The magnetic field of an infinite straight wire carrying a steady current I can be obtained from the displacement current term in the Ampère/Maxwell law, as follows: Picture the current as consisting of a uniform line charge λ moving along the z axis at speed υ (so that I = λυ), with a

Question

The magnetic field of an infinite straight wire carrying a steady current I can be obtained from the displacement current term in the Ampère/Maxwell law, as follows: Picture the current as consisting of a uniform line charge λ moving along the z axis at speed υ (so that I = λυ), with a tiny gap of length ε , which reaches the origin at time t = 0. In the next instant (up to t =ε/υ) there is no real current passing through a circular Amperian loop in the xy plane, but there is a displacement current, due to the “missing” charge in the gap

(a) Use Coulomb’s law to calculate the z component of the electric field, for points in the xy plane a distance s from the origin, due to a segment of wire with uniform density -λ extending from [latex]z_{1}=v t-\epsilon ext { to } z_{2}=v t[/latex] .

(b) Determine the flux of this electric field through a circle of radius a in the xy plane.

(c) Find the displacement current through this circle. Show that [latex]I_{d}[/latex] is equal to I , in the limit as the gap width (ε) goes to zer[latex]o .^{35}[/latex]

Accepted Answer

[latex] ext { (a) } d E_{z}=\frac{1}{4 \pi \epsilon_{0}} \frac{(-\lambda) d z}{ᴫ^{2}} \sin heta[/latex]

[latex]\sin heta=\frac{-z}{ᴫ} ; ᴫ=\sqrt{z^{2}+s^{2}}[/latex]

[latex]E_{z}=\frac{\lambda}{4 \pi \epsilon_{0}} \int \frac{z d z}{\left(z^{2}+s^{2} ight)^{3 / 2}}=\left.\frac{\lambda}{4 \pi \epsilon_{0}}\left[\frac{-1}{\sqrt{z^{2}+s^{2}}} ight] ight|_{v t-\epsilon} ^{v t}[/latex]

[latex]E_{z}=\frac{\lambda}{4 \pi \epsilon_{0}}\left\{\frac{1}{\sqrt{(v t-\epsilon)^{2}+s^{2}}}-\frac{1}{\sqrt{(v t)^{2}+s^{2}}} ight\}[/latex] .

(b) [latex]\Phi_{E}=\frac{\lambda}{4 \pi \epsilon_{0}} \int_{0}^{a}\left\{\frac{1}{\sqrt{(v t-\epsilon)^{2}+s^{2}}}-\frac{1}{\sqrt{(v t)^{2}+s^{2}}} ight\} 2 \pi s d s=\left.\frac{\lambda}{2 \epsilon_{0}}\left[\sqrt{(v t-\epsilon)^{2}+s^{2}}-\sqrt{(v t)^{2}+s^{2}} ight] ight|_{0} ^{a}[/latex]

[latex]=\frac{\lambda}{2 \epsilon_{0}}\left[\sqrt{(v t-\epsilon)^{2}+a^{2}}-\sqrt{(v t)^{2}+a^{2}}-(\epsilon-v t)+(v t) ight][/latex] .

[latex] ext { (c) } I_{d}=\epsilon_{0} \frac{d \Phi_{E}}{d t}=\frac{\lambda}{2}\left\{\frac{v(v t-\epsilon)}{\sqrt{(v t-\epsilon)^{2}+a^{2}}}-\frac{v(v t)}{\sqrt{(v t)^{2}+a^{2}}}+2 v ight\}[/latex] .

[latex] ext { As } \epsilon ightarrow 0, v t<\epsilon ext { also } ightarrow 0, ext { so } I_{d} ightarrow \frac{\lambda}{2}(2 v)=\lambda v=I[/latex] . With an infinitesimal gap we attribute the magnetic field to displacement current, instead of real current, but we get the same answer. qed

Question 7.61: The magnetic field of an infinite straight wire carrying a s...

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