physics_electromagneticwaves Electromagnetic Waves
 Displacement Current: Displacement current is produced when electric field or electric flux varies with time. Maxwell showed that change in electric field also produces magnetic field. Stationary charges produces electric field Charges in uniform motion produces magnetic field Accelerated charges radiate electromagnetic waves EMF is induced in circuit whenever the amount of magnetic flux linked with the circuit changes. Maxwell experiment1. Time varying electric field generate magnetic field2. Time varying magnetic field generate electric field (Faraday Lenz Law).3. If Faraday’s law is true then the vice-versa should also be true, i.e. A time varying electric field should also be able to generate a magnetic field. Faraday Lenz Law: An electric current, induced by a source, such as a changing magnetic field, always creates a counter force opposing the force inducing it. Maxwell’s correction to Ampere’s law:1. Ampere’s law is true only for steady currents.2. Maxwell found the shortcoming in Ampere’s law and he modified Ampere’s law to include time-varying electric fields.3. At the time of charging and discharging or charging the capacitor current will be generated. Then there is time varying electric field between the plates.aE=QA𝜖0⟹EA=Q𝜖0 ⟹𝜙E=Q𝜖0 ⟹d𝜙Edt=1𝜖0dQdt=1𝜖0Id ⟹Id=𝜖0d𝜙Edt $\begin{array}{l}E=\frac{Q}{A{𝜖}_{0}}\\ ⟹EA=\frac{Q}{{𝜖}_{0}}\phantom{\rule{0.22em}{0ex}}\\ ⟹{𝜙}_{E}=\frac{Q}{{𝜖}_{0}}\phantom{\rule{0.22em}{0ex}}\\ ⟹\frac{d{𝜙}_{E}}{dt}=\frac{1}{{𝜖}_{0}}\frac{dQ}{dt}=\frac{1}{{𝜖}_{0}}{I}_{d}\phantom{\rule{0.22em}{0ex}}\\ ⟹{I}_{d}={𝜖}_{0}\frac{d{𝜙}_{E}}{dt}\phantom{\rule{0.22em}{0ex}}\end{array}$ Ampere-Maxwell Law a∮B.dl=𝜇0ic+𝜇0𝜖0d𝜙Edt $\begin{array}{l}\oint B.dl={𝜇}_{0}{i}_{c}+{𝜇}_{0}{𝜖}_{0}\frac{d{𝜙}_{E}}{dt}\phantom{\rule{0.22em}{0ex}}\\ \phantom{\rule{0.22em}{0ex}}\end{array}$ Electromagnetic Wave1. Time varying electric and magnetic field that propagate in space. Nature of electromagnetic wavesElectric and magnetic fields in an electromagnetic wave are perpendicular to each other, and to the direction of propagation.aEx=E0sin((kz-𝜔t)) By=B0sin((kz-𝜔t)) where, k=2𝜋𝜆=wave vector 𝜔=c k c0=1𝜇0𝜖0 $\begin{array}{l}{E}_{x}={E}_{0}sin\left(kz-𝜔t\right)\phantom{\rule{0.22em}{0ex}}\\ {B}_{y}={B}_{0}sin\left(kz-𝜔t\right)\phantom{\rule{0.44em}{0ex}}\\ \phantom{\rule{1.1em}{0ex}}where,\phantom{\rule{0.22em}{0ex}}k=\frac{2𝜋}{𝜆}=wave\phantom{\rule{0.22em}{0ex}}vector\phantom{\rule{0.22em}{0ex}}\\ \\ 𝜔=c\phantom{\rule{0.22em}{0ex}}k\\ \\ {c}_{0}=\frac{1}{\sqrt{{𝜇}_{0}{𝜖}_{0}}}\phantom{\rule{0.22em}{0ex}}\end{array}$ B0=E0c${B}_{0}=\frac{{E}_{0}}{c}$ Electromagnetic waves carries energies and momentum like other waves Electromagnetic waves also exerts pressure called radiation