1.Explain how Maxwell modified Ampere's law for logical consistency.
Ans: Maxwell noticed Ampere's law ($\oint \vec{B} \cdot d\vec{l} = \mu_0 I$) failed across the insulating gap of a charging capacitor because no conduction current flows there. He reasoned that the changing electric field in the gap must produce a magnetic field just like a real current. He introduced a "displacement current" term $I_d = \epsilon_0 \frac{d\Phi_E}{dt}$, modifying the law to $\oint \vec{B} \cdot d\vec{l} = \mu_0 (I_c + I_d)$, ensuring continuous total current.
2.Prove mathematically that displacement current maintains continuity.
Ans: Let $q$ be the charge on the capacitor. The conduction current in the connecting wires is $I_c = \frac{dq}{dt}$.
The electric field between the plates is $E = \frac{q}{A\epsilon_0}$. Electric flux $\Phi_E = E \cdot A = \frac{q}{\epsilon_0}$.
Displacement current $I_d = \epsilon_0 \frac{d\Phi_E}{dt} = \epsilon_0 \frac{d}{dt}\left(\frac{q}{\epsilon_0}\right) = \frac{dq}{dt}$.
Since $I_c = dq/dt$ and $I_d = dq/dt$, $I_c = I_d$. The current is perfectly continuous.
3.Derive expression for induced magnetic field at $r < R$ (inside gap).
Ans: Consider a circular Amperian loop of radius $r < R$ between the plates.
From Ampere-Maxwell law: $\oint \vec{B} \cdot d\vec{l} = \mu_0 I_d$.
The displacement current passing through area $\pi r^2$ is a fraction of the total $I_d$: $I_{d,enclosed} = I_d \frac{\pi r^2}{\pi R^2} = I \frac{r^2}{R^2}$ (since total $I_d = I_c = I$).
$\oint B dl = B(2\pi r) = \mu_0 I \frac{r^2}{R^2}$.
Therefore, $B = \frac{\mu_0 I r}{2\pi R^2}$.
4.Derive magnetic field at $r > R$ (outside gap).
Ans: Consider a circular Amperian loop of radius $r > R$. The loop encloses the entire electric flux between the plates.
Enclosed displacement current is the total $I_d$, which equals $I_c = I$.
By Ampere-Maxwell law: $\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{d,enclosed}$.
$B(2\pi r) = \mu_0 I$.
Therefore, $B = \frac{\mu_0 I}{2\pi r}$. (This perfectly matches the field of a long straight wire).
5.Capacitor $C$ charged by $V(t) = V_0 \sin(\omega t)$. Find $I_d$.
Ans: The conduction current $I_c = \frac{dq}{dt} = \frac{d(CV)}{dt} = C \frac{dV}{dt}$.
$I_c = C \frac{d}{dt}(V_0 \sin\omega t) = C V_0 \omega \cos(\omega t)$.
Since displacement current is always equal to conduction current in a capacitor circuit ($I_d = I_c$),
$I_d = C V_0 \omega \cos(\omega t)$.
6.$C = 1.0\text{ }\mu\text{F}$, $dV/dt = 10^4\text{ V/s}$. Find $I_d$.
Ans: $I_d = I_c = C \frac{dV}{dt} = (1.0 \times 10^{-6}\text{ F}) \times (10^4\text{ V/s}) = 10^{-2}\text{ A} = 0.01\text{ A}$.
7.Electric field changing at $dE/dt$, Area $A$. Expression for $I_d$.
Ans: Electric flux $\Phi_E = E \cdot A$.
$I_d = \epsilon_0 \frac{d\Phi_E}{dt} = \epsilon_0 \frac{d(EA)}{dt} = \epsilon_0 A \frac{dE}{dt}$.
8.Distinguish conduction and displacement current by origin.
Ans: Conduction current originates from the actual physical flow of charged particles (e.g., electrons drifting in a wire) under an applied potential difference. Displacement current does not involve the motion of charges; it originates purely from a time-varying electric field (or changing electric flux) in a region of space.
