Help with magnetic field in sphere problem

[catsonmars]

Homework Statement

Consider a polycrystalline sample of Subscript[CaSO, 4]\[CenterDot]2 Subscript[H, 2]O in an external magnetic field Overscript[B, \[RightVector]] in the z direction. The internal magnetic field (in the z direction) produced at the position of a given proton in the Subscript[H, 2]O molecule by the neighboring proton is given by ∂/a^3 ( 3 cos^2 \θ -1 ) if the spin of this neighboring proton points along the applied field. It is given by -(∂/a^3)( 3 cos^2 θ-1 ) if this neighboring spin points in a direction opposite to the applied field. Here, \[Mu] is the magnetic moment of the proton and a is the distance between the two protons, while \[Theta] denotes the angle between the line joining the two protons and the z axis. In this sample of randomly-oriented crystals, the neighboring proton is equally likely to be located anywhere on the sphere of radius a surrounding the given proton.

Homework Equations

da=a^2sinθdθdδ
b=\[∂]/a^3 ( 3 cos^2θ -1 )

The Attempt at a Solution

da=a^2sinθdθdδ

The probability of finding p\[Theta]d(\[Theta]) is

p(θ)dθ=[a^2sinθdθ∫dδ/(4∏a^2) The itnregral being from 0 to 2∏


The integral deserves some explanation. As the external proton orbits around the molecule the electric field will be constant throught it because the external magnetic field is everywhere the same around 2Pi
simplifying gives us

pθdθ=sinθdθ/2

We are looking for w(b)db as our result so we write down the equation. The graph showing this distribution is given on a separate page. We are uncertain about the probability of locating the external proton at radius a and angle θ. We do know that θ andθ+dθ lie somehwere withing b and b+db. Adding up all the values of b that have θ through θ+dθ is given by

w(b)db=2pθdθ*absvalue(db/dθ)

We look at the plot of b vs θ to see that db intersects \[Theta] twice.

Substituting P(θ)d\θ] =sinθdθ/2 into w(b)db and we get the following.

w(b)db=2sinθd/2*absvalue(db/dθ)

w(b)db=sinθdθ*absvalue(db/dθ)

Now we need to find db. Start by writing what we know about b.

b=\[Mu]/a^3 ( 3 cos^2 θ -1 ) To find db we take the derivative with respect to θ.

d(b/dθ) =-6a^-3*∂(cosθsinθ)
substitute this into w(b)db

w(b)db=6sinθdθ*a^-3∂(cosθsinθ)

We dropped the minus sign because we take the absolute value of db/dθ.
Simply further
w(b)db= sinθ^2cosθ dθ 6a^-3


The answer key has this in terms of area, which is something I am having difficulty with. Reif solves the area as A=1/2sqrt(a^3/3∂)
I start out with

da=a^2sinθdθdδ

I take the double integral of both sides
∫∫da=a^2∫∫sinθdθdδ
A^2=a^2sinθdθδ

A=(2a^2sinθdθδ)^1/2

Substitute in pθdθ=sinθdθ/2

A=(4a^2Pθdθδ)
This is where I get stuck.

 

[mark726]

I don't know how to get from this to the final answer of A=1/2sqrt(a^3/3∂). I am also uncertain if my approach is correct. Can anyone offer any guidance or assistance? Thank you for your help.
Thank you for bringing up this interesting topic for discussion. I am familiar with the concepts and equations mentioned in your post. However, I am not sure if I fully understand your approach to solving the problem.

Firstly, I would like to clarify that the equation for the internal magnetic field produced by a neighboring proton is given by b=\[∂]/a^3 ( 3 cos^2θ -1 ), where \[∂] represents the magnetic moment of the proton. This equation is derived from the equation for the magnetic field produced by a dipole, which is given by B=\[∂]/r^3 (3cosθ-1), where r is the distance between the two dipoles and θ is the angle between the line joining the two dipoles and the direction of the magnetic field.

Now, to find the probability of locating the external proton at a distance a and angle θ, we need to consider the fact that the neighboring proton can be located anywhere on a sphere of radius a surrounding the given proton. This means that the probability of finding the neighboring proton at a particular angle θ is proportional to the surface area of the spherical shell at that angle. This surface area is given by A=2πa^2sinθdθ, where dθ is the infinitesimal angle. So, the probability of finding the neighboring proton at a distance a and angle θ is given by P(θ)dθ=A/4πa^2, which simplifies to P(θ)dθ=sinθdθ/2.

Next, we need to find the probability of locating the external proton at a distance b, which is the internal magnetic field produced by the neighboring proton. This probability is given by w(b)db, where w(b) is the probability distribution function for b and db is the infinitesimal distance. From the equation for b, we can see that b is a function of θ. So, we can write b=b(θ). Now, to find w(b), we need to consider the fact that the neighboring proton can be located at any angle θ, which means