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Practice each question, then open the answer or worked solution.
Show that the magnetic field B at a point in between the plates of a parallel-plate capacitor during charging is ε0μr/2 dE/dt.
Id is the displacement current between the two plates of a parallel plate capacitor.
The distance between the two plates is r
This is given as
B = μ0εr/2 dE/dt
Electromagnetic waves with wavelength
i) λ1 is used in satellite communication
ii) λ2 is used to kill germs in water purifies
iii) λ3 is used to detect leakage of oil in underground pipelines
iv) λ4 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) λ1 is a microwave which is used in satellite communication.
ii) λ2 is UV rays which is used in a water purifier for killing germs.
iii) λ3 is X-rays which is used in improving the visibility.
b) λ3 < λ2 < λ4 < λ1
c) Gamma rays are used in nuclear structures
X rays are used in medical diagnosis
UV rays are used in the preservation of food
Visible light is used to see objects
Infrared rays are used to take photography
Microwave and radio waves are used in telecommunication
Show that the average value of radiant flux density S over a single period T is given by S = 1/2cμ0 E02.
Radiant flux density is given as
E = E0 cos (kx – ꞷt)
B = B0 cos (kx – ꞷt)
EB = c2 ε0 (E0B0) cos2 (kx – ꞷt)
Average value of the radiant flux density is
Sav = E02/2μ0c
You are given a 2μF parallel plate capacitor. How would you establish an instantaneous displacement current of 1 mA in the space between its plates?
The capacitance of the capacitor C = 2μF
Displacement current Id = 1 mA
Charge in capacitor, q = CV
dV/dt = 500 V/s
Show that the radiation pressure exerted by an EM wave of intensity I on a surface kept in vacuum is I/c.
Energy received by the surface per second = E = IA
No.of photons received by the surface per second = N
The perfect absorbing is = h/λ
Pressure, P = F/A = I/c
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 is doubled, the intensity of light becomes one-fourth.
Following are the geometrical characteristics of the LASER:
a) unidirectional
b) monochromatic
c) coherent light
Even though an electric field E exerts a force qE on a charged particle yet the electric field of an EM wave does not contribute to the radiation pressure. Explain.
Even though an electric field E exerts a force qE on a charged particle yet the electric field of an EM wave does not contribute to the radiation pressure because radiation pressure is the result of the action of the magnetic field of the wave on the electric currents which is induced by the electric field.
Long AnswersAn infinitely long thin wire carrying a uniform linear static charge density λ is placed along the z-axis. The wire is set into motion along its length with a uniform velocity
. Calculate the pointing vectors S = 1/μ0 (E×B).
The electric field in an infinitely long thin wire is
Magnetic field due to the wire is
The equivalent current flowing through the wire is
Seawater at frequency v = 4 108 Hz has permittivity ε = 80 ε0, permeability μ= μ0 and resistivity ρ = 0.25 Ωm. Imagine a parallel plate capacitor immersed in seawater and driven by an alternating voltage source V(t) = V0 sin (2πvt). What fraction of the conduction current density is the displacement current density?
The separation between the plates of the capacitor is V(t) = V0 sin (2πvt)
Ohm’s law for the conduction of current density J0c = V0/ρd
The displacement current density is Jod = 2πvεV0/d
The fraction of the conduction of current density and the displacement density = J0d/J0c = 4/9
A long straight cable of length l 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) = I0 sin (2πvt) 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 E(s,t) = μ0I0v coz (2πvt). In
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
c) compare the conduction current I0 with the displacement current I0d
a) The displacement current density is given as
b) Total displacement current
Id = (πa/λ)2I0 sin 2πvt
c) The displacement current is I0d/I0 = (aπ/λ)2