NCERT Solutions Class 12 Physics Chapter 8

Q.1 Figure 8.6 shows a capacitor made of two circular plates each of radius 12 cm and separated by 5 cm. The capacitor is being charged by an external source (not shown in the figure). The charging current is constant and equal to 0.15 A.

(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 valid at each plate of the capacitor Explain?

Ans.

$\begin{array}{l}\text{a) As capacitance C of the capacitor is, C =}\frac{{\text{ε}}_{\text{0}}\text{A}}{\text{d}}\text{thus, we get}\\ \text{C =}\frac{{\text{(8.854×10}}^{\text{-12}}{\text{Fm}}^{-1}{\text{×πr}}^{\text{2}}\text{)}}{\text{d}}\\ \text{C=}\frac{{\text{8.854×10}}^{\text{-12}}{\text{Fm}}^{-1}\text{× 3.14 ×}{\left({\text{12×10}}^{\text{-2}}\mathrm{m}\right)}^{\text{2}}}{{\text{5×10}}^{\text{-2}}\mathrm{m}}\\ \text{C = 8 pF}\\ \text{As,\hspace{0.33em}V=}\frac{\text{Q}}{\text{C}}\text{\hspace{0.33em}thus,\hspace{0.33em}}\frac{\text{dV}}{\text{dt}}\text{=}\frac{\text{dQ}}{\text{Cdt}}\\ \frac{\text{dV}}{\text{dt}}\text{=}\frac{\text{I}}{\text{C}}\text{=}\frac{\text{0.15 A}}{{\text{8×10}}^{\text{-12}}\mathrm{F}}{\text{\hspace{0.33em}= 1.875×10}}^{\text{10}}{\text{Vs}}^{\text{-1}}\\ \left(\text{b}\right)\text{Displacement current}\\ {\text{I}}_{\text{d}}{\text{= ε}}_{\text{0}}\frac{\text{dΦ}}{\text{dt}}{\text{= ε}}_{\text{0}}\frac{\text{d}\left(\text{EA}\right)}{\text{dt}}{\text{= ε}}_{\text{0}}\text{A}\frac{\text{dE}}{\text{dt}}\\ {\text{I}}_{\text{d}}{\text{= ε}}_{\text{0}}\text{A}\frac{\text{dV}}{\text{d}\left(\text{dt}\right)}\text{= C}\frac{\text{dV}}{\text{dt}}\\ {\text{Thus, I}}_{\text{d}}\text{=}\left({\text{8×10}}^{\text{-12}}\mathrm{F}\right)\left({\text{1.875×10}}^{\text{10}}{\mathrm{Vs}}^{-1}\right)\text{=0.15 A}\\ \text{Displacement current is equal to conduction current.}\\ \left(\text{c}\right)\text{Yes, Kirchhoff’s first rule is valid at each plate of the capacitor as}\\ \text{conduction current entering one plate is equal to the displacement current leaving that plate.}\end{array}$

Q.2 A parallel plate capacitor (shown in figure) made of circular plates each of radius R = 6 cm has a capacitance C = 100 pF. The capacitor is connected to a 230 V a.c supply with an angular frequency of 300 rads^-1.
(a) What is the rms value of the conduction current?
(b) Is the conduction current equal to the displacement current?
(c) Determine the amplitude of B at a point 3 cm from the axis between the plates.

Ans.

$\begin{array}{l}\left(\text{a}\right){\text{Here, V\hspace{0.17em}}}_{\text{rms}}{\text{= 230 V , C =100 pF = 10}}^{\text{-10}}{\text{F , ω = 300 rad s}}^{\text{-1}}\\ {\text{As, I\hspace{0.17em}}}_{\text{rms}}\text{=}\frac{{\text{V\hspace{0.17em}}}_{\text{rms}}}{{\text{X\hspace{0.17em}}}_{\text{c}}}\text{=}\frac{{\text{V\hspace{0.17em}}}_{\text{rms}}}{\text{C\hspace{0.17em}ω}}\\ {\text{Thus, I\hspace{0.17em}}}_{\text{rms}}{\text{= 230 V × 10}}^{\text{-10}}{\text{F × 300 rad s}}^{\text{-1}}\text{=\hspace{0.33em} 6.9 µA}\end{array}$ $\begin{array}{l}\left(\text{b}\right)\text{Yes, the conduction current is equal to the displacement current because}\\ {\text{I}}_{\text{d}}{\text{= ε}}_{\text{0}}\frac{\text{dΦ}}{\text{dt}}{\text{= ε}}_{\text{0}}\frac{\text{d}\left(\text{EA}\right)}{\text{dt}}\\ {\text{Thus, I}}_{\text{d}}{\text{= ε}}_{\text{0}}\text{A}\frac{\text{dV}}{\text{d}\left(\text{dt}\right)}\text{= C}\frac{\text{dV}}{\text{dt}}\\ \text{As,\hspace{0.33em}}\frac{\text{dV}}{\text{dt}}\text{=}\frac{{\text{I}}_{\text{C}}}{\text{C}}\\ {\text{Thus,\hspace{0.33em}I}}_{\text{d}}\text{=}\frac{{\text{CI}}_{\text{C}}}{\text{C}}{\text{= I}}_{\text{C}}\left(\text{conduction current}\right)\end{array}$

