Given, \(\lambda\)_mT = 0.29 cm-K
\(\Rightarrow \quad \lambda_m=\frac{0.29}{T} \mathrm{~cm}\)
Let us take, \(\lambda\)_m = 10^-6 m = 10^-4 cm
Required absolute temperature,
\(T=\frac{0.29}{10^{-4}}=2900 \mathrm{~K}\)
Let us take, \(\lambda\)_m = 5 \(\times\)10^-5 cm for visible region.
Required absolute temperature,
\(T=\frac{0.29}{5 \times 10^{-5}}=5800 \mathrm{~K} \approx 6000 \mathrm{~K}\)
Hence, we can find the temperature for other parts of the electromagnetic spectrum in the same way. So, these numbers tell us about the temperature ranges for which atomic vibrations can produce these parts of electromagnetic waves.

Power rating of bulb, P = 100 W
It is given that about 5% of its power is converted into visible radiation.
∴ Power of visible radiation,
\(P I=\frac{5}{100} \times 100=5 W\)
Hence, the power of visible radiation is 5W.
(a) Distance of a point from the bulb, d = 1 m
Hence, intensity of radiation at that point is given as:
\(I=\frac{P \prime}{4 \pi d^{2}}\)
\(=\frac{5}{4 \pi(1)^{2}}=0.398 \frac{W}{m^{2}}\)
(b) Distance of a point from the bulb, d₁ = 10 m
Hence, intensity of radiation at that point is given as:
\(I=\frac{P \prime}{4 \pi d_{1}^{2}}\)
\(=\frac{5}{4 \pi(10)^{2}}=0.00398 \frac{W}{m^{2}}\)

(i) The given equation signifies that the electromagnetic wave is moving along Y-axis and also in negative direction, so it moves in - \(\hat{j}\) direction.
(ii) The electric part of electromagnetic wave in vacuum.
E = [3.1 cos{1.8 y + (5.4 \(\times\)10⁶ t)}] \(\hat{i}\)
Comparing with standard equation,
E = E₀ cos (ky + \(\omega\)t), we get
Angular frequency, \(\omega\) = 5.4 \(\times\)10⁶ rad/s
Wave number, k = 1.8 rad/m
The amplitude of the electric field part of the wave,
E₀ = 3.1 N/C
\(\begin{array}{rlrl} \lambda & =\frac{2 \pi}{k}=\frac{2 \pi}{1.8}=3.491 \mathrm{~m} \end{array}\)
\(\begin{array}{rlrl} \Rightarrow & \lambda =3.5 \mathrm{~m} \end{array}\)
(iii) Angular frequency, \(\omega\) = 2\(\pi\)v
\(v=\frac{\omega}{2 \pi}=\frac{5.4 \times 10^6 \times 7}{2 \times 22}\)
= 0.86 \(\times\)10⁶ Hz
(iv) As, \(c=\frac{E_0}{B_0}\)
Amplitude of magnetic field,
\(\begin{aligned} B_0 & =\frac{E_0}{c}=\frac{3.1}{3 \times 10^8} \end{aligned}\)
\(\begin{aligned} =1.03 \times 10^{-8} \mathrm{~T} \end{aligned}\)
(v) Expression for the magnetic field part of wave,
B = B₀ cos (ky + \(\omega\)t) \(\hat{k}\)
B = 1.03 \(\times\)10^-8 cos (1.8 y + 5.4 \(\times\)10⁶ t) \(\hat{k}\)

Given, v = 2 x 10⁸ m / s and \(\mu _{ r }\)
The speed of electromagnetic waves in  medium is given by 
v = \(\frac { 1 }{ \sqrt { \mu \varepsilon } } \)
Where, \(\mu \) and \(\varepsilon \) are absolute permeability and absolute permittivity of the medium.
Now,  \(\mu \) = \(\mu _{ 0 }\mu _{ r }\)
and  \(\varepsilon =\varepsilon _{ 0 }\varepsilon _{ r }\)
Eq. (i) becomes, v = \(\frac { 1 }{ \sqrt { \mu _{ 0 }\mu _{ r }\varepsilon _{ 0 }\varepsilon _{ r } } } \)
= \(\frac { 1 }{ \sqrt { \mu _{ 0 }\varepsilon _{ 0 } } } \) x  \(\frac { 1 }{ \sqrt { \mu _{ r }\varepsilon _{ r } } } \)
v = \(\frac { c }{ \sqrt { \mu _{ r }\varepsilon _{ r } } } \)  \(\left[ c=\frac { 1 }{ \sqrt { \mu _{ 0 }\varepsilon _{ 0 } } } \right] \)
On squaring both sides, we get 
\(\varepsilon _{ r }=\frac { c^{ 2 } }{ v^{ 2 }\mu _{ r } } \)
= \(\frac { (3\times10^{ 8 })^{ 2 } }{ (2\times10^{ 8 })^{ 2 }\times1 } \) = 2.25

