Question 5.25: If B is uniform, show that A(r) = −1/2 (r × B) works. That i...

If B is uniform, show that A(r) = – \frac{1}{2} (r × B) works. That is, check that · A = 0 and × A = B. Is this result unique, or are there other functions with the same divergence and curl? 

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\nabla \cdot A =-\frac{1}{2} \nabla \cdot( r \times B )=-\frac{1}{2}[ B \cdot( \nabla \times r )- r \cdot( \nabla \times B )]=0, \text { since } \nabla \times B = 0 (B is uniform) and \nabla \times r = 0 (Prob. 1.63). \nabla \times A =-\frac{1}{2} \nabla \times( r \times B )=-\frac{1}{2}[( B \cdot \nabla ) r -( r \cdot \nabla ) B + r ( \nabla \cdot B )- B ( \nabla \cdot r )] . But ( r \cdot \nabla ) B = 0 \text { and } \nabla \cdot B =0 (since B is uniform), and \nabla \cdot r =\frac{\partial x}{\partial x}+\frac{\partial y}{\partial y}+\frac{\partial z}{\partial z}=1+1+1=3 . Finally, ( B \cdot \nabla ) r =\left(B_{x} \frac{\partial}{\partial x}+B_{y} \frac{\partial}{\partial y}+B_{z} \frac{\partial}{\partial z}\right)(x \hat{ x }+y \hat{ y }+z \hat{ z })=B_{x} \hat{ x }+B_{y} \hat{ y }+B_{z} \hat{ z }= B . \text { So } \nabla \times A =-\frac{1}{2}( B -3 B )= B . qed

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