$\begin{array}{l}\\ \left(\text{c}\right)\text{As per the figure given below, using Ampere’s circuital law we get,}\end{array}$

$\begin{array}{l}{\text{B = ε}}_{\text{0}}{\text{\hspace{0.17em}µ}}_{\text{0\hspace{0.17em}}}\text{r}\frac{\text{dE}}{\text{2 dt}}\\ \text{As, E =}\frac{\text{q}}{{\text{A\hspace{0.17em}ε}}_{\text{0}}}\\ \text{Thus,}\frac{\text{dE}}{\text{dt}}\text{=}\frac{\left(\frac{\text{dq}}{\text{dt}}\right)}{{\text{A\hspace{0.17em}ε}}_{\text{0}}}\\ \frac{\text{dE}}{\text{dt}}\text{=}\frac{{\text{I\hspace{0.17em}}}_{\text{rms}}}{\mathrm{\pi }{\text{\hspace{0.17em}R}}^{\text{2}}{\text{\hspace{0.17em}ε}}_{\text{0}}}\\ \text{Hence, B becomes; B =}\left(\frac{{\text{ε}}_{\text{0\hspace{0.17em}}}{\text{µ\hspace{0.17em}}}_{\text{0}}\text{r}}{\text{2}}\right)\text{x}\left(\frac{{\text{I}}_{\text{\hspace{0.17em}rms}}}{\mathrm{\pi }{\text{\hspace{0.17em}R\hspace{0.17em}}}^{\text{2}}{\text{\hspace{0.17em}ε\hspace{0.17em}}}_{\text{0}}}\right)\text{=}\frac{{\text{µ}}_{\text{0\hspace{0.17em}}}{\text{I\hspace{0.17em}}}_{\text{rms}}\text{\hspace{0.17em}r}}{\text{2}\mathrm{\pi }{\text{\hspace{0.17em}R}}^{\text{2}}}\\ {\text{Therefore, amplitude of magnetic field will be; B}}_{\text{0}}\text{=}\frac{{\text{µ}}_{\text{0}}{\text{\hspace{0.17em}I}}_{\text{0}}\text{\hspace{0.17em}r}}{\text{2}\mathrm{\pi }{\text{\hspace{0.17em}R}}^{\text{2}}}\\ {\text{or, B}}_{\text{0}}\text{=}\frac{{\text{µ}}_{\text{0}}\sqrt{\text{2}}{\text{\hspace{0.17em}I\hspace{0.17em}}}_{\text{rms}}}{\text{2\hspace{0.17em}}\mathrm{\pi }{\text{\hspace{0.17em}R}}^{\text{2}}}\\ {\text{B}}_{\text{0}}\text{=}\frac{\left(\text{4\hspace{0.17em}}\mathrm{\pi }{\text{×10}}^{\text{-7}}{\text{NA}}^{\text{-2}}\right)\left({\text{3×10}}^{\text{-2}}\text{m}\right)\left(\text{1.414}\right)\left({\text{6.9 ×10}}^{\text{-6}}\text{A}\right)}{\text{2\hspace{0.17em}}\mathrm{\pi }\text{×}{\left({\text{6 ×10}}^{\text{-2}}\text{m}\right)}^{\text{2}}}\\ \Rightarrow {\text{B}}_{\text{0}}{\text{= 1.63 ×10}}^{\text{-11}}\text{Tesla}\end{array}$

Q.3 What physical quantity is the same for X-rays of wavelength 10^-10 m, red light of wavelength 6800 Angstrom and radio waves of wavelength 500 m?