9.Show $I_c$ and $I_d$ are discontinuous, but sum is continuous.
Ans: In the connecting wires, $I_c = I$ and $I_d = 0$ (electric field is roughly zero inside a good conductor). In the vacuum gap between capacitor plates, $I_c = 0$ (no charges cross the vacuum) but $I_d = I$ (due to changing E-field). Thus, individually, $I_c$ abruptly stops at the plate, and $I_d$ abruptly starts at the plate. However, their sum $(I_c + I_d)$ equals $I$ everywhere, making the total effective current continuous throughout the entire circuit.
10.DC $200\text{V}$ applied to $10\text{ }\mu\text{F}$. At instant when $I_c = 0.2\text{ A}$, what is $I_d$?
Ans: By Maxwell's continuity principle, the displacement current across the plates at any exact instant is identically equal to the conduction current flowing in the wires leading to the plates. Therefore, $I_d = I_c = 0.2\text{ A}$.
11.Charging current is $0.25\text{ A}$. What is displacement current?
Ans: Based on the continuity equation ($I_d = I_c$), the displacement current is exactly $0.25\text{ A}$.
12.Example where both $I_c$ and $I_d$ exist simultaneously.
Ans: A leaky capacitor (a capacitor filled with a dielectric medium that is not a perfect insulator) being charged. The changing electric field creates displacement current ($I_d$), while the imperfect dielectric allows some physical charge carriers to flow across the gap, creating a small simultaneous conduction current ($I_c$).
13.Area $2.0\text{ m}^2$, $dE/dt = 1.0 \times 10^5\text{ V/(m}\cdot\text{s)}$. Calculate $I_d$.
Ans: $I_d = \epsilon_0 A \frac{dE}{dt} = (8.85 \times 10^{-12}) \times (2.0) \times (1.0 \times 10^5) = 17.7 \times 10^{-7}\text{ A} = 1.77\text{ }\mu\text{A}$.
14.Why is magnetic field inside gap non-zero despite no physical charges moving?
Ans: Because as the capacitor charges, the electric field ($E$) between the plates is changing over time. According to Maxwell, this time-varying electric field acts as a "displacement current", which generates a surrounding magnetic field identically to how a physical current of moving charges would.
15.Write generalized Ampere-Maxwell law and define terms.
Ans: $\oint \vec{B} \cdot d\vec{l} = \mu_0 I_c + \mu_0 \epsilon_0 \frac{d\Phi_E}{dt}$.
$\oint \vec{B} \cdot d\vec{l}$ is the line integral of magnetic field around a closed loop.
$\mu_0$ is the permeability of free space.
$I_c$ is the physical conduction current crossing the surface bounded by the loop.
$\epsilon_0$ is the permittivity of free space.
$\Phi_E$ is the electric flux through the surface.
16.State the four fundamental Maxwell's equations in integral form.
Ans: 1. Gauss's Law (E): $\oint \vec{E} \cdot d\vec{A} = \frac{Q_{enclosed}}{\epsilon_0}$
2. Gauss's Law (B): $\oint \vec{B} \cdot d\vec{A} = 0$
3. Faraday's Law: $\oint \vec{E} \cdot d\vec{l} = -\frac{d\Phi_B}{dt}$
4. Ampere-Maxwell Law: $\oint \vec{B} \cdot d\vec{l} = \mu_0 I_c + \mu_0 \epsilon_0 \frac{d\Phi_E}{dt}$
17.How does $\oint \vec{B} \cdot d\vec{A} = 0$ contradict magnetic monopoles?
Ans: The equation states that the net magnetic flux through *any* closed surface is identically zero. This means that every magnetic field line that enters the surface must also exit it. Therefore, you cannot enclose an isolated "source" or "sink" of magnetic field lines (a magnetic monopole). Magnetic poles always exist in North-South pairs.
18.Which equation proves a time-varying magnetic field produces an electric field? Name it.
