STEM Problem Detail

Problem Statement

Prompt

Give step by step solution of Problem 7.59

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**Problem 7.59**

1. **((a))** The charge flowing into $$dz$$ in time $$dt$$ is
\[
dq = I(z)\,dt - I(z+dz)\,dt
= -\frac{dI}{dz}\,dt\,dz.
\]
But also
\[
dq = \lambda(t+dt)\,dz - \lambda(t)\,dz
= \frac{d\lambda}{dt}\,dt\,dz,
\]
so
\[
\frac{d\lambda}{dt} = -\frac{dI}{dz}.
\;\checkmark
\]

Since the left side is a function only of $$t$$, and the right side is a function
only of $$z$$, they must both be constant; call it $$k$$:
\[
\frac{d\lambda}{dt} = k
\;\Rightarrow\;
\lambda(t) = kt + C_1;
\qquad
\frac{dI}{dz} = -k
\;\Rightarrow\;
I(z) = -kz + C_2.
\]
If $$\lambda(0)=0$$ and $$I(0)=0$$, the constants $$C_1$$ and $$C_2$$ must both be zero:
\[
\lambda(t) = kt,
\qquad
I(z) = -kz.
\;\checkmark
\]
2. **((b))** In the quasistatic approximation,
\[
\mathbf{E}
= \frac{\lambda}{2\pi\varepsilon_0 s}\,\hat{\mathbf{s}}
= \frac{kt}{2\pi\varepsilon_0 s}\,\hat{\mathbf{s}},
\qquad
\mathbf{B}
= \frac{\mu_0 I}{2\pi s}\,\hat{\boldsymbol{\phi}}
= -\frac{\mu_0 k z}{2\pi s}\,\hat{\boldsymbol{\phi}}.
\]

\[
\nabla\cdot\mathbf{E}
= \frac{1}{s}\frac{\partial}{\partial s}
\left(s\frac{kt}{2\pi\varepsilon_0 s}\right)
= 0,
\;\checkmark
\]

\[
\nabla\cdot\mathbf{B}
= \frac{1}{s}\frac{\partial}{\partial\phi}
\left(-\frac{\mu_0 kz}{2\pi s}\right)
= 0.
\;\checkmark
\]

\[
\nabla\times\mathbf{E}
= \frac{\partial E_s}{\partial z}
- \frac{1}{s}\frac{\partial E_z}{\partial\phi}
= 0
= -\frac{\partial\mathbf{B}}{\partial t}.
\;\checkmark
\]

\[
\nabla\times\mathbf{B}
= -\frac{\partial B_\phi}{\partial z}\,\hat{\mathbf{s}}
+ \frac{1}{s}\frac{\partial}{\partial s}(sB_\phi)\,\hat{\mathbf{z}}
= \frac{\mu_0 k}{2\pi s}\,\hat{\mathbf{s}}
= \mu_0\varepsilon_0\frac{\partial\mathbf{E}}{\partial t}.
\;\checkmark
\]

For a Gaussian cylinder of radius $$s$$ about the $$z$$ axis, at height $$z$$ and
width $$dz$$:
\[
\oint \mathbf{E}\cdot d\mathbf{a}
= \frac{kt}{2\pi\varepsilon_0 s}(2\pi s)\,dz
= \frac{kt}{\varepsilon_0}dz
= \frac{\lambda\,dz}{\varepsilon_0}
= \frac{Q_{\text{enc}}}{\varepsilon_0}.
\;\checkmark
\]

\[
\oint \mathbf{B}\cdot d\mathbf{a} = 0.
\;\checkmark
\]

For a circular amperian loop of radius $$s$$ about the $$z$$ axis, at height $$z$$:
\[
\oint \mathbf{E}\cdot d\mathbf{l} = 0
= -\frac{d}{dt}\int \mathbf{B}\cdot d\mathbf{a}.
\;\checkmark
\]

\[
\oint \mathbf{B}\cdot d\mathbf{l}
= -\frac{\mu_0 kz}{2\pi s}(2\pi s)
= \mu_0 I_{\text{enc}}
+ \mu_0\varepsilon_0
\frac{d}{dt}\int \mathbf{E}\cdot d\mathbf{a}.
\;\checkmark
\]

(Note that $$\int \mathbf{B}\cdot d\mathbf{a}$$ and
$$\int \mathbf{E}\cdot d\mathbf{a}$$ are zero through this loop.)

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Problem #25