Ans.

$\begin{array}{l}\text{The speed of light (i}\text{.e}\text{.) c = 3}\times {\text{10}}^{\text{8}}{\text{ms}}^{\text{-1}}\text{remains the same in a vacuum for given X rays, red light and}\\ \text{radio waves}\text{.}\end{array}$

Q.4 A plane electromagnetic wave travels in vacuum along z direction. What can you say about the directions of its electric and magnetic field vectors? If the frequency of the wave is 30 MHz, what is its wavelength?

Ans.

The electric field vector E and magnetic field vector B are perpendicular to each other and they lie in x-y plane.

$\begin{array}{l}\text{If\hspace{0.33em}the\hspace{0.33em}frequency\hspace{0.33em}of\hspace{0.33em}wave},\mathrm{\nu }=30\mathrm{MHz}\\ \mathrm{\lambda }=\frac{\text{c}}{\mathrm{\nu }}=\frac{\text{3}\times \text{1}{0}^{\text{8}}{\mathrm{ms}}^{-1}}{\text{3}0\times \text{1}{0}^{\text{6}}\text{Hz}}=\text{1}0\text{m}\end{array}$

Q.5 A radio can tune into any station in the 7.5 MHz to 12 MHz. What is the corresponding wavelength band?

Ans.

$\begin{array}{l}\text{As,\hspace{0.33em}λ =}\frac{\text{c}}{\mathrm{\nu }}\text{=}\frac{{\text{3 ×10}}^{\text{8}}{\mathrm{ms}}^{-1}}{{\text{7.5 × 10}}^{\text{6}}\mathrm{Hz}}\text{= 40 m}\\ {\text{Similarly λ}}^{\text{’}}\text{=}\frac{\text{c}}{{\mathrm{\nu }}^{\text{’}}}\text{=}\frac{{\text{3 ×10}}^{\text{8}}{\mathrm{ms}}^{-1}}{{\text{12 ×10}}^{\text{6}}\mathrm{Hz}}\text{= 25 m}\\ \text{Thus, the corresponding wavelength band is 40 m to 25 m.}\end{array}$

Q.6 A charged particle oscillates about its mean equilibrium position with a frequency of 10⁹ Hz. What is the frequency of the electromagnetic waves produced by the oscillator?

Ans.

Frequency of the electromagnetic waves produced by the oscillator will be same as the frequency of oscillating charged particle (i.e.) 10⁹ Hz.

Q.7 The amplitude of the magnetic field part of a harmonic electromagnetic wave in vacuum is B₀ = 510 nT. What is the amplitude of electric field part of the wave?

Ans.

$\begin{array}{l}\text{As},\text{c}=\frac{{\text{E}}_0}{{\text{B}}_0}\text{thus}\\ {}_{}{\text{E}}_0={\text{cB}}_0\\ =\text{3}\times \text{1}{0}^{\text{8}}{\text{ms}}^{\text{-1}}\times \text{51}0\times \text{1}{0}^{-\text{9}}\text{T}\\ ={\text{153 NC}}^{\text{-1}}\end{array}$

Q.8 Suppose that the electric field amplitude of an electromagnetic wave is E₀ = 120 N/C and that its frequency is ν = 50.0 MHz. (a) Determine, B₀, ω, k, and λ. (b) Find expressions for E and B.

Ans.

$\begin{array}{l}\text{(a)}\left(\text{i}\right)\text{As, c =}\frac{{\text{E}}_{\text{0}}}{{\text{B}}_{\text{0}}}\\ \therefore {\text{B}}_{\text{0}}\text{=}\frac{{\text{E}}_{\text{0}}}{\text{c}}\\ {\text{B}}_{\text{0}}\text{=}\frac{{\text{120 NC}}^{\text{-1}}}{{\text{3×10}}^{\text{8}}{\text{ms}}^{\text{-1}}}\text{= 400 nT}\\ \left(\text{ii}\right)\text{Angular\hspace{0.17em}frequency\hspace{0.17em}is\hspace{0.17em}given\hspace{0.17em}by\hspace{0.17em}ω = 2}\mathrm{\pi \nu }\\ \mathrm{\omega }{\text{= 3.14×50×10}}^{\text{6}}{\text{rad/s = 3.14×10}}^{\text{8}}\text{\hspace{0.17em}rad/s}\\ \left(\text{iii}\right)\text{Propagation\hspace{0.17em}constant,\hspace{0.17em}k =}\frac{\text{2}\mathrm{\pi }}{\text{λ}}\\ \text{k =}\frac{\text{2×3.14}}{\text{6 m}}\text{= 1.05 rad/m}\\ \left(\text{iv}\right)\text{λ =}\frac{\text{c}}{\mathrm{\nu }}\\ \text{λ =}\frac{{\text{3×10}}^{\text{8}}{\text{ms}}^{\text{-1}}}{{\text{50×10}}^{\text{6}}{\text{s}}^{\text{-1}}}\text{= 6 m}\end{array}$

