1.Magnetic field B as function of r in cylindrical capacitor.
Ans: For $r \leq a$, the enclosed displacement current $i_d' = i (r^2/a^2)$. From $\oint B dl = \mu_0 i_d' \implies B(2\pi r) = \mu_0 i (r^2/a^2) \implies B = \frac{\mu_0 i r}{2\pi a^2}$.
For $r > a$, the entire current $i$ is enclosed: $B = \frac{\mu_0 i}{2\pi r}$.
2.Show B is identical at edge using $i$ and $i_d$.
Ans: At $r=a$, from wire side $B = \frac{\mu_0 i}{2\pi a}$. From gap side, total displacement current $i_d = i$, so $B = \frac{\mu_0 i_d}{2\pi a} = \frac{\mu_0 i}{2\pi a}$. Both results coincide, maintaining field continuity.
3.Max displacement current density for $E = E_0 \sin(\omega t)$.
Ans: $J_d = \epsilon_0 \frac{dE}{dt} = \epsilon_0 \omega E_0 \cos(\omega t)$. The maximum value is $J_{d,max} = \epsilon_0 \omega E_0$.
4.B-field for $r < R$ if $dE/dt = K$.
Ans: Enclosed flux $\Phi_E = E \cdot \pi r^2$. $I_{d,enclosed} = \epsilon_0 \pi r^2 \frac{dE}{dt} = \epsilon_0 \pi r^2 K$.
$B(2\pi r) = \mu_0 \epsilon_0 \pi r^2 K \implies B = \frac{1}{2} \mu_0 \epsilon_0 r K$.
5.Prove $i_d = i_c$ in dielectric medium.
Ans: $E = \frac{\sigma}{\epsilon_0 \epsilon_r} = \frac{q}{A \epsilon}$. $\Phi_E = E A = q/\epsilon$.
$i_d = \epsilon \frac{d\Phi_E}{dt} = \epsilon \frac{d}{dt}(q/\epsilon) = \frac{dq}{dt} = i_c$. The factor of $\epsilon_r$ cancels out.
6.$i_d$ as function of $t$ for RC charging.
Ans: Conduction current $i_c(t) = \frac{V}{R} e^{-t/RC}$. Since $i_d = i_c$ in a series circuit, $i_d(t) = \frac{V}{R} e^{-t/RC}$.
7.Does accelerating electron produce displacement current?
Ans: Yes. As the electron moves, the electric field at any fixed point in space changes over time. An accelerating electron changes the field configuration even more dynamically, leading to non-zero $d\Phi_E/dt$, thus displacement current.
8.Ampere-Maxwell for $i_c = 0$.
Ans: $\oint \vec{B} \cdot d\vec{l} = \mu_0(0 + i_d) = \mu_0 \epsilon_0 \frac{d\Phi_E}{dt}$. (Correct as requested).
9.$i_d$ if $q(t) = q_0 e^{-t/\tau}$.
Ans: $i_d = dq/dt = -\frac{q_0}{\tau} e^{-t/\tau}$. The magnitude is $\frac{q_0}{\tau} e^{-t/\tau}$.
10.Ratio $J_c/J_d$ for $E = E_0 \sin\omega t$.
Ans: $J_c = \sigma E_0 \sin\omega t$. $J_d = \epsilon \omega E_0 \cos\omega t$.
Magnitude ratio $\frac{|J_c|}{|J_d|} = \frac{\sigma}{\epsilon \omega}$.
11.Derive wave equation for E.
Ans: Take curl of Faraday's Law: $\nabla \times (\nabla \times \vec{E}) = -\frac{\partial}{\partial t} (\nabla \times \vec{B})$.
Identity: $\nabla(\nabla \cdot E) - \nabla^2 E = -\mu_0 \epsilon_0 \frac{\partial^2 E}{\partial t^2}$.
Since $\nabla \cdot E = 0$ in vacuum, $\nabla^2 E = \mu_0 \epsilon_0 \frac{\partial^2 E}{\partial t^2}$.
