﻿ The Derivation of the Magnetic Field Associated with a Spinning Spherical Surface of Charge
San José State University

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The Derivation of the Magnetic
Field Associated with a Spinning
Spherical Surface of Charge

## Background

Although this webpage is devoted to the derivation of the magnetic field associated with an electrical field it equally well applies to any field generated by any type of charge which has two values. In particular it applies to the force associated with the interaction of nucleons. This force is usually included in what is conventionally called the nuclear strong force. The more important component of the strong force is the exclusive force involved in nucleonic spin pair formation. Exclusive means a nucleon can form a spin pair with one nucleon of the same type and one with a nucleon of the opposite type. The force associated with the nonexclusive interaction of nucleons is such that like-nucleons repel each other and unlike ones attract.

Special Relativity requires the existence of the magnetic field associated with amoving electric field. The same argument applies to any field with positive or negative values.

## Scalar and Vector Potential Functions

The vector of electric field intensities E may be derived from the gradient of a scalar potential function V(r) according to the equation

#### E = − ∇V

The magnetic field B is derived from the curl of the vector potential A as

## Derivation of the Vector Potential for a Magnetic Field

#### A(P) = (μ0/4π)∫ (J/ρ)dS

where P is an arbitrary point, ρ is the distance from the point P to the surface element dS of the surface S, J is the current vector in the surface element dS and μ0 is the permeability of free space.

## Evaluation of the Surface Integral for a Spinniing Sphere of Charge

Because of the spherical symmetry spherical coordinates (r, φ, θ) are the obvious choice. The sphere is spinning at a rate ω about a spin axis. A natural choice of a coordinate system would make the spin axis correspond to φ=0, but that turns out not to be the the most convenient for evaluating the integral to obtain A(P). Instead the vector from O, the center of the sphere, to the point P is the better choice for φ=0. It is also convenient to have a Cartesian coordinate system with the z-axis corresponding to φ=0.

The spin axis is tilted at an angle ψ; i.e., φ=ψ. It is also convenvenient for the Cartesian coordinate system to be oriented so the spin axis lies in the xz-plane.

The sphere has a radius of R. Let the length of OP be denoted as z. Consider a surface element on the sphere at (R, φ,θ). The area of the surface element is dS=R²sin(φ)dφdθ.

The current in the surface element is σv where σ is the surface charge density and v is the velocity vector of the charge in the surface element. The velocity vector v is given by

#### v = Ω×r

where r is the position vector of the surface element; i.e., the vector from the center of the sphere to the surface element. It is given by

#### r = Rsin(φ)cos(θ)i^ + Rsin(φ)sin(θ)j^ + Rcos(φ)k^

where i^, j^ and k^ are the unit vectors in the x, y and z directions, respectively. The spin axis vector is given by

where ω=|Ω|

Thus

#### | i^j^k^          | v = | ωsin(ψ)0ωcos(ψ) | | Rsin(φ)cos(θ) Rsin(φ)sin(θ) Rcos(φ) |

This evaluates to:

#### v = Rω[−(cos(ψ)sin(φ)sin(θ)i^ + (cos(ψ)sin(φ)cos(θ)−sin(ψ)cos(φ))j^ + (sin(ψ)sin(φ)sin(θ))k^]

All but one of the terms of v has a factor of sin(θ) or cos(θ). Since

#### ∫02πsin(θ)dθ = 0 and ∫02πcos(θ)dθ = 0

the only term that remains is

#### A(P) = −½(μ0R³σωsin(ψ)[∫0πsin(φ)cos(φ)dφ/ρ]j^

where ρ = (R²+z²−Rz·cos(φ))½.

Letting cos(φ) equal u the integral becomes

#### ∫−1+1(u/ρ)du = {−(R²+z²+Rzu)(R²+z²−2Rzu)½/(3R²z²)}−1+1which evaluates to −(1/(3R²z²))[(R²+z²+Rz)|(R−z)| − (R²+z²−Rz)(R+z)]

Now consider a point P within the sphere; i.e., z<R. The above integral thus reduces to (2z/3R²). If P is outside of the sphere then z>R and the above integral reduces to (2R/3z²).

Let Z denote the vector OP. Then (Ω×Z) is equal to −ωz·sin(ψ)j^. Then the vector potential A at P is given by

#### A(P) = (μ0Rσ/3)(Ω×Z) for P inside the sphere (z<R) A(P) = (μ0R4σ/3z²)(Ω×Z) for P outside the sphere (z>R)

Now having derived the vector potential it is desirable to transform the coordinate system to one in which the spin axis coincides with the z-axis and φ=0. The coordinates of P are the r=z, φ=ψ and θ. Then

#### A(r, φ, θ) = (μ0Rωσ/3)r·sin(φ)θ^ for r<R A(r, φ, θ) = (μ0R4ωσ/3r²)r·sin(φ)θ^ for r>R

where θ^ is the unit vector in the direction of increasing φ.

The magnetic field B within the sphere is given by

#### B = ∇ × A = (2μ0Rωσ/3)(cos(φ)r^ − sin(φ)φ^) = (2μ0Rωσ/3)z^ = (2μ0Rσ/3)Ω

where r^, φ^ and z^ are the unit vectors in the directions of increasing r, φ and z, respectively. Ω is the spin vector.

Thus the magnetic field B within the sphere is uniform and parallel to the spin axis vector.

Sources:

David Griffiths, Introduction to Electrodynamics, Prentice-Hall Inc., Englewood Cliffs, New Jersey, 1981.

Edward M.Purcell and David J. Morin, Electricity and Magnetism, Cambridge University Press, 2013.