Ans: $\oint \vec{E} \cdot d\vec{l} = -\frac{d\Phi_B}{dt}$. This is Faraday's Law of Electromagnetic Induction. The $-d\Phi_B/dt$ term represents a time-varying magnetic field, which induces a circulating electric field ($\oint \vec{E} \cdot d\vec{l}$).
19.How do Maxwell's equations predict EM waves?
Ans: Faraday's law shows that a changing magnetic field creates a spatially varying electric field. The Ampere-Maxwell law shows that this changing electric field, in turn, creates a spatially varying magnetic field. This creates a continuous, self-sustaining feedback loop where oscillating E and B fields regenerate each other, propagating through space as an electromagnetic wave independent of the original source charges.
20.Write Maxwell's equations for a region of pure free space.
Ans: In free space, charge $Q=0$ and conduction current $I_c=0$.
1. $\oint \vec{E} \cdot d\vec{A} = 0$
2. $\oint \vec{B} \cdot d\vec{A} = 0$
3. $\oint \vec{E} \cdot d\vec{l} = -\frac{d\Phi_B}{dt}$
4. $\oint \vec{B} \cdot d\vec{l} = \mu_0 \epsilon_0 \frac{d\Phi_E}{dt}$
21.Conservation law inherent in Gauss's Law for electricity?
Ans: It is fundamentally linked to the Law of Conservation of Charge. (Changes in the enclosed charge precisely match the current flowing through the boundary surface).
22.Equation that encapsulates electromagnetic induction.
Ans: Faraday's Law: $\oint \vec{E} \cdot d\vec{l} = -\frac{d\Phi_B}{dt}$.
23.How does Ampere-Maxwell law link dynamic electricity and magnetism?
Ans: By introducing the displacement current term ($\mu_0 \epsilon_0 \frac{d\Phi_E}{dt}$), it shows that a dynamically changing electric field ($\Phi_E$) directly generates a circulating magnetic field ($\oint \vec{B} \cdot d\vec{l}$), perfectly complementing Faraday's law where a changing magnetic field creates an electric field.
24.State Lorentz force. Why grouped with Maxwell's equations?
Ans: $\vec{F} = q(\vec{E} + \vec{v} \times \vec{B})$. Maxwell's equations only describe how charges and currents *create* electric and magnetic fields. The Lorentz force equation is the required complementary half of electrodynamics that dictates how those created fields *exert force* back on moving charges.
25.If net electric flux is zero, is electric field zero everywhere on surface?
Ans: No. Gauss's Law ($\oint \vec{E} \cdot d\vec{A} = 0$) only implies that the *net* enclosed charge is zero. There could be an electric dipole inside, or an external electric field passing through the surface. In both cases, electric field $\vec{E}$ is non-zero on the surface, but the total integral (flux entering equals flux exiting) evaluates to zero.
26.If net magnetic flux is invariably zero, what is inferred about B field lines?
Ans: It implies that magnetic field lines are always continuous, closed loops with no beginning or end. They do not originate from or terminate on point charges like electric field lines do.
27.Fundamental source of EM waves. How it creates E and B fields.
Ans: The fundamental source is an accelerating (or oscillating) electric charge. An oscillating charge produces a periodically oscillating electric field. By the Ampere-Maxwell law, this changing E-field creates an oscillating magnetic field. By Faraday's law, the changing B-field recreates the oscillating E-field. This coupled disturbance propagates outward as an EM wave.
28.Propagating along +z. E is along +x. Direction of B?
Ans: The direction of propagation $\vec{c}$ is given by the cross product $\vec{E} \times \vec{B}$.
We need $\hat{i} \times \text{(Direction of B)} = \hat{k}$.
Since $\hat{i} \times \hat{j} = \hat{k}$, the magnetic field must be oriented along the **positive y-axis**.
29.Derive $c = E_0 / B_0$ using dimensional analysis.
Ans: Lorentz force on a moving charge: $\vec{F} = q(\vec{E} + \vec{v} \times \vec{B})$. For the electric and magnetic forces to have the same dimensional units, the dimension of $E$ must equal the dimension of $(v \times B)$.