pressure. In the region of free space, with electric field E, the energy density is UU ((energy density per unit volume))=12𝜖0E2 $U\phantom{\rule{0.22em}{0ex}}\left(energy\phantom{\rule{0.22em}{0ex}}density\phantom{\rule{0.22em}{0ex}}per\phantom{\rule{0.22em}{0ex}}unit\phantom{\rule{0.22em}{0ex}}volume\right)=\frac{1}{2}{𝜖}_{0}{E}^{2}\phantom{\rule{0.22em}{0ex}}$ In the region of free space, with magnetic field B, the energy density is UU ((energy density per unit volume))=12𝜇0B2 $U\phantom{\rule{0.22em}{0ex}}\left(energy\phantom{\rule{0.22em}{0ex}}density\phantom{\rule{0.22em}{0ex}}per\phantom{\rule{0.22em}{0ex}}unit\phantom{\rule{0.22em}{0ex}}volume\right)=\frac{1}{2}{𝜇}_{0}{B}^{2}\phantom{\rule{0.22em}{0ex}}$ If total energy transfered to surface in time t is U then total momentum delivered to this surface for complete absorption is pp=Uc$p=\frac{U}{c}$ Radio Wave: Radio waves are produced by the accelerated motion of charges in conductingwires.MicrowavesInfrared wavesVisible raysUltraviolet raysX-raysGamma rays
Example 8.2 (NCERT)A plane electromagnetic wave of frequency 25 MHz travels in free space along the x-direction. At a particular point in space and time, $E=6.3\phantom{\rule{0.22em}{0ex}}\mathrm{ĵ}\phantom{\rule{0.22em}{0ex}}V/m$. What is B at this point?Answer.Example 8.2$B=\frac{E}{c}=\frac{6.3}{3×{10}^{8}}=2.1×{10}^{-8}\phantom{\rule{0.22em}{0ex}}T\phantom{\rule{0.22em}{0ex}}in\phantom{\rule{0.22em}{0ex}}z-direction$ Q 8.1: A capacitor made of two circular plates each of radius 12 cm, and separated by 5.0 cm. 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 potential difference between the plates.(b) Obtain the displacement current across the plates.(c) Is Kirchhoff’s first rule (junction rule) valid at each plate of the capacitor? Explain. Answer : 8.1 $\begin{array}{l}\left(a\right)\\ C=\frac{{𝜖}_{0}A}{d}\phantom{\rule{0.22em}{0ex}}=\frac{\phantom{\rule{0.22em}{0ex}}8.85⨯{10}^{-12}⨯𝜋⨯\left(0.12{\right)}^{2}}{0.05}=8⨯{10}^{-12}\phantom{\rule{0.22em}{0ex}}F\phantom{\rule{0.22em}{0ex}}\\ \\ C=\frac{Q}{V}\\ ⟹CV=Q\\ ⟹C\phantom{\rule{0.22em}{0ex}}\frac{dV}{dt}=\frac{dQ}{dt}\phantom{\rule{0.22em}{0ex}}=I\\ ⟹8⨯{10}^{-12}⨯\frac{dV}{dt}\phantom{\rule{0.22em}{0ex}}=\phantom{\rule{0.22em}{0ex}}0.15\\ ⟹\frac{dV}{dt}=1.875⨯{10}^{10}\phantom{\rule{0.22em}{0ex}}V/s\\ \\ \\ \left(b\right)\\ Displacement\phantom{\rule{0.22em}{0ex}}current\phantom{\rule{0.22em}{0ex}}={I}_{d}={𝜖}_{0}\frac{d{𝜙}_{E}}{dt}\\ ⟹{I}_{d}={𝜖}_{0}\frac{d\left(EA\right)}{dt}\\ ⟹{I}_{d}={𝜖}_{0}A\frac{d\left(E\right)}{dt}\\ ⟹{I}_{d}={𝜖}_{0}A\frac{d\left(\frac{𝜎}{{𝜖}_{0}}\right)}{dt}\\ \\ ⟹{I}_{d}={𝜖}_{0}A\frac{d\left(\frac{Q}{A{𝜖}_{0}}\right)\right)}{dt}\\ \\ ⟹{I}_{d}=\frac{dQ}{dt}\\ \\ ⟹{I}_{d}=0.15\phantom{\rule{0.22em}{0ex}}A\\ \\ \end{array}$