Question 5.53: Another way to fill in the “missing link” in Fig. 5.48 is to...

(a)  For uniform \text { B, } \int_{0}^{ r }( B \times d l )= B \times \int_{0}^{ r } d l = B \times r \neq A =-\frac{1}{2}( B \times r ) .

(b)   B =\frac{\mu_{0} I}{2 \pi s} \hat{\phi}, \text { so } \oint B \times d l =\left(\frac{\mu_{0} I}{2 \pi a} \hat{ s }-\frac{\mu_{0} I}{2 \pi b} \hat{ s }\right) w=\frac{\mu_{0} I w}{2 \pi}\left(\frac{1}{a}-\frac{1}{b}\right) \hat{ s } \neq 0 .

(c)   A =- r \times B \int_{0}^{1} \lambda d \lambda=-\frac{1}{2}( r \times B ) .

(d)   B =\frac{\mu_{0} I}{2 \pi s} \hat{\phi} ; B (\lambda r )=\frac{\mu_{0} I}{2 \pi \lambda s} \hat{\phi} ; \quad A =-\frac{\mu_{0} I}{2 \pi s}( r \times \hat{\phi}) \int_{0}^{1} \lambda \frac{1}{\lambda} d \lambda=-\frac{\mu_{0} I}{2 \pi s}( r \times \hat{\phi}) . But r here is the vector from the origin—in cylindrical coordinates r =s \hat{ s }+z \hat{ z } . \quad \text { So } A =-\frac{\mu_{0} I}{2 \pi s}[s(\hat{ s } \times \hat{ \phi })+z(\hat{ z } \times \hat{ \phi })] , and

(\hat{ s } \times \hat{ \phi })=\hat{ z },(\hat{ z } \times \hat{ \phi })=-\hat{ s } . \text { So } \mathbf { A } =\frac{\mu_{0} I}{2 \pi s}(z \hat{ s }-s \hat{ z } .

The examples in (c) and (d) happen to be divergenceless, but this is not in general the case. For (letting L \equiv \int_{0}^{1} \lambda B (\lambda r ) d \lambda , for short) \nabla \cdot A =- \nabla \cdot( r \times L )=-[ L \cdot( \nabla \times r )- r \cdot( \nabla \times L )]= r \cdot( \nabla \times L ) , and  \nabla \times L =\int_{0}^{1} \lambda[ \nabla \times B (\lambda r )] d \lambda=\int_{0}^{1} \lambda^{2}\left[ \nabla _{\lambda} \times B (\lambda r )\right] d \lambda=\mu_{0} \int_{0}^{1} \lambda^{2} J (\lambda r ) d \lambda, \text { so } \nabla \cdot A =\mu_{0} r \cdot \int_{0}^{1} \lambda^{2} J (\lambda r ) d \lambda , and it vanishes in regions where J = 0 (which is why the examples in (c) and (d) were divergenceless). To construct an explicit counterexample, we need the field at a point where J≠ 0—say, inside a wire with uniform current.

Here \text { Ampére }’s law gives B 2 \pi s=\mu_{0} I_{ enc }=\mu_{0} J \pi s^{2} \Rightarrow B =\frac{\mu_{0} J}{2} s \hat{\phi} , so 

A =- r \times \int_{0}^{1} \lambda\left(\frac{\mu_{0} J}{2}\right) \lambda s \hat{\phi} d \lambda=-\frac{\mu_{0} J}{6} s( r \times \hat{ \phi })=\frac{\mu_{0} J s}{6}(z \hat{ s }-s \hat{ z }).

\nabla \cdot A =\frac{\mu_{0} J}{6}\left[\frac{1}{s} \frac{\partial}{\partial s}\left(s^{2} z\right)+\frac{\partial}{\partial z}\left(-s^{2}\right)\right]=\frac{\mu_{0} J}{6}\left(\frac{1}{s} 2 s z\right)=\frac{\mu_{0} J z}{3} \neq 0 .

Conclusion: (ii) does not automatically yield ∇ · A = 0.