$[E] = [v][B] \implies [E]/[B] = [v]$.
For an EM wave in vacuum, the velocity $v$ is the speed of light $c$. Hence, $c = E_0/B_0$.
30.$B_0 = 510\text{ nT}$. Calculate $E_0$.
Ans: $E_0 = c \times B_0 = (3 \times 10^8\text{ m/s}) \times (510 \times 10^{-9}\text{ T}) = 1530 \times 10^{-1} = 153\text{ V/m}$.
31.$E_0 = 120\text{ N/C}, \nu = 50.0\text{ MHz}$. Determine $B_0, \omega, k, \lambda$.
Ans: $B_0 = E_0/c = 120 / (3 \times 10^8) = 40 \times 10^{-8} = 4.0 \times 10^{-7}\text{ T}$.
$\omega = 2\pi\nu = 2 \times 3.14 \times (50 \times 10^6) = 3.14 \times 10^8\text{ rad/s}$.
$\lambda = c/\nu = (3 \times 10^8) / (50 \times 10^6) = 300 / 50 = 6.0\text{ m}$.
$k = 2\pi/\lambda = 2 \times 3.14 / 6.0 = 6.28 / 6.0 \approx 1.05\text{ rad/m}$.
32.Standard expressions for E and B propagating along +x.
Ans: If propagating along +x, the fields oscillate in y and z directions.
$\vec{E} = E_0 \sin(kx - \omega t) \hat{j}$
$\vec{B} = B_0 \sin(kx - \omega t) \hat{k}$
33.Show average energy density of E equals average energy density of B.
Ans: $u_E = \frac{1}{2}\epsilon_0 E_{rms}^2$. Since $E_{rms} = c B_{rms}$, substitute this: $u_E = \frac{1}{2}\epsilon_0 (c B_{rms})^2 = \frac{1}{2}\epsilon_0 c^2 B_{rms}^2$.
We know $c^2 = \frac{1}{\mu_0\epsilon_0}$. Substitute $c^2$:
$u_E = \frac{1}{2}\epsilon_0 \left(\frac{1}{\mu_0\epsilon_0}\right) B_{rms}^2 = \frac{B_{rms}^2}{2\mu_0}$.
Since $\frac{B_{rms}^2}{2\mu_0}$ is the exact formula for magnetic energy density $u_B$, we proved $u_E = u_B$.
34.Define intensity. Express in terms of $E_0$.
Ans: Intensity ($I$) is the energy crossing a unit area per unit time perpendicular to the direction of propagation.
$I = \text{Total energy density} \times \text{speed} = u \times c$.
Since $u = \frac{1}{2}\epsilon_0 E_0^2$ (using peak value), $I = \frac{1}{2} c \epsilon_0 E_0^2$.
35.Calculate average intensity for $E_0 = 60\text{ V/m}$.
Ans: $I = \frac{1}{2} c \epsilon_0 E_0^2 = \frac{1}{2} (3 \times 10^8) (8.85 \times 10^{-12}) (60)^2$.
$I = 0.5 \times 3 \times 8.85 \times 10^{-4} \times 3600 = 1.5 \times 8.85 \times 3.6 \approx 4.78\text{ W/m}^2$.
36.What is radiation pressure? Derived momentum for reflecting surface.
Ans: Radiation pressure is the force exerted by EM waves on a unit area of a surface due to the transfer of momentum.
Energy transferred is $U = I A t$. Momentum of wave is $p = U/c$.
For a perfectly reflecting surface, the wave bounces back, so the change in momentum is double: $\Delta p = p_{final} - p_{initial} = (-U/c) - (U/c) = -2U/c$. Momentum transferred to surface is $2U/c$.
37.Beam $20\text{ W/cm}^2$ on $10\text{ cm}^2$ absorbing surface. Force?
