HW10 answer

1 textbook

3.2, 3.6, 3.10, 3.11, 5.3

推迟势为，

A = A_{z} = \frac{μ_{0}}{4 π} \int_{- \infty}^{+ \infty} \frac{I (t_{r})}{R} d z, R = \sqrt{s^{2} + z^{2}}

代入 $I (t_{r}) = k t_{r} \cdot Θ (t - \frac{\sqrt{s^{2} + z^{2}}}{c})$ 得到，

A_{z} (s, t) = \frac{μ_{0} k}{4 π} \cdot \int_{- z_{max}}^{z_{max}} [\frac{t}{\sqrt{s^{2} + z^{2}}} - \frac{1}{c}] d z, z_{max} = \sqrt{(c t)^{2} - s^{2}}

⟹ A_{z} (s, t) = \frac{μ_{0} k}{2 π} [t arccosh (\frac{c t}{s}) - \frac{\sqrt{(c t)^{2} - s^{2}}}{c}] Θ (c t - s)

因为电势 $φ = c o n s t$ ，所以在 $c t > s$ 时电磁场为

E = - \frac{\partial A_{z}}{\partial t} \hat{z} = - \frac{μ_{0} k}{2 π} arccosh (\frac{c t}{s}) \hat{z}

B = \nabla \times A = - \frac{\partial A_{z}}{\partial s} \hat{ϕ} = \frac{μ_{0} k t}{2 π s} \sqrt{1 - {(\frac{s}{c t})}^{2}} \hat{ϕ}

在 $s ≪ c t$ 时，保留零阶项 $\sqrt{(c t)^{2} - s^{2}} \approx c t$ ，由 $\ln (\frac{c t + \sqrt{(c t)^{2} - s^{2}}}{s}) = arccosh (\frac{c t}{s})$ 得到

A_{z} \approx \frac{μ_{0} k}{2 π} [t \ln (\frac{2 c t}{s}) - t] = \frac{μ_{0} k t}{2 π} [\ln (\frac{2 c t}{s}) - 1]

E (s, t) \approx - \frac{μ_{0} k}{2 π} \ln (\frac{2 c t}{s}) \hat{z}

B (s, t) \approx \frac{μ_{0} I (t)}{2 π s} \hat{ϕ}

t \ln (\frac{c t + \sqrt{(c t)^{2} - s^{2}}}{c t - \sqrt{(c t)^{2} - s^{2}}}) - 2 \frac{\sqrt{(c t)^{2} - s^{2}}}{c}