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

$${B_y} = 2 \times {10^{ - 7}}\sin \left( {0.5 \times {{10}^3}z + 1.5 \times {{10}^{11}}t} \right)T$$
The electric vector is perpendicular to $$B$$ as well as direction of propagation of electromagnetic wave.
Therefore $${E_x}$$ has to be taken.
Further, $${E_0} = {B_0} \times c$$
$$\eqalign{ & = 2 \times {10^{ - 7}} \times 3 \times {10^8}\,V/m \cr & {E_0} = 2 \times {10^{ - 7}} \times 3 \times {10^8} = 60\,V/m \cr} $$
$$\therefore $$ The corresponding value of the electric field is
$${E_x} = 60\sin \left( {0.5 \times {{10}^3}z + 1.5 \times {{10}^{11}}t} \right)V/m$$

Depends on the magnitude of frequency

$$\eqalign{ & {\text{Here}}\,c = 3 \times {10^8}m/s,v = 2 \times {10^{10}}Hz,{E_0} = 48\,V/m \cr & \lambda = ?,{B_0} = ?,\overrightarrow {{u_E}} = ?,\overrightarrow {{u_B}} = ? \cr & \because c = v\lambda \cr & \Rightarrow \lambda = \frac{c}{v} = \frac{{3 \times {{10}^8}}}{{2 \times {{10}^{10}}}} = 1.5 \times {10^{ - 2}}m \cr & \frac{{{E_0}}}{{{B_0}}} = c \Rightarrow {B_0} = \frac{{48}}{{3 \times {{10}^8}}} = 16 \times {10^{ - 8}}T. \cr} $$
Energy density due to electric field and magnetic field
$$\eqalign{ & {u_E} = \frac{1}{2}{\varepsilon _0}{E^2}\,\& \,{u_B} = \frac{1}{2}{\mu _0}{B^2} \cr & {\text{We}}\,{\text{have}}\,E = cB\,{\text{and}}\,{C^2} = \frac{1}{{{\mu _0}{\varepsilon _0}}}\,{\text{so,}} \cr & {U_E} = \frac{1}{2}{\varepsilon _0}{\left( {cB} \right)^2} = \frac{1}{2}{\varepsilon _0}\left( {\frac{1}{{{\mu _0}{\varepsilon _0}}}} \right){B^2} = \frac{1}{{2{\mu _0}}}{B^2} = {U_B}. \cr} $$

A stationary charge produces electric field only; a uniformly moving charge produces localised electromagnetic field.
An accelerated charge produces electromagnetic radiations. The reason is that due to accelerated charges, magnetic field is produced around accelerating charges. As the velocity of charge changes, the magnetic field produced due to it also changes with time. This varying magnetic field produces the electric field.
The electric field so produced also changes with time. These two varying fields are mutually perpendicular and also perpendicular to the direction of propagation of wave and both the fields are in same phase and of same frequency. The frequency of these fields is same as the frequency of oscillations of the charged particle. The wave associated with these oscillations is called the electromagnetic wave.

As we know, $$\overrightarrow E \cdot \overrightarrow B = 0\,\,\,\,\because \left[ {\overrightarrow E \bot \overrightarrow B } \right]$$
and $$\overrightarrow E \times \overrightarrow B $$   should be along $$Z$$ direction
As $$\left( { - 2\hat i - 3\hat j} \right) \times \left( {3\hat i - 2\hat j} \right) = 5\hat k$$
Hence option (B) is the correct answer.

Ultraviolet radiations are used in the detection of invisible writing, forged documents, finger prints in forensic lab. While microwaves are used in microwave oven.

Electromagnetic wave equation is given by
$$E = {E_0}\cos \left( {kz - \omega t} \right)\,......\left( {\text{i}} \right)$$
Speed of electromagnetic wave, $$v = \frac{\omega }{k}$$
Given, equation is
$$E = \hat i40\cos \left( {kz - 6 \times {{10}^8}t} \right)\,.....\left( {{\text{ii}}} \right)$$
Comparing Eqs. (i) and (ii), we get
$$\omega = 6 \times {10^8}$$
and $${E_0} = 40\,\hat i$$
Here, wave factor, $$k = \frac{\omega }{v} = \frac{{6 \times {{10}^8}}}{{3 \times {{10}^8}}} = 2\,{m^{ - 1}}$$

The power per unit area carried by an $$E.M.$$   wave i.e., energy transported per unit time across a unit cross-section area is perpendicular to the direction in which the wave is travelling.

Comparing the given equation
$${E_y} = 2.5\frac{N}{C}\cos \left[ {\left( {2\pi \times {{10}^6}\frac{{rad}}{m}} \right)t - \left( {\pi \times {{10}^{ - 2}}\frac{{rad}}{{\sec }}} \right)x} \right]$$
With the standard equation
$${E_y} = {E_0}\cos \left( {\omega t - kx} \right)$$
we get $$\omega = 2\pi f = 2\pi \times {10^6}$$
$$\therefore f = {10^6}\,Hz$$
Moreover, we know that
$$\eqalign{ & \frac{{2\pi }}{\lambda } = k = \pi \times {10^{ - 2}}\;{m^{ - 1}} \cr & \Rightarrow \lambda = 200\,m \cr} $$
As direction of field $$E$$ of electromagnetic wave is in $$y$$ direction so, the wave is moving along positive $$x$$-direction with frequency $${10^6}\,Hz$$  and wavelength $$200\,m.$$

The time varying electric and magnetic fields are mutually perpendicular to each other and also perpendicular to the direction of propagation.