Ans: Intensity $I = 20\text{ W/cm}^2 = 20 \times 10^4\text{ W/m}^2$. Area $A = 10\text{ cm}^2 = 10 \times 10^{-4}\text{ m}^2$.
Total Power $P = I \times A = (20 \times 10^4) \times (10 \times 10^{-4}) = 200\text{ W}$.
Force exerted on perfectly absorbing surface $F = \frac{\text{Power}}{c} = \frac{200}{3 \times 10^8} = 6.67 \times 10^{-7}\text{ N}$.
38.Speed in liquid with $\epsilon_r = 2.25, \mu_r = 1$.
Ans: $v = \frac{c}{\sqrt{\mu_r \epsilon_r}} = \frac{3 \times 10^8}{\sqrt{1 \times 2.25}} = \frac{3 \times 10^8}{1.5} = 2.0 \times 10^8\text{ m/s}$.
39.Phase difference between E and B? What does it imply?
Ans: The phase difference is zero. This implies that the electric and magnetic fields reach their maximum values (peaks) and zero values simultaneously at the same point in space.
40.Conceptually why EM waves exert pressure. Which property causes this?
Ans: EM waves exert pressure because they carry **linear momentum** ($p = U/c$). When an EM wave strikes a surface and is absorbed or reflected, its momentum changes. According to Newton's second law, a rate of change of momentum imparts a continuous force on the surface, which manifests as radiation pressure.
41.Radio wave $f = 30.0\text{ MHz}$. Calculate $\lambda$.
Ans: $\lambda = \frac{c}{f} = \frac{3 \times 10^8}{30 \times 10^6} = \frac{300}{30} = 10\text{ meters}$.
42.$E_y = 0.5 \cos[2\pi \times 10^8 (t - x/c)]\text{ V/m}$. Write equation for B.
Ans: The wave is propagating along +x (because of the $t - x/c$ term). $E$ is along y-axis ($\hat{j}$). Since $\hat{j} \times \hat{k} = \hat{i}$, the B field must be along the z-axis ($\hat{k}$).
Amplitude $B_0 = E_0/c = 0.5 / (3 \times 10^8) = 1.67 \times 10^{-9}\text{ T}$.
Equation: $B_z = 1.67 \times 10^{-9} \cos[2\pi \times 10^8 (t - x/c)]\text{ T}$.
43.Prove theoretically that EM waves carry linear momentum.
Ans: Consider a charge $q$ moving with velocity $v$ in an EM wave. The electric field $\vec{E}$ accelerates it, doing work and transferring energy. Simultaneously, the magnetic field $\vec{B}$ exerts a Lorentz force $\vec{F}_m = q(\vec{v} \times \vec{B})$. Since $\vec{v}$ is produced by $\vec{E}$, $\vec{F}_m$ points in the direction of wave propagation ($\vec{E} \times \vec{B}$). This continuous pushing force in the direction of travel proves the wave is transferring linear momentum.
44.EM wave in medium $n = 1.50$. Find speed.
Ans: $v = \frac{c}{n} = \frac{3 \times 10^8}{1.50} = 2.0 \times 10^8\text{ m/s}$.
45.Two characteristics distinguishing EM waves from mechanical waves.
Ans: 1. EM waves do not require any physical material medium to propagate (they can travel through a perfect vacuum). Mechanical waves absolutely require a medium. 2. EM waves travel at the ultimate speed limit of the universe ($3 \times 10^8\text{ m/s}$ in vacuum), which is vastly faster than any mechanical wave.
46.EM spectrum definition. Longest to shortest order.
Ans: The continuous distribution of all electromagnetic waves sorted by their wavelength or frequency.
Order (Longest $\lambda$ to Shortest $\lambda$): Radio waves $\to$ Microwaves $\to$ Infrared $\to$ Visible light $\to$ Ultraviolet $\to$ X-rays $\to$ Gamma rays.
47.Part of spectrum produced by bombarding metal with high-speed electrons.
Ans: X-rays. (Produced in a Coolidge tube when fast electrons are suddenly decelerated by a heavy metal target, converting their kinetic energy into high-energy X-ray photons).
