4 \pi r^3##?
 
  • #6
gabriellelee said:
But then how would I get rid of the unit vector if I replace ##3\mathbf{r}## with ##3\mathbf{z}##? And it's actually Bz(r). Does that change anything?
it isn't right to get rid of ##\hat r## inside the dot product. What is right is to replace ##3\hat r## by its z component which is ##3\frac{z}{|r|}=3\frac{|r|\cos\theta}{|r|}=3\cos\theta##
where ##\theta## the inclination angle.

m is a constant vector not a scalar constant so you can't do what you say to the dot product ##\hat r\cdot m##.
 
  • #7
This is getting confusing, you want to get the z component in cartesian coordinates or in spherical coordinates?
 
  • #8
Delta2 said:
This is getting confusing, you want to get the z component in cartesian coordinates or in spherical coordinates?

In the cartesian coordinate
 
  • #9
Ok so we looking for the function ##B_z(x,y,z)## (and not for the function ##B_z(|r|,\theta,\phi##))

First answer what is the z component of the ##\hat r##?
 
  • #10
Delta2 said:
Ok so we looking for the function ##B_z(x,y,z)## (and not for the function ##B_z(|r|,\theta,\phi##))

First answer what is the z component of the ##\hat r##?

Isn't it just ##\hat r##?
 
  • #11
gabriellelee said:
Isn't it just ##\hat r##?
er NO try again
 
  • #12
Isnt the z-component of ##\hat r## equal to ##\frac{z}{|r|}\hat z##?
 
  • #13
Delta2 said:
er NO try again

##z \hat z/ r##?
 
  • #14
gabriellelee said:
##z \hat z/ r##?
ok correct, put that in place of ##\hat r## in the original expression, replacing only the first ##\hat r## and not the second inside the dot product. And then we are almost finished, we just need to express everything in terms of cartesian coordinates or spherical coordinates.
 
  • #15
Delta2 said:
ok correct, put that in place of ##\hat r## in the original expression, replacing only the first ##\hat r## and not the second inside the dot product. And then we are almost finished, we just need to express everything in terms of cartesian coordinates or spherical coordinates.

This is what I got.
Screen Shot 2019-09-21 at 11.25.46 PM.png
 
  • #16
Well there is a slight mistake, the dipole moment ##m## is a constant vector so it will have a z component ##m_z\hat z## put that instead of the second m in the above expression but leave the m inside the dot product unharmed.
 
  • #17
Delta2 said:
Well there is a slight mistake, the dipole moment ##m## is a constant vector so it will have a z component ##m_z\hat z## put that instead of the second m in the above expression but leave the m inside the dot product unharmed.

Screen Shot 2019-09-21 at 11.38.15 PM.png

I still don't get why only the first ##\hat r## is replaced with ##\hat z## and same for m for the second one.
 
  • #18
Because the dot product is a scalar, it depends on the coordinates (x,y,z) (or ##|r|,\theta,\phi)## but its a scalar, it is a quantity that "propagates" as a multiplicative constant to all three components ##B_x,B_y,B_z## of the magnetic field.

Now it is correct to write it more properly with the unit z vector ##\hat z## at the end it will be
$$B_z(r)=\frac{3\mu_0}{4\pi}\frac{z(m\cdot \hat r)-m_z|r|}{|r|^4}\hat z$$
 
  • #19
Delta2 said:
Because the dot product is a scalar, it depends on the coordinates (x,y,z) (or ##|r|,\theta,\phi)## but its a scalar, it is a quantity that "propagates" as a multiplicative constant to all three components ##B_x,B_y,B_z## of the magnetic field.

Now it is correct to write it more properly with the unit z vector ##\hat z## at the end it will be
$$B_z(r)=\frac{3\mu_0}{4\pi}\frac{z(m\cdot \hat r)-m_z|r|}{|r|^4}\hat z$$

Oh, okay. Thank you so much!
 
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  • #20
Sometimes it's helpful to rely on the math. To find the (scalar) ##z##-component of ##\vec B##, you calculate
$$B_z(\vec r) = \hat z \cdot \vec B = \hat z \cdot \left[\frac {\mu_0}{4\pi} \frac{3\hat r (\hat r \cdot \vec m)-\vec m}{r^3}\right] = \frac {\mu_0}{4\pi} \frac{3(\hat z \cdot \hat r) (\hat r \cdot \vec m)-\hat z \cdot\vec m}{r^3},$$ evaluating the dot products that you can. There's no need to guess if ##\hat r## is replaced by ##\hat z##, etc. Just follow what the math tells you. (If you want the vector component, ##\vec B_z(\vec r)##, tack on the ##\hat z## that @Delta2 did in post 18.)

By the way, the factor of 3 is in the wrong spot in both posts 17 and 18.
 
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What is the meaning of "Find B in the z-direction"?

"Find B in the z-direction" refers to finding the value of the magnetic field (B) in the z-direction, which is the vertical direction in a three-dimensional coordinate system. This can also be interpreted as finding the strength and direction of the magnetic field at a specific point in the z-axis.

What is the importance of finding B in the z-direction?

Finding B in the z-direction is important in understanding the behavior of magnetic fields in three-dimensional space. It allows us to determine the strength and direction of the magnetic field at a specific point, which is crucial in many scientific and technological applications such as in electromagnetism, particle accelerators, and MRI machines.

What are the different methods for finding B in the z-direction?

There are several methods for finding B in the z-direction, including using the Biot-Savart law, Ampere's law, or the right-hand rule. In some cases, it may also be possible to use mathematical calculations or experimental measurements to determine the value of B in the z-direction.

How does B in the z-direction affect charged particles?

B in the z-direction has a significant impact on charged particles, as it determines the direction and magnitude of the force exerted on the particles. This force, known as the Lorentz force, is perpendicular to both the magnetic field and the velocity of the charged particle and can cause the particle to move in a circular or helical path.

Can B in the z-direction be manipulated or controlled?

Yes, B in the z-direction can be manipulated and controlled by using various methods such as changing the current in an electromagnet, altering the shape or orientation of a permanent magnet, or using ferromagnetic materials. This manipulation of B in the z-direction is essential in many technological devices, including motors, generators, and magnetic storage devices.

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