8.1 Figure 8.5 shows a capacitor made of two circular plates each of radius 12 cm, and separated by 5.0 cm. The capacitor is being charged by an external source (not shown in the figure). The charging current is constant and equal to 0.15A.
(a) Calculate the capacitance and the rate of change of potential difference between the plates.
(b) Obtain the displacement current across the plates.
c) Is Kirchhoff's first rule (junction rule) valid at each plate of the capacitor? Explain
Solution :
(a)Using
$$C = \frac{\epsilon_{0} A}{d}$$
$$C = \frac{8.854 \times 10^{-12} \times \pi r^{2}}{d}$$
$$C = \frac{8.854 \times 10^{-12} \times 3.14 \times (12\times 10^{-2})^{2}}{5\times 10^{-2}}$$
$$C = 800.7 \times 10^{-14}$$
$$C = 8 pF$$
Now using
$$V = \frac{Q}{C}$$
$$\frac{dV}{dt} = \frac{1}{C}\frac{dQ}{dt}$$
$$or \frac{dV}{dt} = \frac{I}{C} = \frac{0.15}{8 \times 10^{-12}}$$
$$\frac{dV}{dt}= 1.875 \times 10^{10}VS^{-1}$$
(b) Displacement Current
$$I_{d} = \epsilon_{0}\frac{d\phi_{E}}{dt}$$
$$I_{d}=\epsilon_{0}\frac{d(EA)}{dt}$$
$$I_{d} = \epsilon_{0}A\frac{dE}{dt}$$
$$\because E = \frac{V}{d}$$
$$I_{d}= \epsilon_{0}\frac{A}{d}\frac{dV}{dt}$$
$$(\because C = \frac{\epsilon_{0}A}{d})$$
$$= C\frac{dV}{dt}$$
$$= 8 \times 10^{-12} \times 1.875 \times 10^{10}$$
$$I_{d}= 0.15 A$$
(c) Yes, Because conduction current entering one plate is equal to the displacement current leaving that plate.
8.2 A parallel plate capacitor (shown in figure) made of circular plates each of radius R = 6.0 cm has a capacitance C = 100 pF. The capacitor is connected to a 230 V ac supply with a (angular) frequency of 300 rad/s.
(a) What is the rms value of the conduction current?
(b) Is the conduction current equal to the displacement current?
(c) Determine the amplitude of B at a point 3.0 cm from the axis between the plates.
Solution :
Here $$V_{rms} = 230V$$
$$C = 100pF = 100 \times 10^{-12}$$ $$C=10^{-10}F$$
$$\omega = 300 rad s^{-1}$$
(a) Using
$$I_{rms} = \frac{V_{rms}}{X_{C}}$$
$$I_{rms} = \frac{V_{rms}}{\frac{1}{C\omega}} = V_{rms} \times C\omega$$
$$= 230 \times 10^{-10} \times 300$$
$$= 69 \times 10^{-7}$$
$$I_{rms} = 6.9 \times 10^{-6}$$
$$I_{rms}=6.9 \mu A$$
(b) Yes
(c) consider the figure given below :
$$\therefore B_{0} = \frac{\mu_{0}r I_{0}}{2 \pi R^{2}}$$
$$\because I_{0} = \sqrt{2}I$$
$$B_{0} = \frac{\mu_{0}r\sqrt{2}I}{2 \pi R^{2}}$$
$$B_{0} = \frac{(4 \pi \times 10^{-7})(3 \times 10^{-2})1.414 (6.9 \times 10^{-6})}{2 \pi \times (6 \times 10^{-2})^{2}}$$
$$B_{0} = 1.626 \times 10^{-11}T$$
8.3 What physical quantity is the same for X-rays of wavelength $10^{-10}$ m, red light of wavelength 6800 Å and radiowaves of wavelength 500m?
Solution :
All e.m. Waves travel with same speed $c=3 \times 10^{8}$ m/s in vacuum. Therefore, x-rays, red light and radiowaves have same speed.
8.4 A plane electromagnetic wave travels in vacuum along z-direction. What can you say about the directions of its electric and magnetic. field vectors? If the frequency of the wave is 30 MHz, what is its wavelength?
Solution :
The electric field vector E and magnetic field vector B are in xy plane. They are normal to each other.
Using $$c =\lambda \times \upsilon$$
We get $$\lambda = \frac{c}{\upsilon }$$
i.e $$\lambda = \frac{3 \times 10^{8}}{30 \times 10^{6}}$$
$$\lambda =10 m$$
8.5 A radio can tune in to any station in the 7.5 MHz to 12 MHz band. What is the corresponding wavelength band?
Solution :
Using
For $$\nu =7.5 MHz = 7.5 \times 10^{6}Hz$$
$$\lambda = \frac{3 \times 10^{8}}{7.5 \times 10^{6}} = 40 m$$
For $$\nu = 12 MHz = 12 \times 10^{6}Hz$$
$$\lambda = \frac{3 \times 10^{8}}{12 \times 10^{6}} = 25m$$
The corresponding wavelength band is 40m to 25m
8.6 A charged particle oscillates about its mean equilibrium position with a frequency of $10^{9}Hz$. What is the frequency of the electromagnetic waves produced by the oscillator?
