Physics Ncert Solutions Class 12th

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

Since, electric field of an EM wave is an oscillating field and so is the electric force caused by it on a charged particle. This electric force averaged over an integral number of cycles is zero, since its direction changes every half cycle.

Hence, electric field is not responsible for radiation pressure.

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

As the distance is doubled, the area of spherical region ( 4πr² )

will become four times, so the intensity becomes one fourth the initial value but in case of laser it does not spread, so its intensity remain same.

Geometrical characteristic of LASER beam which is responsible for the constant intensity are as following

(i) Unidirectional

(ii) Monochromatic

(iii) Coherent light

(iv) Highly collimated

These characteristic are missing in the case of light from the bulb.

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

Pressure = force/area

Force =dp/dt

U=p.C or p= U/c

Pressure = $\frac{1}{A}.\frac{U}{C.dt}$

Pressure = I/C

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

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