Homework Solutions

229

{\bar{℘}}_{p r i m} = \bar{ℰ_{i n} ℐ_{1}} = \frac{ℰ_{i n}^{2}}{2 (R_{1} + R_{2} {(\frac{N_{1}}{N_{2}})}^{2})}, {\bar{℘}}_{R_{1}} = \bar{R_{1} ℐ_{1}^{2}} = \frac{ℰ_{i n}^{2} R_{1}}{2 {(R_{1} + R_{2} {(\frac{N_{1}}{N_{2}})}^{2})}^{2}}

{\bar{℘}}_{R_{2}} = \bar{R_{2} ℐ_{2}^{2}} = \frac{ℰ_{i n}^{2} R_{2} {(\frac{N_{1}}{N_{2}})}^{2}}{2 {(R_{1} + R_{2} {(\frac{N_{1}}{N_{2}})}^{2})}^{2}}, r a t i o = {(\frac{N_{1}}{N_{2}})}^{2} \frac{R_{2}}{R_{1}}

230
From the previous problem

{\bar{℘}}_{p r i m} = {\bar{℘}}_{R_{1}} + {\bar{℘}}_{R_{2}}, {\bar{℘}}_{R_{1}}, {\bar{℘}}_{R_{2}} \geq 0

231
Just follow the example in the text.

232
At any time the charge on the capacitor plates, fed by $I (t)$ , asuming $Q (t = 0) = 0$ will be

Q (t) = C V (t) = \int_{0}^{i} I (t) d t

resulting in an electric ﬁeld between the plates (suppose that the plates are arranged with normals $\pm i$ ) $E (t) = \frac{Q (t)}{A ε_{0}} i$ . With no fringing, the total displacement current density between the plates is $J_{d} = ε_{0} \frac{d}{d t} E = \frac{\dot{Q} (t)}{A} i = \frac{I (t)}{A} i = \frac{C \frac{d V (t)}{d t}}{A} i$ with the total displacement current spread over the entire plate area $A$ . This means that the total displacement current between the plates equals the total conduction currents in the wires feeding the plates $I_{d} = J_{d} \cdot i A = C \frac{d V (t)}{d t} = I (t)$ .

233
Just use the previous problem and the AC-Kirchoff rules

ℰ_{0} = ℐ (ω) \frac{- i}{ω C}, ℐ (ω) = i ω C ℰ_{0} = ω C ℰ_{0} e^{i \frac{π}{2}}, I (t) = I m (ω C ℰ_{0} e^{i ω t + i \frac{π}{2}})

I (t) = ω C ℰ_{0} sin (ω t + \frac{π}{2})

Since the maximal displacement curren equals the maximal conduction current

I_{d, m a x} = I_{c o n d, m a x} = 7.6 \times 1 0^{- 6} A = \frac{π {(0.18 m)}^{2} ε_{0}}{d} (130 \frac{r a d}{s}) (220 V)

This gives us $d$ , and the maximal conduction current supplied by the generator.

234

\frac{1}{\sqrt{L C}} = \frac{1}{\sqrt{L \cdot 17 \times 1 0^{- 12} F}} = ω = 2 π f = 2 π \frac{c}{λ} = 2 π \frac{3 \times 1 0^{8} \frac{m}{s}}{5.5 \times 1 0^{- 7} m}

235
This wave travels in the $x$ -direction at speed $v = \frac{c}{1} = c$ , with polarization in the $z$ -direction, so we must have $E_{z} z \times B \approx i$ , so $B$ points in the $- j$ direction;

B = \frac{2.0 \frac{V}{m}}{3 \times 1 0^{8} \frac{m}{s}} (- j) cos (π \times 1 0^{15} \frac{1}{s} (t - \frac{x}{c}))

236

E = - \frac{\partial}{\partial t} A = \frac{2 π c}{λ} A_{0} p cos (\frac{2 π}{λ} (n \cdot r - c t))

B = \nabla \times A = \frac{2 π}{λ} A_{0} n \times p cos (\frac{2 π}{λ} (n \cdot r - c t))

Now check for

\nabla \times (\nabla \times A) = - ε_{0} μ_{0} \frac{d^{2}}{d t^{2}} A,

(\frac{2 π}{λ})^{2} A_{0} n \times (n \times p) sin (\frac{2 π}{λ} (n \cdot r - c t)) = - μ_{0} ε_{0} (\frac{2 π c}{λ})^{2} A_{0} p sin (\frac{2 π}{λ} (n \cdot r - c t))

which is true if

n \times (n \times p) = - μ_{0} ε_{0} c^{2} p

requiring $μ_{0} ε_{0} c^{2} = 1$ and $n \times (n \times p) = (n \cdot p) n - p (n \cdot n) = - p$ , and you can see that this requires $n \cdot p = 0$ .
For this reason, light waves in a transfparent medium (real index of refraction) are called transverse. Light traveling through a conductor (a material with a complex index of refraction) are still transverse, electric and magnetic ﬁelds are $⊥$ to each other and to the propagation direction, but they are not in phase.

237
The wave equation is

v^{2} \frac{\partial^{2}}{\partial x^{2}} f - \frac{\partial^{2}}{\partial t^{2}} f = 0

substitution of the proposes solution

\frac{\partial^{2}}{\partial x^{2}} f = k^{2} f^{''}, \frac{\partial^{2}}{\partial t^{2}} f = ω^{2} f^{''}

results in $v^{2} k^{2} = ω^{2}$ , so if $k$ and $ω$ are chosen so that this is so, we have a solution.

238
Simply follow through the previous two exercises with $ε \to κ ε_{0}$ and you will ﬁnd

B = B_{y} (z, t) j = \frac{\sqrt{κ} E_{0}}{c} j sin (\frac{2 π}{λ} (z - v t))