17-In a plane electromagnetic wave, the electric field oscillates sinusoidally at a frequency of $2.0\times {10}^{10}$ Hz and amplitude 48 V/m. 

i) What is the wavelength of the wave? 

ii) What is the amplitude of the oscillating magnetic field? 

iii) Show that, the average energy density of the E field equals the average energy density of the B field. (Take, $\mathrm{c}=3\times {10}^8\mathrm{m}/\mathrm{s}$)
                                                           OR
(i) Which segment of electromagnetic waves has highest frequency? How are these waves produced? Give one use of these waves. 

(ii) Which EM waves lie near the high frequency end of visible part of EM spectrum?
Give its one use. In what way, this component of light has harmful effects
on humans?

$$\begin{gathered} \mathrm{Given}, \mathrm{frequency} \mathrm{of} \mathrm{oscillation}=2\times {10}^{10}\mathrm{Hz}, \\ \mathrm{Speed} \mathrm{if} \mathrm{wave}, \mathrm{c}=3\times {10}^8 \mathrm{m}/\mathrm{s} \\ \mathrm{and} \mathrm{electric} \mathrm{field} \mathrm{amplitude}, {\mathrm{E}}_0=48 \mathrm{V}/\mathrm{m} \\ \left(\mathrm{i}\right) \mathrm{Wavelength} \mathrm{of} \mathrm{waves}, \mathrm{\lambda }=\frac{\mathrm{c}}{\text{⨍}}=\frac{3\times {10}^8}{2\times {10}^{10}}=1.5\times {10}^{-2}\mathrm{m} \\ \left(\mathrm{ii}\right) \mathrm{Using} \mathrm{the} \mathrm{formula}, \mathrm{c}=\frac{{\mathrm{E}}_0}{{\mathrm{B}}_0} \\ \mathrm{The} \mathrm{amplitude} \mathrm{of} \mathrm{the} \mathrm{oscillating} \mathrm{magnetic} \mathrm{field}, \\ {\mathrm{B}}_0=\frac{{\mathrm{E}}_0}{\mathrm{c}}=\frac{48}{3\times {10}^8}=1.6\times {10}^{-7}\mathrm{T} \\ \left(\mathrm{iii}\right) \mathrm{The} \mathrm{average} \mathrm{energy} \mathrm{density} \mathrm{of} \mathrm{electric} \mathrm{field}, \\ {\mathrm{\upsilon }}_{\mathrm{E}}=\frac{1}{4}{\mathrm{\epsilon }}_0{\mathrm{E}}_0^2 \dots \left(\mathrm{i}\right) \\ \mathrm{We} \mathrm{know} \mathrm{that}, \frac{{\mathrm{E}}_0}{{\mathrm{B}}_0}=\mathrm{c} \\ \mathrm{Putting} \mathrm{in} \mathrm{Eq}. \left(\mathrm{i}\right), \mathrm{we} \mathrm{get} \\ \therefore {\mathrm{\upsilon }}_{\mathrm{E}}=\frac{1}{4}{\mathrm{\epsilon }}_0.{\mathrm{c}}^2{\mathrm{B}}_0^2 \dots \left(\mathrm{ii}\right) \end{gathered}$$OR

$$\begin{gathered} \mathrm{Speed} \mathrm{of} \mathrm{electromagnetic} \mathrm{waves}, \mathrm{c}=\frac{1}{\sqrt{{\mathrm{\mu }}_0{\mathrm{\epsilon }}_0}} \\ \mathrm{Putting} \mathrm{in} \mathrm{Eq}. \left(\mathrm{ii}\right), \mathrm{we} \mathrm{get} \\ {\mathrm{\upsilon }}_{\mathrm{E}}=\frac{1}{4}{\mathrm{\epsilon }}_0{\mathrm{B}}_0^2.\frac{1}{{\mathrm{\mu }}_0{\mathrm{\epsilon }}_0} \\ {\mathrm{\upsilon }}_{\mathrm{E}}=\frac{1}{4}.\frac{{\mathrm{B}}_{\mathbf{0}}^{\mathbf{2}}}{{\mathrm{\mu }}_0} \\ \mathrm{We} \mathrm{may} \mathrm{express} \mathrm{the} \mathrm{average} \mathrm{energy} \mathrm{density} \mathrm{in} \mathrm{EM} \mathrm{waves} \\ \overline{\mathrm{\upsilon }}=\frac{1}{2}{\mathrm{\epsilon }}_0{\mathrm{E}}_0^2=\frac{{\mathrm{B}}_0^2}{2{\mathrm{\mu }}_0} \end{gathered}$$Thus, the average energy density of the E field equals the

Average energy density of B field.

(i) Gamma rays has the highest frequency in the

     Electromagnetic waves. These rays are of the nuclear

Origin and are produced in the disintegration of

 Radioactive atomic nuclei and in the decay of certain

  Sub-atomic particles. They are used in the treatment of

  Cancer an tumours,

(ii) Ultraviolet rays lie near the high frequency end of

     Visible part of EM spectrum. These rays are used to

     Preserve food stuff. The harmful effect from exposure

     To ultraviolet (UV) radiation can be life threatening and

Include premature aging of the skin, suppression of the

Immune systems, damage to the eyes and skin cancer.