Electromagnetic Waves

A long straight cable of length I is placed symmetrically along the z-axis and has radius a. The cable consists of a thin wire and a co-axial conducting tube. An alternating current I(t)=I_{0} sin (2\pi vt) flows down the central thin wire and returns along the co-axial conducting tube. The induced electric field at a distance s from the wire inside the cable is \mathbf{E}(\mathrm{s}, \mathrm{t})=\mu_{0} \mathrm{l}_{0} \mathrm{~V} cos (2\pi vt). In \left(\frac{s}{a}\right) \hat{k},
compare the conduction current 10 with the displacement current I_{0}^{\mathrm{d}}

The displacement will be, $I_{0}^{\mathrm{d}} / \mathrm{I}_{0}=(\mathrm{am} / \lambda)^{2}$

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A long straight cable of length I is placed symmetrically along the z-axis and has radius a. The cable consists of a thin wire and a co-axial conducting tube. An alternating current I(t)=I_{0} sin (2\pi vt) flows down the central thin wire and returns along the co-axial conducting tube. The induced electric field at a distance s from the wire inside the cable is \mathbf{E}(\mathrm{s}, \mathrm{t})=\mu_{0} \mathrm{l}_{0} \mathrm{~V} cos (2\pi vt). In \left(\frac{s}{a}\right) \hat{k},
a) calculate the displacement current density inside the cable
b) integrate the displacement current density across the cross-section of the cable to find the total displacement current I

a) The displacement current density is given as $\vec{J}_{d}=\frac{2 \pi I_{0}}{\lambda^{2}} \ln \frac{a}{s} \sin 2 \pi v t \hat{k}$ b) Total displacement current will be, $I^{d}=\int J_{d} 2 \pi s...

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Seawater at frequency v=4 \times 10^8 \mathrm{~Hz} has permittivity \varepsilon=80 \varepsilon_{0}, permeability \mu=\mu_{0} and resistivity \rho= 0.25 \Omega \mathrm{m}. Imagine a parallel plate capacitor immersed in seawater and driven by an alternating voltage source V(t)=V_{0} \sin (2 \pi v t) . What fraction of the conduction current density is the displacement current density?

The separation between the plates of the capacitor is given as $V(t)=V_{0} \sin (2 \pi v t)$ Ohm's law for the conduction of current density is given as $\mathrm{J}_{0}{...

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What happens to the intensity of light from a bulb if the distance from the bulb is doubled? As a laser beam travels across the length of a room, its intensity essentially remains constant. What geometrical characteristics of the LASER beam is responsible for the constant intensity which is missing in the case of light from the bulb?

When the distance between two points is doubled, the intensity of light is reduced by one-fourth. Geometrical characteristics of the LASER are: a) unidirectional b) monochromatic c) coherent...

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Electromagnetic waves with wavelength
i) \lambda_{1} is used in satellite communication
ii) \lambda_{2} is used to kill germs in water purifies
iii) \lambda_{3} is used to detect leakage of oil in underground pipelines
iv) \lambda_{4} is used to improve visibility in runaways during fog and mist conditions
a) identify and name the part of the electromagnetic spectrum to which these radiations belong
b) arrange these wavelengths in ascending order of their magnitude
c) write one more application of each

a) i) $\lambda_{1}$ is a microwave, used in satellite communication. ii) $\lambda_{2}$ is UV rays, used in a water purifier for killing germs. iii) $\lambda_{3}$ is X-rays, used in improving the...

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An EM wave of intensity I falls on a surface kept in vacuum and exerts radiation pressure p on it. Which of the following are true?
a) radiation pressure is \mathrm{I} / \mathrm{c} if the wave is totally absorbed
b) radiation pressure is I / c if the wave is totally reflected
c) radiation pressure is 2 \mathrm{I} / \mathrm{c} if the wave is totally reflected
d) radiation pressure is in the range I / c<p<2 I / c for real surface

The correct options are: a) radiation pressure is $\mathrm{l} / \mathrm{c}$ if the wave is totally absorbed c) radiation pressure is $2 \mathrm{l} / \mathrm{c}$ if the wave is totally reflected d)...

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An electromagnetic wave travelling along the z-axis is given as: E=E_{0} \cos (k z-\omega t). Choose the correct options from the following a) the associated magnetic field is given as B=\frac{1}{c} \hat{k} \times E=\frac{1}{\omega}(\hat{k} \times E)
b) the electromagnetic field can be written in terms of the associated magnetic field asE=c(B \times \hat{k})
c)\hat{k} \cdot E=0, \hat{k} \cdot B=0
d)\hat{k} \times E=0, \hat{k} \times B=0

a) the associated magnetic field is given as $B=\frac{1}{c} \hat{k} \times E=\frac{1}{\omega}(\hat{k} \times E)$ b) the electromagnetic field can be written in terms of the associated magnetic field...

