Question 7.51: An infinite wire carrying a constant current I in the zˆ dir...

In the quasistatic approximation the magnetic field of the wire is  B =\left(\mu_{0} I / 2 \pi s\right) \hat{ \phi } , or, in Cartesian coordinates,

B =\frac{\mu_{0} I}{2 \pi s}(-\sin \phi \hat{ x }+\cos \phi \hat{ y })=\frac{\mu_{0} I}{2 \pi s}\left(-\frac{y}{s} \hat{ x }+\frac{x}{s} \hat{ y }\right)=\frac{\mu_{0} I}{2 \pi} \frac{(-y \hat{ x }+x \hat{ y })}{x^{2}+y^{2}} ,

where x and y are measured from the wire. To convert to the stationary coordinates in the diagram, y \rightarrow y-v t :

B =\frac{\mu_{0} I}{2 \pi} \frac{[-(y-v t) \hat{ x }+x \hat{ y }]}{x^{2}+(y-v t)^{2}} .

Faraday’s law says

\nabla \times E =-\frac{\partial B }{\partial t}=-\frac{\mu_{0} I}{2 \pi}\left\{\frac{v \hat{ x }}{x^{2}+(y-v t)^{2}}-\frac{[-(y-v t) \hat{ x }+x \hat{ y }]}{\left[x^{2}+(y-v t)^{2}\right]^{2}}(-2 v)(y-v t)\right\} .

At t = 0, then,

\nabla \times E =-\frac{\mu_{0} I v}{2 \pi}\left\{\frac{\hat{ x }}{x^{2}+y^{2}}+2 \frac{\left[-y^{2} \hat{ x }+x y \hat{ y }\right]}{\left[x^{2}+y^{2}\right]^{2}}\right\}=-\frac{\mu_{0} I v}{2 \pi}\left\{\frac{\left.\left(x^{2}-y^{2}\right) \hat{ x }+2 x y \hat{ y }\right]}{\left[x^{2}+y^{2}\right]^{2}}\right\}=-\frac{\mu_{0} I v}{2 \pi s^{2}}(\cos \phi \hat{ s }+\sin \phi \hat{ \phi }) .

Our problem is to find a vector function of s and \phi (it obviously doesn’t depend on z) whose divergence is zero, whose curl is as given above, that goes to zero at large s:

E (s, \phi)=E_{s}(s, \phi) \hat{ s }+E_{\phi}(s, \phi) \hat{ \phi }+E_{z}(s, \phi) \hat{ z } ,

with

\nabla \cdot E =\frac{1}{s} \frac{\partial}{\partial s}\left(s E_{s}\right)+\frac{1}{s} \frac{\partial E_{\phi}}{\partial \phi}=0 ,

(\nabla \times E )_{\phi}=-\frac{\partial E_{z}}{\partial s}=-\frac{\mu_{0} I v}{2 \pi s^{2}} \sin \phi ,

(\nabla \times E )_{z}=\frac{1}{s}\left[\frac{\partial}{\partial s}\left(s E_{\phi}\right)-\frac{\partial E_{s}}{\partial \phi}\right]=0 .

The first and last of these are satisfied if  E_{s}=E_{\phi}=0 , the middle two are satisfied by  E_{z}=-\frac{\mu_{0} I v}{2 \pi s} \sin \phi .

vidently E =-\frac{\mu_{0} I v}{2 \pi s} \sin \phi \hat{ z } . The electric and magnetic fields ride along with the wire.