# Concept Of Physics Electromagnetic Waves

#### H C Verma

1.   Show that the dimensions of the displacement current $\epsilon_0\dfrac{d\psi_k}{dt}$ are that of an electric current.

$\dfrac{\epsilon_{0}\ d\phi_{E}}{dt} = \dfrac{\epsilon_{0}\ EA}{dt\ 4\pi\ \epsilon_{0}r^2}$

$=\dfrac{M^{-1}L^{-3}T^4 A^2}{M^{-1}L^{-3}A^2} \times \dfrac{A^1T^1}{L^2} \times \dfrac{L^2}{T} = A^1$

$= (Current) \ \ \ \ \ \ \ \ \ \ \ \ \ (proved).$

2.   A point charge is moving along a straight line with a constant velocity $v$. Consider a small area A perpendicular to the direction of motion of the charge (figure 40-E1). Calculate the displacement current through the area when its distance from the charge is $x$. The value of $x$ is not large so that the electric field at any instant is essentially given by Coulomb's law.

$E = \dfrac{Kq}{x^2}$, [from coulomb’s law] $\\$ $\phi_E = EA = \dfrac{KqE}{x^2}$ $\\$ $I_d = \epsilon_0\ \dfrac{d\phi E}{dt} = \epsilon_0\ \dfrac{d}{dt} \dfrac{KqA}{x^2} = \epsilon_0\ KqA = \dfrac{d}{dt}x^2$ $\\$ $= \epsilon_0 \times \dfrac{1}{4\pi\ \epsilon_0} \times q \times A \times -2 \times x^{-3} \times \dfrac{dx}{dt} = \dfrac{qAv}{2\pi x^3}.$

3.   A parallel-plate capacitor having plate-area $A$ and plate separation $d$ is joined to a battery of $emf \sum$ and internal resistance $R$ at $t = 0$. Consider a plane surface of area $A/2$, parallel to the plates and situated symmetrically between them. Find the displacement current through this surface as a function of time.

$E = \dfrac{Q}{\epsilon_0\ A}$ (Electric field) $\\$ $\phi = E.A. =\dfrac{Q}{\epsilon_0\ A} \dfrac{A}{2} = \dfrac{Q}{\epsilon_0\ 2}$ $\\$ $i_0 = \epsilon_0 \dfrac{d\phi\ E}{dt} = \epsilon_0 \dfrac{d}{dt}\Big(\dfrac{Q}{\epsilon_0\ 2}\Big) = \dfrac{1}{2}\Big(\dfrac{dQ}{dt}\Big)$ $\\$ $= \dfrac{1}{2}\dfrac{d}{dt}(EC_e^{-t/RC}) = \dfrac{1}{2}EC - \dfrac{1}{RC}e^{-t/RC} = \dfrac{-E}{2R}e^{\dfrac{-td}{RE_0\lambda}}$

4.   Consider the situation of the previous problem. Define displacement resistance $R_d - V/i_d$ of the space between the plates where $V$ is the potential difference between the plates and $i_d$ is the displacement current. Show that $R_d$ varies with time as $\\$ $$R_{d} = R(e^{1/T} - 1)$$

$E = \dfrac{Q}{\epsilon_0\ A}$ (Electric field) $\\$ $\phi = E.A. =\dfrac{Q}{\epsilon_0\ A} \dfrac{A}{2} = \dfrac{Q}{\epsilon_0\ 2}$ $\\$ $i_0 = \epsilon_0 \dfrac{d\phi\ E}{dt} = \epsilon_0 \dfrac{d}{dt}\Big(\dfrac{Q}{\epsilon_0\ 2}\Big) = \dfrac{1}{2}\Big(\dfrac{dQ}{dt}\Big)$ $\\$

5.   Using $B = \mu_{0}H$ find the ratio $E_{0}/H_{0}$ for a plane electromagnetic wave propagating through vacuum. Show that it has the dimensions of electric resistance. This ratio is a universal constant called $the\ impedance\ of\ free\ sapce$

$B = \mu_0 H$ $\\$ $\Rightarrow H = \dfrac{B}{\mu_0}$ $\\$ $\dfrac{E_0}{H_0} = \dfrac{B_0/ (\mu_0\ \epsilon _0\ C)}{B_0/ \mu_0} = \dfrac{1}{\epsilon_0\ C}$ $\\$ $= \dfrac{1}{8.85 \times 10^{-12} \times 3 \times 10^{8}} = 376.6 \Omega = 377 \Omega$ $\\$ Dimension $\dfrac{1}{\epsilon_0\ C } = \dfrac{1}{M^{-1}L^{-3}T^4A^2} = \dfrac{1}{M^1L^2T^3A^2 } = M^1L^2T^{-3}A^{-2} = [R]$

6.   The sunlight reaching the earth has maximum electric field of $810\ V/m$. What is the maximum magnetic field in this light ?

