Obvious Magnetic potential question

[bayners123]

Homework Statement

I've been given a scalar magnetic potential of \phi = D cos (\theta) and asked to prove that this corresponds to a constant magnetic field. It's obvious that it does but I keep running into walls!

Homework Equations

\bar{H} = - \nabla \phi
\bar{H} = -D cos (\theta) \hat{r} + D sin (\theta) \hat{\theta}

The Attempt at a Solution

I tried to find \left (\frac{\partial \bar{H}}{\partial z} \right)_{x,y}
and \left (\frac{\partial \bar{H}}{\partial x} \right)_{z} but it came out non-zero (and I'm pretty sure I did it wrong).

I also did \frac{\partial H}{\partial r} and \frac{\partial H}{\partial \theta} and got 0 for R, but non zero for theta. Doing \frac{\partial (H^2)}{\partial r} gives 0 but that's not proof..

 

[Rayquesto]

hey dude! I'm no expert and I'm not in an advance physics course, but this looks loike really hard stuff. WHen it comes to partial derivatives, that's going to be me in like 2 years. I am way behind it all. So, I am curious, what level of physics are you in?

 

[bayners123]

hey dude! I'm no expert and I'm not in an advance physics course, but this looks loike really hard stuff. WHen it comes to partial derivatives, that's going to be me in like 2 years. I am way behind it all. So, I am curious, what level of physics are you in?

It's not as bad as it looks, I'm just having trouble with something that should be simple. I'm a second year undergrad in the UK.

 

[I like Serena]

I've been given a scalar magnetic potential of \phi = D cos (\theta) and asked to prove that this corresponds to a constant magnetic field. It's obvious that it does but I keep running into walls!

Hi bayners, welcome to PF!

Just an observation, but if the field were not constant, shouldn't phi have a dependency on time?

(Btw, I'm impressed that you obviously know how to calculate a gradient in spherical coordinates! )

 

[bayners123]

Hi bayners, welcome to PF!

Just an observation, but if the field were not constant, shouldn't phi have dependency on time?

(Btw, I'm impressed that you obviously know how to calculate a gradient in spherical coordinates! )

Hi ILSe! I've been a long time lurker but only just started posting

Sorry, I've been a bit unclear: I meant a uniform field, not constant. If the field varied on time then I suppose it wouldn't be valid to use the scalar potential anyway.

The larger question this comes from is a ferromagnetic sphere inserted into what was previously a uniform magnetic field. I've done Laplace's equation and narrowed it down to the D cos(\theta) potential inside the sphere but I'm trying to show that this potential is that of another uniform H field, parallel to the original field but weaker.

 

[I like Serena]

Hi ILSe! I've been a long time lurker but only just started posting

That sounds nice.
I hope you've seen only good things from me then. ;)

Sorry, I've been a bit unclear: I meant a uniform field, not constant. If the field varied on time then I suppose it wouldn't be valid to use the scalar potential anyway.

Mmh, as I've learned it, a uniform magnetic field is a field that has the same strength and direction everywhere.
It seems to me that your field does not have this property.

The larger question this comes from is a ferromagnetic sphere inserted into what was previously a uniform magnetic field. I've done Laplace's equation and narrowed it down to the D cos(\theta) potential inside the sphere but I'm trying to show that this potential is that of another uniform H field, parallel to the original field but weaker.

I'm a little rusty here.
Could you show some results?
Then perhaps I might be able to be of some assistance.

 

[bayners123]

Mmh, as I've learned it, a uniform magnetic field is a field that has the same strength and direction everywhere.
It seems to me that your field does not have this property.

I'm a little rusty here.
Could you show some results?
Then perhaps I might be able to be of some assistance.

The question is effectively the same as a dielectric sphere in a uniform electric field (I think): there, the potential V = - z E_o = - r cos (\theta) E_o . That's going from a uniform field to a potential in the form of A cos(\theta) which I suppose is proof enough, but I'd rather work back from the given potential and find that the field does not vary in any Cartesian direction. The rest of the question is just using the spherical solutions to Laplace's equation and fitting the boundary conditions to it:

• \lim_{r \to +\infty} \bar{H} = - E_o r cos(\theta)
• \lim_{r \to 0} \bar{H} is finite
• at r=a, (B_r)_{in} = (B_r)_{out}
• at r=a, (H_\theta)_{in} = (H_\theta)_{out} since there are no surface currents

 

[I like Serena]

The question is effectively the same as a dielectric sphere in a uniform electric field (I think): there, the potential V = - z E_o = - r cos (\theta) E_o . That's going from a uniform field to a potential in the form of A cos(\theta) which I suppose is proof enough, but I'd rather work back from the given potential and find that the field does not vary in any Cartesian direction. The rest of the question is just using the spherical solutions to Laplace's equation and fitting the boundary conditions to it:

• \lim_{r \to +\infty} \bar{H} = - E_o r cos(\theta)
• \lim_{r \to 0} \bar{H} is finite
• at r=a, (B_r)_{in} = (B_r)_{out}
• at r=a, (H_\theta)_{in} = (H_\theta)_{out} since there are no surface currents

Hmm, well I know a sphere in an static electric field is different from a sphere in a static magnetic field.
The first behaves like a cage of Faraday, while the second does not.
I'd have to rework Maxwell's equations to remember why that is though.
But I think we should be able to find that on e.g. wikipedia...

It's not enough to solve Laplace's equation for the field outside the sphere though...

Either way, I'll have to admit that your field is uniform after all.

A field of the type E=E0, will indeed have a potential of :

V = - z E_o = - r cos (\theta) E_o

Something similar holds for a magnetic field (vector potential!).

But still, this is outside the sphere...

I'd have to work on this a bit more to be of more help.

Or if you'd like I can see if I can call in more expert help...?

 

[bayners123]

Hmm, well I know a sphere in an static electric field is different from a sphere in a static magnetic field.
The first behaves like a cage of Faraday, while the second does not.
I'd have to rework Maxwell's equations to remember why that is though.
But I think we should be able to find that on e.g. wikipedia...

I think you're thinking of a conductive shell there (or sphere for that matter): if you put an insulating dielectric sphere in an electric field you do get a similar uniform electric field.

I know in general it's a vector potential, but since \nabla \times H = \bar{J} + \frac{\partial D}{\partial t} = 0 we know a closed line integral of H . dl would always be 0 so we can also use a scalar potential. You can get to the field inside being only dependant on an r cos(theta) term by making the field finite at r=0, eliminating the r^{-2} term, but that's actually given in the question. I was just wondering if there was a nifty way of showing that the A r cos(\theta) potential corresponds to a uniform field, because it's not immediately obvious.

 

[bayners123]

And of course, it's blindingly obvious once you notice it! Just looked at your last post again: all you do is say \phi = A r cos(\theta) = A z therefore:
\bar{H}_z = -\frac{\partial \phi}{\partial z} = -A, \frac{\partial \phi}{\partial x} = \frac{\partial \phi}{\partial y} = 0

You just need to convert to cartesians at the \bar{H} = -\nabla \phi stage rather than once you've found the field in polars.
Thanks so much for all your help!

 

[I like Serena]

And of course, it's blindingly obvious once you notice it! Just looked at your last post again: all you do is say \phi = A r cos(\theta) = A z therefore:
\bar{H}_z = -\frac{\partial \phi}{\partial z} = -A, \frac{\partial \phi}{\partial x} = \frac{\partial \phi}{\partial y} = 0

You just need to convert to cartesians at the \bar{H} = -\nabla \phi stage rather than once you've found the field in polars.
Thanks so much for all your help!

That looks right!

And you're welcome!