(b) Suppose a plane electromagnetic wave is propagating in X direction. Then, the electric field vector will be in Y-direction and magnetic field will be in Z-direction. This is because all three vectors are mutually perpendicular to each other.

$\begin{array}{l}{\text{Electric\hspace{0.17em}field\hspace{0.17em}vector,\hspace{0.17em}E = E}}_{\text{0}}\text{sin}\left(\text{kx – ωt}\right)\widehat{\mathrm{j}}\\ {\text{E = (120\hspace{0.17em} NC}}^{\text{-1}}\text{)\hspace{0.17em}sin\hspace{0.17em}}\left({\text{(1.05 \hspace{0.17em}radm}}^{\text{-1}}\text{)}\mathrm{x}{\text{– (3.14×10}}^{\text{8}}{\text{\hspace{0.17em}rads}}^{\text{-1}}\text{)\hspace{0.17em}t}\right)\widehat{\mathrm{j}}\text{and}\\ {\text{Magnetic\hspace{0.17em}field\hspace{0.17em}vector,\hspace{0.17em}B = B}}_{\text{0}}\text{sin}\left(\text{kx – ωt}\right)\widehat{\mathrm{k}}\\ {\text{B = (400×10}}^{\text{-7}}\text{\hspace{0.17em}T)\hspace{0.17em}sin\hspace{0.17em}}\left({\text{(1.05\hspace{0.17em} radm}}^{\text{-1}}{\text{)x – (3.14×10}}^{\text{8}}{\text{\hspace{0.17em}rads}}^{\text{-1}}\text{)\hspace{0.17em}t}\right)\widehat{\mathrm{k}}\text{\hspace{0.17em} with usual units.}\end{array}$

Q.9 The terminology of different parts of the electromagnetic spectrum is given in the text. Use the formula

\text{E}=\text{h}\nu

(for energy of a quantum of radiation : photon) and obtain the photon energy in units of eV for different parts of electromagnetic spectrum. In what way are the different scales of photon energies that you obtain related to the sources of electromagnetic radiation?

Ans.

