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Numerical on magnetic field

  • #1
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Homework Statement



Consider a solid sphere of radius r and mass m which has a charge q distributed uniformly over its volume. The sphere is rotated about the diameter with an angular speed ω. Show that magnetic moment μ and the angular momentum l of the sphere is related as,

μ=ql/2m

Homework Equations





The Attempt at a Solution



I know that μ=ql/2m is a general result and q/2m is gyromagnetic ratio and is same for every body of given charge distribution, mass and angular speed and is independent of the shape.

But how to derive it for solid sphere ?

Angular momentum=l=Iω=2mr2ω/5

Now to find magnetic moment:
μ=ieqA

How to proceed after this ??

Please help !!

Thanks in advanced... :)
 

Answers and Replies

  • #2
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Start by calculating magnetic moment of small rings and then integrate. Not sure but this could be a way.
 
  • #3
785
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Start by calculating magnetic moment of small rings and then integrate. Not sure but this could be a way.
I already tried that approach but to no avail.

V=4πr3/3

I took a small strip at a distance x from the centre. Then volume of strip=4πx2dx.

dq = qdV/V = 3qx2dx/r3

di=dq/T= 3qωx2dx/2πr3
dμ= 6qωx4dx/r3

On integrating R.H.S under the limit x=0 to x=r

μ = 6qωr2/5

On dividing this by angular momentum, does the yield the correct answer ... :(
 
  • #4
3,812
92
I already tried that approach but to no avail.

V=4πr3/3

I took a small strip at a distance x from the centre. Then volume of strip=4πx2dx.

dq = qdV/V = 3qx2dx/r3

di=dq/T= 3qωx2dx/2πr3
dμ= 6qωx4dx/r3

On integrating R.H.S under the limit x=0 to x=r

μ = 6qωr2/5

On dividing this by angular momentum, does the yield the correct answer ... :(
μ=iA can be directly applied only to closed loops. To solve this problem, you can also start by using the fact that a solid sphere is made of many hollow spheres. Find magnetic moment due to a hollow sphere and integrate. If you want to go by selecting rings, you may have to deal with double integrals.

EDIT: I see that you have already selected a hollow sphere. What is the magnetic moment of a hollow sphere?
 
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  • #5
785
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μ=iA can be directly applied only to closed loops. To solve this problem, you can also start by using the fact that a solid sphere is made of many hollow spheres. Find magnetic moment due to a hollow sphere and integrate. If you want to go by selecting rings, you may have to deal with double integrals.

EDIT: I see that you have already selected a hollow sphere. What is the magnetic moment of a hollow sphere?
I do not have to remember magnetic moment of hollow sphere for doing this, I hope. This question is 61st and 60th question was to find the same for hollow sphere. I can verify this for a ring or a disc but not for spheres. Usual integration is not yielding correct answer. What you said, I have already done. What is wrong in my approach ? Point it out please.

Edit: BTW, are you giving a second try for JEE ?
 
  • #6
3,812
92
I do not have to remember magnetic moment of hollow sphere for doing this, I hope.
Yes, you don't need to. You can find it by using μ=(q/2m)l or through integration.
This question is 61st and 60th question was to find the same for hollow sphere.
Umm....what are you talking about? :uhh:
I can verify this for a ring or a disc but not for spheres. Usual integration is not yielding correct answer. What you said, I have already done. What is wrong in my approach ? Point it out please.
The error is here
di=dq/T= 3qωx2dx/2πr3
dμ= 6qωx4dx/r3
As I said before, you cannot directly μ=iA. This is applicable only to closed loops. For objects as presented in the above question, you need to go through the process of integration. For example, you will have to select closed rings in case of a rotating charged disk.
 
  • #7
785
15
Yes, you don't need to. You can find it by using μ=(q/2m)l
That will be using the answer to the question itself in order to find the very answer.

or through integration.
Do you know how long will the solution become then ? :uhh:

Umm....what are you talking about? :uhh:
Nevermind.