12.Speed of EM waves comparison.
Ans: Comparing $\nabla^2 E = \mu_0 \epsilon_0 \frac{\partial^2 E}{\partial t^2}$ with $\nabla^2 E = \frac{1}{v^2} \frac{\partial^2 E}{\partial t^2}$ gives $1/v^2 = \mu_0 \epsilon_0 \implies v = 1/\sqrt{\mu_0 \epsilon_0} = c$.
13.Prove Transverse Nature.
Ans: For a wave along x, $E = E_0 \sin(kx - \omega t)$. From Gauss's Law $\nabla \cdot E = 0 \implies \partial E_x / \partial x = 0 \implies E_x$ must be constant or zero. For a radiation wave, it's zero. Hence E is perpendicular to x. Same for B.
Ans: From Faraday's law $\oint E dl = -d\Phi_B/dt \implies E_0 k = \omega B_0$.
$E_0/B_0 = \omega/k = c$.
15.Amplitudes in medium $n$.
Ans: Intensity $I \propto n E_0^2$. If $n$ increases, $E_0$ must decrease as $1/\sqrt{n}$.
Since $B_0 = n E_0 / c$, $B_0$ actually increases as $\sqrt{n}$.
16.Energy density in terms of $B_0$.
Ans: $u = u_E + u_B$. Since $u_E = u_B$, $u = 2 u_B = 2 \times \frac{B_0^2}{4 \mu_0} = \frac{B_0^2}{2 \mu_0}$.
17.Energy in cylinder $10\text{ cm}^2, 50\text{ cm}$.
Ans: $u_{avg} = \frac{1}{2} \epsilon_0 E_0^2 = \frac{1}{2} (8.85 \times 10^{-12}) (60)^2 \approx 1.59 \times 10^{-8} \text{ J/m}^3$.
Volume $V = 10 \times 50 \times 10^{-6} \text{ m}^3 = 5 \times 10^{-4} \text{ m}^3$.
Energy $U = u_{avg} V \approx 7.9 \times 10^{-12}\text{ Joules}$.
18.Poynting Vector time-average.
Ans: $\vec{S}_{avg} = \frac{1}{2 \mu_0} E_0 B_0 = \frac{E_0^2}{2 \mu_0 c} = I$ (Intensity).
19.Radiation pressure reflecting surface $= 2I/c$.
Ans: Momentum of incident wave $p = U/c$. Momentum of reflected wave $p' = -U/c$.
$\Delta p = 2U/c$. Pressure $P = \frac{\Delta p}{A \Delta t} = \frac{2 U / A \Delta t}{c} = \frac{2 I}{c}$.
20.Peak $E_0$ at distance $r$ from station power $P$.
Ans: $I = P / (4\pi r^2)$. Also $I = \frac{1}{2} c \epsilon_0 E_0^2$.
$E_0 = \sqrt{\frac{P}{2\pi \epsilon_0 c r^2}} = \frac{1}{r} \sqrt{\frac{P}{2\pi \epsilon_0 c}}$.
21.Find B for $\vec{E} = E_0 \hat{j} \cos(kz - \omega t)$.
Ans: Wave propagates in +z. $\vec{B}$ must be perpendicular to z and y. $\vec{B} = -\frac{E_0}{c} \hat{i} \cos(kz - \omega t)$.
22.Force on $2.0\text{ m}^2$ panel ($1.35\text{ kW/m}^2$).
Ans: Force $F = \text{Power}/c = (I \times A)/c = (1350 \times 2) / (3 \times 10^8) = 9 \times 10^{-6}\text{ N}$.
23.Pressure shift due to $30^\circ$ tilt.
Ans: Pressure $P = \frac{I \cos^2\theta}{c}$. With $\theta = 30^\circ$, $P_{new} = P_{old} \cos^2 30 = \frac{3}{4} P_{old}$.
24.$B_0$ at $2\text{m}$ for $100\text{W}$ ($5\%$ efficient).