$\begin{array}{l}\text{As per the relation for\hspace{0.33em} photon energy\hspace{0.33em}E = h}\mathrm{\nu }\text{\hspace{0.33em}joules}\\ \text{E =}\frac{\text{h}\mathrm{\nu }}{{\text{1.6 × 10}}^{\text{-19}}}\text{eV}\\ \text{As,}\mathrm{\nu }\text{=}\frac{\text{c}}{\text{λ}}\text{\hspace{0.33em}then}\\ \text{we get,\hspace{0.33em}E =}\frac{\text{h\hspace{0.17em}c}}{\text{λ\hspace{0.17em}e-}}\\ \text{For,\hspace{0.33em}wavelength = λ, we obtain}\\ {\text{E}}_{\text{λ}}\text{=}\frac{\left({\text{6.63×10}}^{\text{-34}}\text{Js}\right){\text{( 3×10}}^{\text{8}}{\text{ms}}^{-1}\text{)}}{\left({\text{λ ×1.6×10}}^{\text{-19}}\mathrm{C}\right)}\\ \text{Similarly wavelengths of different parts of electromagnetic spectrum can be obtained as}\\ {\text{E}}_{\text{λ}}\text{=}\left(\frac{{\text{1.24 × 10}}^{\text{-6}}}{\text{λ}}\right)\text{eV\hspace{0.17em}}\\ \left(\text{a}\right)\text{For radio waves let \hspace{0.33em}λ = 1 km, \hspace{0.33em}then}\\ {\text{E}}_{\text{λ}}\text{=}\left(\frac{{\text{1.24 × 10}}^{\text{-6}}}{\text{1000 m}}\right)\text{eV =}\left({\text{1.24×10}}^{\text{-9}}\right)\text{eV\hspace{0.33em}}\\ \left(\text{b}\right)\text{For microwaves let \hspace{0.33em}λ = 1 cm,\hspace{0.33em} then}\\ {\text{E}}_{\text{λ}}\text{=}\left(\frac{{\text{1.24 × 10}}^{\text{-6}}}{{\text{10}}^{\text{-2}}}\right)\text{eV =}\left({\text{1.24×10}}^{\text{-4}}\right)\text{eV\hspace{0.17em}}\\ \left(\text{c}\right)\text{For visible light let\hspace{0.33em} λ = 1 µm,\hspace{0.33em} then}\\ {\text{E}}_{\text{λ}}\text{=}\left(\frac{{\text{1.24 × 10}}^{\text{-6}}}{{\text{10}}^{\text{-6}}}\right)\text{eV = 1.24 eV\hspace{0.17em}.}\\ \left(\text{d}\right){\text{For X rays let \hspace{0.33em}λ = 1nm = 10}}^{\text{-9}}\text{\hspace{0.17em}m,\hspace{0.33em}then}\\ {\text{E}}_{\text{λ}}\text{=}\left(\frac{{\text{1.24 × 10}}^{\text{-6}}}{{\text{10}}^{\text{-9}}}\right)\text{eV =}\left({\text{1.24 ×10}}^{\text{3}}\right)\text{eV\hspace{0.17em}}\\ \left(\text{e}\right){\text{For gamma rays let λ = 10}}^{\text{-12}}\text{\hspace{0.17em}m,\hspace{0.33em}then}\\ {\text{E}}_{\text{λ}}\text{=}\left(\frac{{\text{1.24 × 10}}^{\text{-6}}}{{\text{10}}^{\text{-12}}}\right)\text{eV =}\left({\text{1.24 ×10}}^{\text{6}}\right)\text{eV\hspace{0.17em}}\end{array}$

Q.10 In a plane electromagnetic wave, the electric field oscillates sinusoidally at a frequency of 2 × 10¹⁰ Hz and amplitude 48 Vm^-1.
(a) What is the wavelength of the wave?
(b) What is the amplitude of the oscillating magnetic field?
(c) Show that the average energy density of the E field equals the average energy density of the B field. (c = 3 x 10⁸ ms^-1).

Ans.

$\begin{array}{l}\left(\text{a}\right)\text{As,wavelength\hspace{0.33em}of\hspace{0.33em}the\hspace{0.17em}wave,\hspace{0.33em}λ =}\frac{\text{c}}{\mathrm{\nu }}\\ \text{λ =}\frac{\text{c}}{\mathrm{\nu }}\\ \text{=}\frac{{\text{3 × 10}}^{\text{8}}{\text{ms}}^{-1}}{{\text{2 ×10}}^{\text{10}}{\text{s}}^{-1}}\\ {\text{= 1.5 ×10}}^{\text{-2}}\text{\hspace{0.17em}m\hspace{0.17em}}\\ \left(\text{b}\right)\text{Using c =}\frac{{\text{E}}_{\text{0}}}{{\text{B}}_{\text{0}}}\text{,}\\ {\text{we get amplitude\hspace{0.33em}of\hspace{0.33em}the\hspace{0.33em}oscillating\hspace{0.33em}magnetic\hspace{0.33em}field,\hspace{0.33em}B}}_{\text{0}}\text{=}\frac{{\text{E}}_{\text{0}}}{\text{c}}\\ {\text{Thus B}}_{\text{0}}\text{=}\frac{{\text{48 Vm}}^{-1}}{\left({\text{3×10}}^{\text{8}}{\text{ms}}^{-1}\right)}\\ {\text{= 1.6 ×10}}^{\text{-7}}\text{T\hspace{0.17em}}\\ \left(\text{c}\right){\text{In vacuum, the average energy density of electric field (U\hspace{0.17em}}}_{\text{E}}\text{),}\\ {\text{U\hspace{0.17em}}}_{\text{E}}\text{=}\frac{\text{1}}{\text{2}}{\text{\hspace{0.17em}ε}}_{\text{0\hspace{0.17em}}}{\text{E}}^{\text{2}}\\ {\text{whereas,\hspace{0.33em}the average energy density of magnetic field (U\hspace{0.17em}}}_{\text{B}}\text{),}\\ {\text{\hspace{0.33em}U\hspace{0.17em}}}_{\text{B}}\text{=}\frac{\text{1}}{\text{2}}\frac{{\text{B}}^{\text{2}}}{{\text{µ}}_{\text{0}}}\\ \text{Thus}\\ \frac{{\text{U\hspace{0.17em}}}_{\text{E}}}{{\text{U}}_{\text{\hspace{0.17em}B}}}\text{=}\frac{{\text{µ}}_{\text{0}}{\text{\hspace{0.17em}ε}}_{\text{0}}{\text{\hspace{0.17em}E}}^{\text{2}}}{{\text{B}}^{\text{2}}}\\ \text{=}\frac{{\text{µ}}_{\text{0\hspace{0.17em}}}{\text{ε}}_{\text{0}}}{{\text{µ}}_{\text{0\hspace{0.17em}}}{\text{ε}}_{\text{0}}}\text{= 1}\\ \text{as,}\frac{\text{E}}{\text{B}}\text{=}\frac{\text{1}}{\sqrt{{\text{µ}}_{\text{0\hspace{0.17em}}}{\text{ε}}_{\text{0}}}}\\ {\text{Hence U\hspace{0.17em}}}_{\text{E}}{\text{= U\hspace{0.17em}}}_{\text{B}}\end{array}$

