# Purcell and the Magnetic Field

When showing that the magnetic field derived from the curl being proportional to current and the divergence being zero is unique, Purcell says that at some sufficiently remote enclosing boundary the difference of two magnetic fields (two possible fields from the curl), D, would take a constant value. Why is this true?

• Seems unlikely. No choice of gauge should affect the magnetic field. More than likely he meant the vector potential could be adjusted by an overall constant in, for example, the Coulomb gauge, since applying the divergence and curl operations send all constants to zero. – Pricklebush Tickletush May 9 '13 at 4:35
• Is this at the end of section 6.2? In the edition of Purcell that I have handy (1963), there is no specific argument about a remote enclosing boundary. – Ben Crowell May 9 '13 at 11:18
• @BenCrowell I think it may only be in the second edition. Purcell makes a similar argument to the one I make below, except he uses $\nabla\times \mathbf D = 0$ to obtain $\mathbf D = \nabla f$ for some scalar $f$, then notes that since $\nabla\cdot \mathbf D = 0$, $f$ satisfies Laplace's equation. He then states "Over a sufficiently remote enclosing boundary, $f$ must take on some constant value $f_0$." – joshphysics May 9 '13 at 16:12
• @Tony I added an addendum to my response that might be helpful. – joshphysics May 9 '13 at 16:56

Suppose that two fields $\mathbf B_1$ and $\mathbf B_2$ both satisfy the magnetostatics equations $$\nabla\times\mathbf B_i = \mu_0\mathbf J, \qquad \nabla\cdot\mathbf B_i = 0, \qquad i = 1,2$$ Let $$\mathbf D= \mathbf B_2 - \mathbf B_1$$ then the curl and divergence conditions give $$\nabla\times \mathbf D= 0, \qquad \nabla\cdot\mathbf D = 0$$ Now your question reduces to what we can say about the vector field $\mathbf D$. Take the curl of both sides of the first equation and use the following vector calculus identity: $$(\nabla\times(\nabla\times \mathbf D))_i = \partial_i(\nabla\cdot \mathbf D) - \nabla^2D_i$$ along with the zero divergence condition, to obtain $$\nabla^2 D_i = 0$$ In other words, each component $D_i$ satisfies Laplace's equation. Now, if we assume that the magnetic field components vanish at infinity (as would be the case for a bounded current distribution) then we recall that the only solution to Laplace's equation that vanishes at infinity is the zero solution. This gives $\mathbf D = 0$ and thus $\mathbf B_1 = \mathbf B_2$.