Ans: $P_{vis} = 5\text{W}$. $I = 5 / (4\pi \times 2^2) = 0.099\text{ W/m}^2$.
$I = \frac{c B_0^2}{2 \mu_0} \implies B_0 = \sqrt{\frac{2 \mu_0 I}{c}} \approx 2.9 \times 10^{-8}\text{ T}$.
25.Show average $u = \frac{1}{2} \epsilon_0 E_0^2$.
Ans: $u = \epsilon_0 E^2 = \epsilon_0 E_0^2 \sin^2\omega t$. Average of $\sin^2$ is $1/2$. Thus $\langle u \rangle = \frac{1}{2} \epsilon_0 E_0^2$.
26.Frequency doubled effect.
Ans: Speed: Unchanged ($c$). Momentum: Doubled (per photon, $p=hf/c$). Radiation Pressure: Unchanged (depends on intensity, not freq).
27.Ratio $F_e/F_m$ for $v=0.1c$.
Ans: $F_e = qE$. $F_m = q v B = q v (E/c)$.
Ratio $F_e/F_m = E / (v E / c) = c/v = 1/0.1 = 10$.
28.$E_0$ for laser $10\text{mW}$ in $10\text{ }\mu\text{m}$ spot.
Ans: $I = 10 \times 10^{-3} / [\pi (5 \times 10^{-6})^2] \approx 1.27 \times 10^8\text{ W/m}^2$.
$E_0 = \sqrt{2 I / \epsilon_0 c} \approx 3.1 \times 10^5\text{ V/m}$.
29.$\lambda = 500\text{nm}$, find $k, \omega$.
Ans: $k = 2\pi / \lambda = 2(3.14) / (5 \times 10^{-7}) = 1.256 \times 10^7\text{ m}^{-1}$.
$\omega = c k = (3 \times 10^8) \times (1.256 \times 10^7) \approx 3.77 \times 10^{15}\text{ rad/s}$.
30.Static B vs oscillating charge.
Ans: EM waves require time-varying fields. A static B field has $d\Phi_B/dt = 0$, so no E-field is induced. An oscillating charge has $dv/dt \neq 0$ and changing position, creating time-varying E and B fields that support each other.
31.$2.48\text{ eV}$ spectrum part.
Ans: $\lambda = hc/E \approx 1240 / 2.48 = 500\text{ nm}$. This is **Visible Light** (Green).
32.Frequency more fundamental than wavelength. Why?
Ans: Frequency is determined by the source oscillation and remains constant regardless of the medium. Wavelength changes because wave speed changes in different materials ($v = f \lambda$).
33.Production of (a) X-rays, (b) Gamma rays.
Ans: (a) Inner shell transitions $\rightarrow$ X-rays ($\sim$ keV). (b) Nuclear transitions $\rightarrow$ Gamma rays ($\sim$ MeV). Gamma rays are typically $10^3$ times more energetic.
34.FM range $2.8\text{m}$ to $3.4\text{m}$.
Ans: $f_{max} = 3 \times 10^8 / 2.8 \approx 107\text{ MHz}$. $f_{min} = 3 \times 10^8 / 3.4 \approx 88\text{ MHz}$. Range: $88 - 108\text{ MHz}$.
35.Microwave diffraction proof.
Ans: Diffraction spreading $\theta \approx \lambda/D$. For microwaves ($\lambda \sim$ cm), spreading is small compared to radio ($\lambda \sim$ m). This allows for tight, narrow beams.
36.Greenhouse gases transparency logic.
Ans: $CO_2$ and $H_2O$ molecules have natural vibrational frequencies in the infrared range. Resonant absorption occurs for IR, but visible light has frequencies much higher than molecular resonances, so it passes through.
37.$\lambda = 10^{-13}\text{ m}$. Find energy in MeV.
Ans: $E = hc/\lambda = (6.63 \times 10^{-34} \times 3 \times 10^8)/10^{-13} \approx 2 \times 10^{-12}\text{ J} \approx 12.4\text{ MeV}$.
