Electromagnetic Waves

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q=CV

And we know q=It

Then I_ddt= cdV

I_d= Cdv/dt

1 $*{10}^{-3}$ = 2 $*$ 10^{-6 $*$} dV/dt

DV/dt= 500V/s, this will produce the desired result.

This is a short answer type question as classified in NCERT Exemplar

As we know radiant flux density S=  $\frac{1}{{\mu }_0}$ (E $*B$ )= c^{2 $?$} ₀(E $*B$ )

Suppose electromagnetic waves be propagating along x-axis. The electric field vector of

Electromagnetic wave be along y-axis and magnetic field vector be along z-axis. Therefore,

E=E₀cos (kx-wt)

B=B₀cos (kx-wt)

E $*$ B= (E₀B₀)cos²(kx-wt)

S= c^{2 ${\epsilon }_0$} (E $*B$ )

Radiant flux density over a complete cycle is

S_av= c^{2 $?$} ₀|E_{0 $*$} B₀| $\frac{1}{T}{\int }_0^T{cos}^2(kx-wt)dt$

     = c^{2 $?$} ₀|E₀B₀| $\frac{1}{T}*\frac{T}{2}$

     = $\frac{C^2}{2}\epsilon$ ₀E₀( $\frac{E_0}{c}$ )

     = $\frac{c}{2}\epsilon$ ₀E₀²= $\frac{c}{2}*$ $\frac{1}{c^2{\mu }_0}$  E₀²

     = $\frac{E_0^2}{2{\mu }_{0}c}$

This is a short answer type question as classified in NCERT Exemplar

Magnetic field induction at a point in a region between two plates of a parallel plate capacitor is

B= $\frac{{\mu }_{0}2I_d}{4\pi r}$ = $\frac{{\mu }_0{I}_d}{2\pi r}$ = $\frac{{\mu }_0}{2\pi r}?$ 0 $\frac{d{\varnothing }_E}{dt}$

= $\frac{{\mu }_{0{?}_0}}{2\pi r}\frac{d}{dt}$  (E $\pi r^2$ ) = $\frac{{\mu }_{0{?}_0}}{2\pi r}\pi r^2\frac{d}{dt}$  (E)

B= $\frac{{\mu }_{0{?}_{0r}}}{2\pi r}\frac{d}{dt}$  (E)

This is a short answer type question as classified in NCERT Exemplar

An electromagnetic wave carries energy and momentum like other waves.

Since, it carries momentum, an electromagnetic wave also exerts pressure called radiation pressure. This property of electromagnetic waves helped professor CV Raman surprised his students by suspending freely a tiny light ball in a transparent vacuum chamber by shining a laser beam on it. The tails of the camets are also due to radiation pressure.

This is a short answer type question as classified in NCERT Exemplar

Consider and electromagnetic waves, let E be varying along y-axis, B is along z-axis and propagation of wave be along x-axis. Then E*B will tell the direction of propagation of

energy flow in electromagnetic wave, along x-axis.

E= E₀sin (wt-kx)j

B=B₀sin (wt=kx)k

So S will become $\frac{E_0{B}_0}{{\mu }_0}$ sin² (wt-kx)i

And its variation with time is given below

This is a short answer type question as classified in NCERT Exemplar

Aaverage magnetic energy density I_av= $\frac{1}{2}\frac{B^2}{{\mu }_0}$ c

From eqn we know that B=12 $*{10}^{-8}$

= $\frac{1}{2}*\frac{({12*{10}^{-8})}^2*3*{10}^8}{4\pi *{10}^{-7}}$ = 1.71W/m²

This is a short answer type question as classified in NCERT Exemplar

X_c=1/2 $\pi fc$

And displacement current is inversely proportional to X_c so when capacitive reactance increases then displacement current will decrease.

This is a short answer type question as classified in NCERT Exemplar

I_d=I_c= dq/dt

And q=q₀ cos 2πvt

By putting this in above equation I_d=I_c= q₀ sin2πvt $*$ 2 $\pi \upsilon$

This is a short answer type question as classified in NCERT Exemplar

Microwave oven heats up the food items containing water molecules most efficiently because the frequency of microwaves matches the resonant frequency of water molecules.