Q.11 Suppose that electric field part of an electromagnetic wave in vacuum is

$\overrightarrow{\text{E}}{\text{= 3.1 \hspace{0.17em}N\hspace{0.17em}C}}^{\text{-1}}\text{cos\hspace{0.17em}}\left[{\text{(1.8 \hspace{0.17em}radm}}^{\text{-1}}{\text{)\hspace{0.17em}y + (5.4 × 10}}^{\text{8}}{\text{\hspace{0.17em}rads}}^{\text{-1}}\text{)\hspace{0.17em}t}\right]\widehat{\mathrm{i}}$

(a) What is the direction of propagation?
(b) What is the wavelength λ?
(c) What is the frequency

$\mathrm{\upsilon }$

?
(d) What is the amplitude of the magnetic field part of the wave?
(e) Write an expression for the magnetic field part of the wave.

Ans.

$\begin{array}{l}\left(\text{a}\right)\text{From the given equation, it is clear that the direction of propagation of the wave is}\\ \text{along negative y-direction}\left(\text{i.e}\right)\text{–j, because coefficient of y is positive.}\\ \left(\text{b}\right)\text{Comparing\hspace{0.33em}the\hspace{0.33em}given\hspace{0.33em}equation\hspace{0.33em}with\hspace{0.33em}the\hspace{0.33em}equation,}\\ {\text{E = E}}_{\text{0}}{\text{\hspace{0.17em}cos\hspace{0.17em}(k\hspace{0.17em}y+ω\hspace{0.17em}t),\hspace{0.33em}we\hspace{0.33em}get\hspace{0.33em}the\hspace{0.33em}value\hspace{0.33em}of\hspace{0.33em}k = 1.8 \hspace{0.17em}radm}}^{\text{-1}}\text{,}\\ {\text{ω = 5.4 × 10}}^{\text{8}}{\text{\hspace{0.17em}rad\hspace{0.17em}s}}^{\text{-1}}\text{,}\\ {\text{E}}_{\text{0}}{\text{= 3.1\hspace{0.17em}NC}}^{\text{-1}}\text{\hspace{0.33em}then}\\ \text{wavelength\hspace{0.33em}is\hspace{0.33em}given\hspace{0.33em}by}\\ \text{\hspace{0.33em}λ =}\frac{\text{2\hspace{0.17em}π}}{\text{k}}\\ \text{=}\frac{\text{2\hspace{0.17em}π}}{\text{1.8}}\text{= 3.5 m\hspace{0.17em}}\\ \left(\text{c}\right)\text{As,\hspace{0.33em}}\mathrm{\nu }\text{=}\frac{\text{ω}}{\text{2\hspace{0.17em}π}}\text{=}\frac{{\text{5.4×10}}^{\text{6}}}{\text{2\hspace{0.17em}π}}\\ \text{= 0.86 MHz\hspace{0.17em}}\\ \left(\text{d}\right)\text{c =}\frac{{\text{E}}_{\text{0}}}{{\text{B}}_{\text{0}}}{\text{\hspace{0.33em}thus,\hspace{0.33em}B}}_{\text{0}}\text{=}\frac{{\text{E}}_{\text{0}}}{\text{c}}\\ \text{=}\frac{\text{3.1}}{\left({\text{3×10}}^{\text{8}}\right)}{\text{=1.03×10}}^{\text{-8}}\text{=10.3 \hspace{0.17em}nT\hspace{0.17em}}\\ \left(\text{e}\right)\text{Using\hspace{0.33em} B =}\left({\text{B}}_{\text{0}}\text{\hspace{0.17em}cos}\left(\text{k\hspace{0.17em}y+ ω\hspace{0.17em}t}\right)\right)\text{,\hspace{0.33em}we get}\\ \text{B =}\left(\text{10.03 nT}\right)\text{cos\hspace{0.17em}}\left[{\text{(1.8\hspace{0.17em}rad\hspace{0.17em}m}}^{\text{-1}}{\text{)\hspace{0.17em}y+(5.4×10}}^{\text{8}}{\text{\hspace{0.17em}rad\hspace{0.17em}s}}^{\text{-1}}\text{)\hspace{0.17em}t}\right]\widehat{\mathrm{k}}\end{array}$

