$$EM$$ waves carry momentum and hence can exert pressure on surfaces. They also transfer energy to the surface so $$p \ne 0$$ and $$E \ne 0.$$
23.
The energy contained in a cylinder of cross-section $$10\,c{m^2}$$ and length $$50\,cm$$ along the $$x$$-axis if electric field in an electromagnetic wave is given by
$$E = 50\sin \omega \left( {t - \frac{x}{c}} \right)$$ is
Energy density $$ = \frac{1}{2}{\varepsilon _0}E_0^2 = 1.1 \times {10^{ - 8}}J{m^{ - 3}}$$
Volume of cylinder $$ = A \times L = 5 \times {10^{ - 4}}{m^3}$$
So, required energy $$ = 1.1 \times {10^{ - 8}} \times 5 \times {10^{ - 4}}$$
$$ = 5.5 \times {10^{ - 12}}J$$
24.
The electric field component of a monochromatic radiation is given by
$$\overrightarrow E = 2{E_0}\hat i\cos kz\cos \omega t$$
Its magnetic field $$\overrightarrow B $$ is then given by :
A
$$\frac{{2{E_0}}}{c}\hat j\sin kz\cos \omega t$$
B
$$ - \frac{{2{E_0}}}{c}\hat j\sin kz\sin \omega t$$
C
$$\frac{{2{E_0}}}{c}\hat j\sin kz\sin \omega t$$
D
$$\frac{{2{E_0}}}{c}\hat j\cos kz\cos \omega t$$
Magnetic field vector, $$\vec B = \frac{{2{E_0}}}{c}\hat j\sin kz\sin \omega t$$
25.
The $$rms$$ value of the electric field of the light coming from the Sun is $$720\,N/C.$$ The average total energy density of the electromagnetic wave is
$${E_{rms}} = 720$$
The average total energy density
$$\eqalign{
& = \frac{1}{2}{ \in _0}E_0^2 = \frac{1}{2}{ \in _0}{\left[ {\sqrt 2 {E_{rms}}} \right]^2} = { \in _0}E_{rms}^2 \cr
& = 8.85 \times {10^{ - 12}} \times {\left( {720} \right)^2} = 4.58 \times {10^{ - 6}}J/{m^3} \cr} $$
26.
An electromagnetic wave with frequency $$\omega $$ and wavelength $$\lambda $$ travels in the $$+y$$ direction. Its magnetic field is along $$+x$$ -axis. The vector equation for the associated electric field (of amplitude $${E_0}$$) is
A
$$\overrightarrow E = - {E_0}\cos \left( {\omega t + \frac{{2\pi }}{\lambda }y} \right)\hat x$$
B
$$\overrightarrow E = {E_0}\cos \left( {\omega t - \frac{{2\pi }}{\lambda }y} \right)\hat x$$
C
$$\overrightarrow E = {E_0}\cos \left( {\omega t - \frac{{2\pi }}{\lambda }y} \right)\hat z$$
D
$$\overrightarrow E = - {E_0}\cos \left( {\omega t + \frac{{2\pi }}{\lambda }y} \right)\hat z$$
Answer :
$$\overrightarrow E = {E_0}\cos \left( {\omega t - \frac{{2\pi }}{\lambda }y} \right)\hat z$$
In an electromagnetic wave electric field and magnetic field are perpendicular to the direction of propagation of wave. The vector equation for the electric field is
$$\vec E = {E_0}\cos \left( {\omega t - \frac{{2\pi }}{\lambda }y} \right)\hat z$$
27.
The decreasing order of wavelength of infrared, microwave, ultraviolet and gamma rays is
The frequency of the charged particles oscillates about its mean equilibrium position
$$ = {10^9}Hz.$$
Vibrating particle produces electric and magnetic field.
So, frequency of electromagnetic waves produced by the charged particle is $$v = {10^9}Hz.$$
Wavelength $$\lambda = \frac{c}{v} = \frac{{3 \times {{10}^8}}}{{{{10}^9}}} = 0.3\,m$$
Since the range of radiowaves is between $$10\,Hz$$ to $${10^12}Hz$$ and hence $${10^9}Hz$$ lies in region of radio waves.
29.
A plane electromagnetic wave travels in free space along $$X$$-direction. If the value of $${\vec B}$$ (in $$tesla$$ ) at a particular point in space and time is $$1.2 \times {10^{ - 8}}\hat k.$$ The value of $${\vec E}$$ (in $$V{m^{ - 1}}$$ ) at that point is
Given : $$\vec B = 1.2 \times {10^{ - 8}}\hat kT$$
$$\vec E = ?$$
From formula,
$$E = Bc = \left( {1.2 \times {{10}^{ - 8}}T} \right)\left( {3 \times {{10}^8}m{s^{ - 1}}} \right) = 3.6\,V{m^{ - 1}}$$
$${\vec B}$$ is along $$Z$$-direction and the wave propagates along $$X$$-direction. Therefore $${\vec E}$$ should be along $$Y$$-direction.
Thus, $$\vec E = 3.6\,\hat j\,V{m^{ - 1}}$$
30.
A plane electromagnetic wave propagating in the $$X$$-direction has wavelength of $$6.0\,mm.$$ The electric field is in the $$Y$$-direction and its maximum magnitude is $$33\,V{m^{ - 1}}.$$ The equation for the electric field as a function of $$x$$ and $$t$$ is