48.Radiations used for (a) water purification, (b) eye surgery.
Ans: Both applications utilize **Ultraviolet (UV) rays**. (UV-C kills bacteria in water; precise UV excimer lasers reshape the cornea in LASIK).
49.EM wave in radar for aircraft. Reason?
Ans: Microwaves. Because of their short wavelengths (high frequencies), they undergo very little diffraction (bending) around obstacles. They travel in straight, tight, highly directional beams that accurately bounce back from small targets (aircraft), making them perfect for radar.
50.Approximate frequency limits of visible light.
Ans: Uses $f = c/\lambda$. For $\lambda = 700\text{ nm}$ (Red), $f \approx 4 \times 10^{14}\text{ Hz}$. For $\lambda = 400\text{ nm}$ (Violet), $f \approx 7.5 \times 10^{14}\text{ Hz}$.
Limits: roughly $4 \times 10^{14}\text{ Hz}$ to $7.5 \times 10^{14}\text{ Hz}$.
51.How are infrared waves produced? Why called heat waves?
Ans: They are produced by the thermal motion (vibrations and rotations) of atoms and molecules in hot bodies. They are called heat waves because they are readily absorbed by water molecules present in most materials, drastically increasing their thermal kinetic energy and heating up the object.
52.Two applications of gamma rays.
Ans: 1. Medical radiotherapy to precisely target and destroy cancerous tumor cells. 2. Sterilization of medical equipment and food products, as their extreme energy completely destroys microbial DNA.
53.Part of EM spectrum absorbed by ozone layer?
Ans: Ultraviolet (UV) radiation (specifically the highly energetic and biologically damaging UV-B and UV-C bands).
54.Wave used for studying crystal structure. Frequency range?
Ans: X-rays. Because their wavelength is on the order of Angstroms ($10^{-10}\text{ m}$), matching the atomic spacing in crystals. Typical frequency range is roughly $10^{16}\text{ Hz}$ to $10^{19}\text{ Hz}$.
55.How are microwaves produced? Main application.
Ans: Produced artificially using specialized vacuum tubes called Klystrons, Magnetrons, and Gunn diodes where electron beams are forced to oscillate at microwave frequencies. Main application: Radar systems and microwave ovens.
56.Wavelength range of X-rays and primary production mechanism.
Ans: Wavelength range: roughly $10^{-8}\text{ m}$ to $10^{-13}\text{ m}$ ($10\text{ nm}$ to $10^{-3}\text{ nm}$). Production: Rapid deceleration of fast-moving electrons when they strike a heavy metal target, or electronic transitions in the inner shell of heavy atoms.
57.EM waves used in cellular mobile phones.
Ans: Microwaves (specifically in the Ultra High Frequency (UHF) radio/microwave boundary band).
58.Ascending order of frequencies: Gamma, Micro, UV, Radio.
Ans: Ascending frequency implies longest to shortest wavelength. Order: Radio waves $<$ Microwaves $<$ Ultraviolet rays $<$ Gamma rays.
59.Why is ozone layer crucial for human survival?
Ans: The ozone layer completely absorbs the highly energetic Ultraviolet (UV-B and UV-C) rays from the solar electromagnetic spectrum. Without it, these rays would reach Earth's surface, causing severe DNA mutations, skin cancer, cataracts, and the destruction of phytoplankton, collapsing the global food chain.
60.Photon energy $10\text{ keV}$. Frequency and spectrum part?
Ans: Energy $E = 10\text{ keV} = 10 \times 10^3 \times 1.6 \times 10^{-19}\text{ J} = 1.6 \times 10^{-15}\text{ J}$.
$E = hf \implies f = E/h = (1.6 \times 10^{-15}) / (6.63 \times 10^{-34}) \approx 2.4 \times 10^{18}\text{ Hz}$.
A frequency of $10^{18}\text{ Hz}$ falls squarely in the **X-ray** region of the electromagnetic spectrum.