Q.12 About 5 % of the power of a 100 W light bulb is converted to visible radiation. What is the average intensity of visible radiation
(a) at a distance of 1 m from the bulb.
(b) at a distance of 10 m.
Assume that the radiation is emitted isotropically and neglect reflection.

Ans.

$\begin{array}{l}\text{Power rating of the bulb, P = 100 W}\\ \text{It is given that 5 \% of its power is convered into visible radiation.}\\ \therefore \text{Power of visible radiation, P’ =}\frac{\text{5}}{\text{100}}\text{× 100 = 5 W}\\ \left(\text{a}\right)\text{As average intensity of radiation at 1 m}\\ \text{distance from the bulb, I =}\frac{\text{Power}}{{\text{4\hspace{0.17em}π\hspace{0.17em}r}}^{\text{2}}}\\ \text{Thus, intensity, I =}\frac{\text{5 W}}{{\text{4×π×1 m}}^{\text{2}}}{{\text{= 0.4 Wm}}^{\text{–}}}^{\text{2}}\text{.}\\ \left(\text{b}\right)\text{Average intensity of radiation at 10 m}\\ \text{Distance from the bulb,I’ =}\frac{\text{5 W}}{{\text{4×π×10}}^{\text{2}}{\text{m}}^{\text{2}}}{\text{= 0.004\hspace{0.33em}Wm}}^{\text{-2}}\end{array}$

Q.13 Use the formula

${\mathrm{\lambda }}_{\text{m}}\text{T}$

= 0.29 cm K to obtain the characteristic temperature ranges for different parts of the electromagnetic spectrum. What do the numbers that you obtain tell you?

Ans.

$\begin{array}{l}{\text{According\hspace{0.33em}to\hspace{0.17em}Wein’s\hspace{0.33em}law,\hspace{0.33em}λ}}_{\text{m}}\text{T = 0.29\hspace{0.17em}cm K, thus\hspace{0.33em}T =}\frac{\text{0.29 cm K}}{{\text{λ}}_{\text{m}}}\\ {\text{which implies that, T = 0.29 K for\hspace{0.33em}λ}}_{\text{m}}\text{= 1\hspace{0.17em}cm\hspace{0.17em}.}\\ \left(\text{a}\right){\text{For\hspace{0.33em}λ}}_{\text{m}}{\text{= 10}}^{\text{-10}}\text{cm\hspace{0.33em}we get,}\\ \text{T =}\frac{\text{0.29 cm K}}{{\text{10}}^{\text{-10}}\text{cm}}\\ {\text{= 2.9×10}}^{\text{9}}\text{K\hspace{0.17em}}\\ \left(\text{b}\right){\text{For\hspace{0.33em}λ}}_{\text{m}}{\text{=10}}^{\text{-6}}\text{\hspace{0.17em}m}\\ {\text{=10}}^{\text{-4}}\text{cm}\\ \text{we get,}\\ \text{T =}\frac{\text{0.29 cmK}}{{\text{10}}^{\text{-4}}\text{cm}}\\ \text{= 2900 \hspace{0.17em}K}\\ \text{Similarly, temperatures for other wavelengths can be found. These numbers tell us the temperature ranges required for obtaining radiations in different}\\ {\text{parts of the electromagnetic spectrum. Thus to obtain visible radiation of wavelength 5 x 10}}^{\text{-7}}\text{m, the source should have a temperature of about}\\ \text{T =}\frac{\text{0.29 cmK}}{{\text{5×10}}^{\text{-5}}\text{cm}}\\ \text{= 5800 K}\\ \text{T}\simeq \text{6000 K.}\end{array}$