$E_0 = 810\ V/m,\ B_0 = ?$ $\\$ We know, $B_0 = \mu_0\ \epsilon_0\ C\ E_0$ $\\$ Putting the values, $\\$ $B_0 = 4\pi \times 10^{–7} \times 8.85 \times 10^{–12} \times 3 \times 10^8 \times 810$ $\\$ $= 27010.9 \times 10^{–10} = 2.7 \times 10^{–6}\ T = 2.7 \mu\ T$

7.   The magnetic field in a plane electromagnetic wave is given by $\\$ $$B = (200 \mu{T}) sin[(4.0 \times 10^15 s^{-1}(t - x/c))].$$ $\\$ Find the maximum electric field and the average energy density corresponding to the electric field.

$B = (200 \mu\ T)\ Sin\ \Big[(4 \times 10^{15}\ 5^{–1})\ (t – \dfrac{x}{C})\Big]$ $\\$ a) $B_0 = 200\ \mu\ T$ $\\$ $E_0 = C \times B_0 = 200 \times 10^{–6} \times 3 \times 10^8 = 6 \times 10^4$

b) Average energy density $= \dfrac{1}{2\mu_0}B_{0}^{2} = \dfrac{(200 \times 10^{-6})^2}{2 \times 4\pi \times 10^{-7}} = \dfrac{4 \times 10^{-8}}{8\pi \times 10^{-7}} \dfrac{1}{20\pi} = 0.0159 = 0.016.$

8.   A laser beam has intensity $2\cdot5 \times 10^{14}\ W/m^2$ Find the amplitudes of electric and magnetic fields in the beam.

$I = 2.5 \times 10^{14}\ W/m^2$ $\\$ We know, $I = \dfrac{1}{2} \epsilon_0\ E_{0}^{2}C$ $\\$ $\Rightarrow E_{0}^{2} = \dfrac{2I}{\epsilon_0\ C} \ \ \ \ \ \ \ \ \ \ or \ \ \ \ \ \ \ E_0 = \sqrt{\dfrac{2I}{\epsilon_0\ C}}$

$E_0 = \sqrt{\dfrac{2 \times 2.5 \times 10^{14}}{8.85 \times 10^{-12} \times 3 \times 10^8}} = 0.4339 \times 10^9 = 4.33 \times 10^8\ N/c.$ $\\$ $B_0 = \mu_0\ \epsilon_0\ C\ E_0$ $\\$ $= 4 \times 3.14 \times 10^{–7} \times 8.854 \times 10^{–12} \times 3 \times 10^8 \times 4.33 \times 10^8 = 1.44\ T.$

9.   The intensity of the sunlight reaching the earth is $1380\ W/m^2$. Assume this light to be a plane, monochromatic wave. Find the amplitudes of electric and magnetic fields in this wave.

Intensity of wave $= \dfrac{1}{2} \epsilon_0\ E_{0}^{2}C$ $\\$ $\epsilon_0 = 8.85 \times 10^{–12} ;\ E_0 = ? ;\ C = 3 \times 10^8 ,\ I = 1380\ W/m^2$ $\\$ $1380 = \dfrac{1}{2} \times 8.85 times 10^{–12} \times E_{0}^{2} \times 3 \times 10^8$ $\\$ $\Rightarrow E_{0}^{2} = \dfrac{2 \times 1380}{8.85 \times 3 \times 10^{-4}} = 103.95 \times 10^4$ $\\$ $\Rightarrow E_{0} = 10.195 \times 10^2 = 1.02 \times 10^3$ $\\$ $E_{0} = B_0C$ $\\$ $\Rightarrow B_0 =\dfrac{E_0}{C} = \dfrac{1.02 \times 10^3}{3 \times 10^8} = 3.398 \times 10^{–5} = 3.4 \times 10^{–5}\ T.$