(a) Equation second shows that the e.m.wave travels along the positive x-axis
(b) Wavelength of wave, \(\lambda =\frac { c }{ v } =\frac { c }{ (\omega /2\pi ) } =\frac { 2\pi c }{ \omega } =\frac { 2\pi \times (3\times { 10 }^{ 8 }) }{ 2\pi \times { 10 }^{ 8 } } =3.0m\)
(c) since the magnetic field is perpendicular to an electric field as well as the direction of propagation of e.m.wave, hence magnetic field must be varying along a z-axis. Therefore
\(\overset { \rightarrow }{ { B }_{ x } } =0; \ \overset { \rightarrow }{ { B }_{ y } } =0; \ and \ \overset { \rightarrow }{ { B }_{ z } } =\frac { 0.5 }{ 3\times { 10 }^{ 8 } } cos\left[ 2\pi \times { 10 }^{ 8 }(t-x/c) \right] \ \therefore [E/B=c]\)

Faraday from his experimental study on electromagnetic induction concluded that a magnetic field changing with time in a region produces an electric field there. Maxwell thought that there is a great symmetry in nature. He from his theoretical study concluded that an electric field changing with time in a region produces a magnetic field there. It means the change in either field (electric or magnetic) with time produces the other field. This idea led maxwell to conclude that the variation in electric and magnetic field vector perpendicular to each other leads to the production of e.m.waves, which can travel in space. 

Light and e.m.waves are of transverse nature and they travel with the same velocity in vacuum. This led maxwell to conclude that light itself is e.m. wave

If \(\theta \) is the angle which the direction of propagation of c.m. wave with y-axis, then
\(cos \ \theta =\frac { \overset { \rightarrow }{ r } .\hat { j } }{ r } =\frac { (\hat { i } +\hat { j } +\hat { k } ) }{ \sqrt { { 1 }^{ 2 }+{ 1 }^{ 2 }+{ 1 }^{ 2 } } } =\frac { 1 }{ \sqrt { 3 } } \ or \ \theta ={ cos }^{ -1 }\left( \frac { 1 }{ \sqrt { 3 } } \right) \)

Average energy density of electricfield
\( { U }_{ E }=\frac { 1 }{ 2 } { \epsilon }_{ 0 }{ E }_{ rms }^{ 2 }=\frac { 1 }{ 2 } { \epsilon }_{ 0 }{ \left( \frac { { \epsilon }_{ 0 } }{ \sqrt { 2 } } \right) }^{ 2 }=\frac { 1 }{ 4 } { \epsilon }_{ 0 }{ E }_{ 0 }^{ 2 }\)
Average energy density of magneticfield
\( \frac { { B }_{ rms }^{ 2 } }{ 2{ \mu }_{ 0 } } =\frac { { ({ B }_{ 0 }/\sqrt { 2 } ) }^{ 2 } }{ 2{ \mu }_{ 0 } } =\frac { 1 }{ 4 } \frac { { B }_{ 0 }^{ 2 } }{ { \mu }_{ 0 } } =\frac { 1 }{ 4 } { \epsilon }_{ 0 }{ E }_{ 0 }^{ 2 }\)

Given, amplitude of the magnetic field part of harmonic electromagnetic wave,
B₀ = 510 nT = 510 \(\times\)10^-9 T
Speed of light in a vacuum, c = 3 × 10⁸ m/s
Amplitude of electric field of the electromagnetic wave is given by the relation,
E = cB₀
= 3 × 10⁸ × 510 × 10⁻⁹ = 153 N/C
Therefore, the electric field part of the wave is 153 N/C.

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