38.Penetrating Power order.
Ans: Gamma $>$ X-ray $>$ UV $>$ Microwave. Penetration increases with energy/frequency as photons are less likely to interact with outer electrons.
39.Remote carrier frequency $38\text{kHz}$ vs EM frequency.
Ans: No. The carrier frequency $38\text{kHz}$ is the rate at which the IR source is blinked. The EM wave frequency is determined by the IR wavelength ($\sim 10^{13}\text{ Hz}$).
40.Physics of fluorescence in UV.
Ans: Molecules absorb high-energy UV photons, lose some energy internally, and then re-emit lower-energy photons which fall in the visible range (Stokes Shift).
41.$i_d = C (dV/dt) = 100 \times 10^{-12} \times 10^9 = 0.1\text{ A}$. 0.1
42.$u = B_0^2 / (2 \mu_0) \implies B_0 = \sqrt{2 \mu_0 u} \approx 2.3 \times 10^{-9}\text{ T}$. 2.3e-9
43.$i_d = \epsilon_0 A (dE/dt) \implies dE/dt = 0.5 / (8.85 \times 10^{-12} \times 0.01) \approx 5.65 \times 10^{12}\text{ V/m}\cdot\text{s}$. 5.65e12
44.$\lambda = v/f = (2 \times 10^8) / (5 \times 10^{14}) = 4 \times 10^{-7}\text{ m} = 400\text{ nm}$. 400
45.$I = \epsilon_0 c E_{rms}^2 = (8.85 \times 10^{-12}) \times (3 \times 10^8) \times 100 \approx 0.265\text{ W/m}^2$. 0.265
46.$I = 180 / (0.5 \times 60) = 6\text{ W/m}^2$. $P = I/c = 6 / (3 \times 10^8) = 2 \times 10^{-8}\text{ Pa}$. 2e-8
47.$B_0 = E_0/c = 450 / (3 \times 10^8) = 1.5 \times 10^{-6}\text{ T} = 1.5\text{ }\mu\text{T}$. 1.5
48.$v = c / \sqrt{\mu_r \epsilon_r} = c / \sqrt{4} = c/2 = 1.5 \times 10^8\text{ m/s}$. 1.5e8
49.$k = 2\pi f / c \implies f = k c / 2\pi = (31.4 \times 3 \times 10^8) / 6.28 = 1.5 \times 10^9\text{ Hz} = 1.5\text{ GHz}$. 1.5
50.$\lambda = c/f = (3 \times 10^8) / 10^9 = 0.3\text{ m} = 30\text{ cm}$. 30
51.$I = 20 / 10^{-6} = 2 \times 10^7\text{ W/m}^2$. $u = I/c \approx 0.067\text{ J/m}^3$. 0.067
52.$E_0 = c B_0 = (3 \times 10^8) \times (20 \times 10^{-9}) = 6\text{ V/m}$. 6
54.Ratio $c/v = n = 2.0$. 2
55.$dE/dt = i_d / (\epsilon_0 A) = 2 / (8.85 \times 10^{-12} \times 0.1) \approx 2.26 \times 10^{12}\text{ V/m}\cdot\text{s}$. 2.26e12
56.$\lambda = (3 \times 10^8) / (300 \times 10^6) = 1\text{ meter}$. 1
57.$E = hc / \lambda = 1240 / 0.1 = 12400\text{ eV} \approx 2 \times 10^{-15}\text{ J}$. 2e-15
58.$I = P / (4\pi r^2) = 1000 / (4 \times 3.14 \times 100) \approx 0.8\text{ W/m}^2$. $E_0 = \sqrt{2 I / \epsilon_0 c} \approx 24.5\text{ V/m}$. 24.5
59.$v = c / \sqrt{81} = c/9 \approx 3.33 \times 10^7\text{ m/s}$. 3.33e7
60.$F = 2 P / c = (2 \times 600) / (3 \times 10^8) = 4 \times 10^{-6}\text{ N}$. 4e-6
