Taking small element for integration purpose in SOLID sphere?

[shivam01anand]

Homework Statement

The Question originally is to find the m of a solid uniformly charged solid sphere which is rotating uniformly with ω

Now

Homework Equations

Now my question to you is how to take the small element?

The Attempt at a Solution

i take a small disc with radius rsinθ.

Now a= ∏(rsinθ)^2

i= q/t= dq/(2∏/w)

where dq= rho times dv= 2∏rsinθ times rdθ.

now this gives the relevant answer but why is it that i can't take a small element of a small sphere?

because i can't decribe m for that too?

It's just that i don't recall taking a disc as a small element in case of solid sphere :~(

 

[rude man]

Please quote the problem verbatim. If it's not in English then I would need a better translation.
What is m? Usually mass, obviously not in this case.

 

[shivam01anand]

I am so sorry for not posting here accurately.


The question is to find the magnetic dipole moment of a solid sphere rotating about its axis with w angular velocity. The total charge on the sphere, Q is uniformly distributed@ rho ρ volumetric distribution.

It's really not the doubt actually.

It's actually about the small element which must be taken which i guess is quite important considering if we want to find the com of a solid hemisphere.right?

 

[rude man]

Looks like a double integration is necessary.

Consider a thin slice dz of the sphere with normal along the spin axis = z axis. The sphere's center is at the origin of an xyz coordinate system. Then consider a thin annular volume of radius r, width dr and thickness dz within this slice.

What is the differential current di due to this annular volume? Then, what is the differential magnetic moment dm due to this di?

dm now needs to be integrated from r = 0 to r = R where R is the radius of the slice, giving a magnetic moment dμ.

Note that R = R(z): R(+/-a) = 0 and R(0) = a where a is the radius of the sphere.

That takes care of the slice of thickess dz and radius R.

Now you have to do a second integration along z from z= -a to z = a, adding all the dμ. Then you have integrated all the dμ moments into one magnetic moment μ.

 

[Nickie]

Taking a small element for integration purposes in a solid sphere is a common practice in many scientific fields, including physics, mathematics, and engineering. The reason for this is that it allows us to break down a complex problem into smaller, more manageable parts. In the case of finding the mass of a rotating charged solid sphere, using a small disc as the element is a valid approach.

The key is to choose a small element that is representative of the entire sphere and can be easily integrated. In this case, using a disc with a radius of rsinθ is a suitable choice. This disc can be thought of as a small slice of the sphere, and by integrating over all slices, we can find the total mass of the sphere.

It is important to note that the choice of the small element may vary depending on the specific problem and the desired outcome. In some cases, a small sphere may be a more suitable element, while in others, a disc may work better. Ultimately, the choice should be based on the properties of the problem and the desired outcome.

I hope this explanation helps to clarify why a small disc is a valid choice for taking a small element in the integration process for a solid sphere. It is a common and effective approach used by many scientists and engineers in various fields.