Q.14 Given below are some famous numbers associated with electromagnetic radiations in different contexts in physics. State the part of the electromagnetic spectrum to which each belongs:

(a) 21 cm (wavelength emitted by atomic hydrogen in interstellar space)

(b) 1057 MHz (frequency of radiation arising from two close energy levels in hydrogen)

(c) 2.7K ( temperature associated with the isotropic radiation filling all space – thought to be a relic of the big-bang origin of the universe)

(d) 5890 A – 5896 A (double lines of sodium)

(e) 14.4 keV (energy of a particular transition in Fe nucleus associated with a famous high resolution spectroscopic method.

Ans.

$\begin{array}{l}\left(\text{a}\right){\text{Short radio wave as wave length is of the order of 10}}^{\text{-2}}\text{m.}\\ \left(\text{b}\right){\text{Short radio wave as frequency is of the order of 10}}^{\text{-9}}\text{Hz.}\\ \left(\text{c}\right){\text{As\hspace{0.33em}λ}}_{\text{m}}\text{T = 0.29 cm K, thus,}\\ {\text{λ}}_{\text{m}}\text{=}\frac{\text{0.29}}{\text{T}}\text{=}\frac{\text{0.29 cm K}}{\text{2.7 K}}\\ \text{= 0.11 cm}\\ \text{Thus wavelength corresponds\hspace{0.33em}to microwave\hspace{0.33em}region\hspace{0.33em}of}\\ \text{e.m.\hspace{0.33em}spectrum\hspace{0.17em}.}\\ \left(\text{d}\right)\text{Visible radiations}\left(\text{Yellow light}\right){\text{as given wavelengths are of the order of 10}}^{\text{-7}}\text{m.}\\ \left(\text{e}\right){\text{E = 14.4\hspace{0.17em}keV = 14.4 × 10}}^{\text{3}}{\text{× 1.6 × 10}}^{\text{-19}}\text{\hspace{0.17em}J}\\ \text{Using ,}\\ \text{E = hν}\\ \text{ν =}\frac{\text{E}}{\text{h}}\\ \text{=}\frac{\left({\text{14.4 × 10}}^{\text{3}}{\text{× 1.6 x 10}}^{\text{-19}}\text{J}\right)}{\left({\text{6.63 ×10}}^{\text{-34}}\mathrm{Js}\right)}\\ {\text{= 3.47 × 10}}^{\text{18}}\text{\hspace{0.17em}Hz\hspace{0.17em}.}\\ \text{This corresponds to X rays.}\end{array}$

Q.15 Answer the following questions:
(a) Long distance radio broadcasts use short wave bands. Why?
(b) It is necessary to use satellites for long distance TV transmission. Why?
(c) Optical and radio telescopes are built on the ground but X-ray astronomy is possible only from satellites orbiting earth. Why?
(d) The small ozone layer on top of the stratosphere is crucial for human survival. Why?
(e) If the earth did not have its atmosphere, would its average surface temperature be higher or lower than what it is now?
(f) 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 of earth. What might be the basis of this prediction?

Ans.

(a) As long distance radio broadcasts make use of sky waves where Ionosphere of earth’s atmosphere reflects the radiations of this range.

(b) For every long distance transmission of TV signals, a very high frequency is required. Waves of this frequency just pass through the ionosphere, without being reflected, so a satellite is required to return the signals to the earth.

(c) As X-rays has smaller wavelength, it can be absorbed by the earth. Hence X ray astronomy is possible only from satellites orbiting the earth but visible waves and radio waves can pass through the atmosphere therefore we can work with visible waves and radio waves on earth’s surface.

(d) The small ozone layer on top of the stratosphere is crucial for human survival as it absorbs harmful radiations (ultraviolet radiations) present in the sunlight and prevents it from reaching the surface of the earth. Ultraviolet radiations are harmful for the life on earth.

(e) In this case, there will be no green house effect. So the earth will be at lower temperature.

(f) In the case of worldwide nuclear war the sky may get overcast with clouds. These clouds will prevent sunlight from reaching many parts of the globe. Thus earth will be as cool as in winter.