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An electromagnetic wave travels in vacuum along z-direction: E=\left(E_{1} \hat{i}+E_{2} \hat{j}\right) \cos (k z-\omega t) \cdot Choose the correct options from the following:
a) the associated magnetic field is given as B=\frac{1}{c}\left(E_{1} \hat{i}-E_{2} \hat{j}\right) \cos (k z-\omega t)
b) the associated magnetic field is given as B=\frac{1}{c}\left(E_{1} \hat{i}-E_{2} \hat{j}\right) \cos (k z-\omega t)
c) the given electromagnetic field is circularly polarised
d) the given electromagnetic waves is plane polarised

a) the associated magnetic field is given as $B=\frac{1}{c}\left(E_{1} \hat{i}-E_{2} \hat{j}\right) \cos (k z-\omega t)$ d) the given electromagnetic waves is plane polarised

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Answer the following questions:
(a) If the earth did not have an atmosphere, would its average surface temperature be higher or lower than what it is now?
(b) Some scientists have predicted that a global nuclear war on the earth would be followed by a severe ‘nuclear winter’ with a devastating effect on life on earth. What might be the basis of this prediction?

(a) There would be no greenhouse effect if there was no atmosphere. As a result, the earth's temperature would plummet. (b) Smoke clouds from a worldwide nuclear war might potentially cover large...

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Answer the following questions:
(a) Optical and radio telescopes are built on the ground but X-ray astronomy is possible only from satellites orbiting the earth. Why?
(b) The small ozone layer on top of the stratosphere is crucial for human survival. Why?

(a) X-rays are absorbed by the atmosphere, although visible and radio waves can get through. (b) The ozone layer absorbs ultraviolet energy from the sun, preventing it from reaching the earth's...

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About 5 \% of the power of a 100 \mathrm{~W} light bulb is converted to visible radiation. What is the average intensity of visible radiation.
(a) at a distance of 1 \mathrm{~m} from the bulb?
(b) at a distance of 10 \mathrm{~m} ? Assume that the radiation is emitted isotropically and neglect reflection.

(a) Average intensity of the visible radiation is given by the expression, $I=P^{\prime} / 4 \pi d^{2}$ So, the power of the visible radiation will be $P^{\prime}=(5 / 100) \times 100=5 \mathrm{~W}$...

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The terminology of different parts of the electromagnetic spectrum is given in the text. Use the formula E=h v (for the energy of a quantum of radiation: photon) and obtain the photon energy in units of eV for different parts of the electromagnetic spectrum. In what way are the different scales of photon energies that you obtain related to the sources of electromagnetic radiation?

The energy of a photon is represented by the expression, $\mathrm{E}=\mathrm{hv}=\frac{h c}{\lambda}$ Where, $\mathrm{h}$ is the planck's constant with a value of$6.6 \times 10^{-34} J_{S}$...

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A parallel plate capacitor made of circular plates each of radius \mathbf{R}=\mathbf{6 . 0} \mathbf{~ c m} has a capacitance C=100 \mathrm{pF}. The capacitor is connected to a 230 \mathrm{~V} ac supply with an (angular) frequency of 300 \mathrm{rad} \mathrm{s}^{-1}.
(a) What is the rms value of the conduction current?
(b) Is the conduction current equal to the displacement current?

Solution: Radius of each circular plate is given as $0.06m$ Capacitance of a parallel plate capacitor is given as $\mathrm{C}=100 \mathrm{pF}=100 \times 10^{-12} \mathrm{~F}$ Supply voltage is given...

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The Figure shows a capacitor made of two circular plates each of radius 12 \mathrm{~cm} and separated by \mathbf{5 . 0} \mathbf{~ c m}. The capacitor is being charged by an external source (not shown in the figure). The charging current is constant and equal to 0.15A. Is Kirchhoff’s first rule (junction rule) valid at each plate of the capacitor? Explain.

Solution: Yes Kirchhoff's first rule holds true for each capacitor plate if we use the total of conduction and displacement to calculate current.

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The Figure shows a capacitor made of two circular plates each of radius 12 \mathrm{~cm} and separated by \mathbf{5 . 0} \mathbf{~ c m}. The capacitor is being charged by an external source (not shown in the figure). The charging current is constant and equal to 0.15A.
(a) Calculate the capacitance and the rate of change of the potential difference between the plates.
(b) Obtain the displacement current across the plates.

The radius of each circular plate $(r)$ is given as $0.12 \mathrm{~m}$ The distance between the plates $(d)$ is given as $0.05 \mathrm{~m}$ The charging current (I) is given as $0.15 \mathrm{~A}$...

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