The error is here

As I said before, you cannot directly μ=iA. This is applicable only to closed loops. For objects as presented in the above question, you need to go through the process of integration. For example, you will have to select closed rings in case of a rotating charged disk.
I did not directly μ=iA. In fact, I did find dμ and integrated it within right limits. Finding di and integrating it first and then applying μ=iA will not yield correct answer because area A is itself function of parameter x.
 
  • #8
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That will be using the answer to the question itself in order to find the very answer.
Its completely ok to do that. You said you already found out the magnetic moment of hollow sphere in one of the previous question, you can use that. Another way is to deal with double integrals to solve the problem.

I did not directly μ=iA. In fact, I did find dμ and integrated it within right limits. Finding di and integrating it first and then applying μ=iA will not yield correct answer because area A is itself function of parameter x.
You agree that μ=iA can be directly applied only to closed loops? What you have here is a hollow sphere not a closed loop like ring. How did you solve the problem of finding out the magnetic moment for hollow sphere then?
 
  • #9
785
15
Its completely ok to do that. You said you already found out the magnetic moment of hollow sphere in one of the previous question, you can use that. Another way is to deal with double integrals to solve the problem.
I would prefer double integrals to do it.

You agree that μ=iA can be directly applied only to closed loops?
Yes..

What you have here is a hollow sphere not a closed loop like ring. How did you solve the problem of finding out the magnetic moment for hollow sphere then?
No I did not. I mistyped before. The previous question was to find the magnetic moment for the disc which I did by taking rings. So I hope that here I will have to take rings and integrate for the disc and then again integrate the disc for the whole sphere. That will be performing "double integration". Am I correct ?

I cannot take a disc for a closed loop, right ?

Edit: Magnetic moment of hollow sphere is qωr2/3 on using the gyromagnetic ratio directly.
 
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  • #10
3,812
92
I cannot take a disc for a closed loop, right ?
Yes. You yourself said that you had to choose rings for a disc and integrate.

Edit: Magnetic moment of hollow sphere is qωr2/3 on using the gyromagnetic ratio directly.
Correct.

I tried the problem in two ways. First is to consider the solid sphere is made of many hollow spheres. Second I tried was to think of the solid sphere made of discs. I couldn't reach the final answer using the second method. I am still trying to find out the error. The first method gives the answer in a few steps given that you know the magnetic moment of a hollow sphere.
 
  • #11
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Yes. You yourself said that you had to choose rings for a disc and integrate.


Correct.

I tried the problem in two ways. First is to consider the solid sphere is made of many hollow spheres. Second I tried was to think of the solid sphere made of discs. I couldn't reach the final answer using the second method. I am still trying to find out the error.
I think your error was to consider disc instead of rings.

The second method is tough.

Either perform double integration by taking rings, i.e. put double integrals with their limits and then integrate the dμ for the ring selected in a solid sphere. This is the integrand.

Either take the rings in the solid sphere and integrate for the disc and then again integrate for the sphere. This one seems to be silly.

Either if I use the magnetic moment directly for the disc in previous question and integrate for the sphere.
(I see, in one reference book they took the parameter angle subtended θ in order to find magnetic moment of a hollow sphere. Obviously they first considered rings.)

The first method gives the answer in a few steps given that you know the magnetic moment of a hollow sphere.
So in first method, If I take a hollow sphere at a distance x from the centre of a solid sphere, its magnetic moment is,

dμ=(dq)ωx2/3

dq= qdV/V = ....

dq= 3qx2dx/r3

Then I found dμ and integrated it for μ. Then I divided it by angular momentum which yielded correct answer...

Thanks. :)

I will give a try on the second method.
 
  • #12
785
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Ahh yes !!!! I did this question directly by transforming double integration into single integration by inserting the parameter angle (theta). Anyone (Pranav ?) interested ?
 

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