Electromagnetic Waves MCQ Questions & Answers in Electrostatics and Magnetism | Physics

$${E_0} = C{B_0}\,{\text{and}}\,C = \frac{1}{{\sqrt {{\mu _0}{\varepsilon _0}} }}$$
Electric energy density $$ = \frac{1}{2}{\varepsilon _0}E_0^2 = {\mu _E}$$
Magnetic energy density $$ = \frac{1}{2}\frac{{B{o^2}}}{{{\mu _0}}} = {\mu _B}$$
Thus, $${\mu _E} = {\mu _B}$$
Energy is equally divided between electric and magnetic field

Intensity of $$EM$$  wave is given by
$$\eqalign{ & I = \frac{P}{{4\pi {R^2}}} = {v_{av}} \cdot c = \frac{1}{2}{\varepsilon _0}E_0^2 \times c \cr & \Rightarrow {E_0} = \sqrt {\frac{P}{{2\pi {R^2}{\varepsilon _0}c}}} \cr & = \sqrt {\frac{{800}}{{2 \times 3.14 \times {{\left( 4 \right)}^2} \times 8.85 \times {{10}^{ - 12}} \times 3 \times {{10}^8}}}} \cr & = 54.77\frac{V}{m} \cr} $$

According to Maxwell, the electromagnetic waves are those waves in which there are sinusoidal variations of electric and magnetic field vectors at right angles to each other as well as right angles to the direction of wave propagation. Both these fields vary with time and space and have the same frequency.
In figure, the electric field vector $$\left( E \right)$$  and magnetic field vector $$\left( B \right)$$  are vibrating along $$y$$ and $$z$$-directions and propagation of electromagnetic wave is shown in $$x$$-direction.

From figure it is clear that electric and magnetic fields oscillate in same phase.

$${v_{\gamma - {\text{rays}}}} > {v_{{\text{visible}}\,{\text{radiation}}}} > {v_{{\text{infrared}}}} > {v_{{\text{Radio}}\,{\text{waves}}}}$$

Radio wave < yellow light < blue light < X-rays
(Increasing order of energy)

Gamma rays < X-rays < Ultra violet < Visible rays < Infrared rays < Microwaves < Radio waves.

$$\eqalign{ & {I_d} = 1\,mA = {10^{ - 3}}A \cr & C = 2\mu F = 2 \times {10^{ - 6}}F \cr & {I_D} = {I_C} = \frac{d}{{dt}}\left( {CV} \right) = C\frac{{dV}}{{dt}} \cr} $$
Therefore, $$\frac{{dV}}{{dt}} = \frac{{{I_D}}}{C} = \frac{{{{10}^{ - 3}}}}{{2 \times {{10}^{ - 6}}}} = 500\,V{s^{ - 1}}$$
Therefore, applying a varying potential difference of $$500\,V{s^{ - 1}}$$  would produce a displacement current of desired value.

Molecular spectra due to vibrational motion lie in the microwave region of $$EM$$ -spectrum. Due to Kirchhoff’s law in spectroscopy the same will be absorbed.

Wavelength range of various waves are as follows

So, radiowaves are the longest waves.

Name	Walenength range (m)
Gamma rays	$$6 \times {10^{ - 14}}$$   to $$1 \times {10^{ - 10}}$$
X-rays	$$1 \times {10^{ - 13}}$$   to $$3 \times {10^{ - 8}}$$
Radio waves	greater than 0.1
Microwaves	$${10^{ - 3}}$$  to 0.3

Given, energy of $$EM$$  waves is of the order of $$15\,keV$$
i.e. $$E = hv = h \times \frac{c}{\lambda }$$
$$\eqalign{ & \Rightarrow \lambda = \frac{{h \times c}}{E} = \frac{{6.624 \times {{10}^{ - 34}} \times 3 \times {{10}^{18}}}}{{15 \times {{10}^3} \times 1.6 \times {{10}^{ - 19}}}} \cr & = \frac{{1.3248 \times {{10}^{ - 29}}}}{{1.6 \times {{10}^{ - 19}}}} = 0.828 \times {10^{ - 10}}m \cr & = 0.828\,\mathop {\text{A}}\limits^ \circ \,\,\left[ {\because 1\mathop {\text{A}}\limits^ \circ = {{10}^{ - 10}}m} \right] \cr & \lambda = 0.828\,\mathop {\text{A}}\limits^ \circ \cr} $$
Thus, this spectrum is a part of X-rays.