find the magnetic flux density when the vector potential is a position vector?
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14The Magnetic Field in Various Situations

14–1The vector potential
In this chapter we continue our discussion of magnetic fields associated with steady currents—the subject of magnetostatics. The magnetic field is related to electric currents by our basic equations∇⋅B=0,c2∇×B=jϵ0.We want now to solve these equations mathematically in a general way, that is, without requiring any special symmetry or intuitive guessing. In electrostatics, we found that there was a straightforward procedure for finding the field when the positions of all electric charges are known: One simply works out the scalar potential ϕ by taking an integral over the charges—as in Eq. (4.25). Then if one wants the electric field, it is obtained from the derivatives of ϕ. We will now show that there is a corresponding procedure for finding the magnetic field [math]B if we know the current density j of all moving charges.
In electrostatics we saw that (because the curl of E was always zero) it was possible to represent E as the gradient of a scalar field ϕ. Now the curl of B is not always zero, so it is not possible, in general, to represent it as a gradient. However, the divergence of B is always zero, and this means that we can always represent B as the curl of another vector field. For, as we saw in Section 2–7, the divergence of a curl is always zero. Thus we can always relate B to a field we will call A byB=∇×A.Or, by writing out the components,Bx=(∇×A)x=∂Az∂y−∂Ay∂z,By=(∇×A)y=∂Ax∂z−∂Az∂x,Bz=(∇×A)z=∂Ay∂x−∂Ax∂y.Writing B=∇×A guarantees that Eq. (14.1) is satisfied, since, necessarily,∇⋅B=∇⋅(∇×A)=0.The field A is called the vector potential.
You will remember that the scalar potential ϕ was not completely specified by its definition. If we have found ϕ for some problem, we can always find another potential ϕ′ that is equally good by adding a constant:ϕ′=ϕ+C.The new potential ϕ′ gives the same electric fields, since the gradient ∇C is zero; ϕ′ and ϕ represent the same physics.
Similarly, we can have different vector potentials A which give the same magnetic fields. Again, because B is obtained from A by differentiation, adding a constant to A doesn’t change anything physical. But there is even more latitude for A. We can add to A any field which is the gradient of some scalar field, without changing the physics. We can show this as follows. Suppose we have an A that gives correctly the magnetic field B for some real situation, and ask in what circumstances some other new vector potential A′ will give the same field B if substituted into (14.3). Then A and A′ must have the same curl:B=∇×A′=∇×A.Therefore∇×A′−∇×A=∇×(A′−A)=0.But if the curl of a vector is zero it must be the gradient of some scalar field, say ψ, so A′−A=∇ψ. That means that if A is a satisfactory vector potential for a problem then, for any ψ, at all,