Solution :
Frequency of the produced E. M. Wave is same as the frequency of oscillating charged particles i. e $10^{9}Hz$
8.7 The amplitude of the magnetic field part of a harmonic electromagnetic wave in vacuum is $B_{0} =510 nT$. What is the amplitude of the electric field part of the wave?
Solution :
Using
$$\frac{E_{0}}{B_{0}}= c$$
We get
$$E_{0} = B_{0}c$$
$$E = (510 \times 10^{-9})(3 \times 10^{8})$$
$$E = 1530 \times 10^{-1}$$
$$E = 153 NC^{-1}$$
8.8 Suppose that the electric field amplitude of an electromagnetic wave Is E = 120 N/C and that its frequency is $\nu$ = 50.0 MHz. (a) Determine $B_{0}$, $\omega$ ,k and $\lambda$ (b) Find expressions for E and B.
Solution :
(a)(i) Using $$\frac{E_{0}}{B_{0}} = c$$
we get
$$B_{0} = \frac{E}{c} = \frac{120}{3 \times 10^{8}}$$
$$B_{0} = 4 \times 10^{-7} T$$
i.e. $$B_{0} = 400nT$$
(ii)
$$\omega = 2 \pi\nu$$
$$\omega = 2 \times \pi \times 50 \times 10^{6}$$
$$= 3.14 \times 10^{8} rad s^{-1}$$
(iii) $$c = \lambda \nu$$
i.e.
$$\lambda = \frac{c}{\nu} = \frac{3 \times 10^{8}}{ 50 \times 10^{6}} = 6m$$
(iv)
$$k = \frac{w}{c} = \frac{3.14 \times 10^{8}}{3 \times 10^{8}}$$
$$k = 1.05 rad m^{-1}$$
(b)
$$\overrightarrow{E} = E_{0}sin(kx-wt)\widehat{j}$$
$$\overrightarrow{E} = 120sin(1.05x-3.14 \times 10^{8} \times t)\widehat{J},$$
$$\overrightarrow{B} = B_{0} sin(kx-wt)\widehat{k}$$
$$\overrightarrow{B}= 400 \times 10^{-7} sin(1.05x-3.14\times 10^{8} \times t)\widehat{k}$$
8.9 The terminology of different parts of the electromagnetic spectrum is given in the text. Use the formula $E = h\nu$ (for energy of a quantum of radiation: photon) and obtain the photon energy in units of eV for different parts of the electromagnetic spectrum. In what way are the different scales of photon energies that you obtain related to the sources of electromagnetic radiation?
Solution :
Using the relation for photon energy,
$$E = h\nu$$
We get
$$E = \frac{hc}{\lambda }$$
For $\lambda = 1 m$, we get
$$E= \frac{(6.63 \times 10^{-34}) (3 \times 10^{8}) }{1 \times 1.6 \times 10^{-19}}$$
$$E= 1.24 \times 10^{-6}eV$$
Energy for other wavelengths can be worked out from the relation
$$E_{\lambda } = \frac{1.24 \times 10^{-6}}{ \lambda }eV$$
(a) For
$$\gamma-rays, let \lambda = 10^{-12}m$$
,then
$$E = \frac{1.24 \times 10^{-6}}{ 10^{-12} }eV$$
$$E=1.24MeV$$
(b) For X-rays, let $\lambda = 1nm = 10^{-9}m$,then
$$E = \frac{1.24 \times 10^{-6}}{ 10^{-9} }eV$$
$$E=1240eV$$
(c) For visible light, let $\lambda = 1\mu m = 10^{-6}m$,then
$$E = \frac{1.24 \times 10^{-6}}{ 10^{-6} }eV$$
$$E=1.24 eV$$
(d) For visible light, let $\lambda = 1cm = 10^{-2}m$,then
$$E = \frac{1.24 \times 10^{-6}}{ 10^{-2} }eV$$
$$E=1.24 \times 10^{-4}eV$$
(e) For radio waves, For visible light, let $\lambda = 1km =1000m$ ,then
$$E = \frac{1.24 \times 10^{-6}}{ 1000} eV$$
$$E=1.24 \times 10^{-9}eV$$
Conclusion : The above result indicates that the different wavelengths in the electromagnetic spectrum can be obtained by multiplying roughly the powers of ten. The visible wavelengths are spaced by a few eV.
The Energy of a photon that a source produces indicates the spacing of relevant energy levels of the source
8.10 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. (a) What is the wavelength of the wave? (b) What is the amplitude of the oscillating magnetic field? (c) Show that the average energy density of the E field equals the average energy density of the B field. [$c=3 \times 10^{8} ms^{-1}$]
Solution :
(a) using $c=\lambda \nu$, we get
$$\lambda = \frac{c}{\nu } = \frac{3 \times 10^{8}}{2 \times 10^{10}}$$
$$\lambda = 1.5 \times 10^{-2}m$$
(b) using
$$c =\frac{E_{0}}{B_{0}}$$
we get
$$B_{0} = \frac{E_{0}}{c} = \frac{48}{3 \times 10^{8}}$$
$$= 1.6 \times 10^{-7}T$$
(c) Average energy density of the electric field, $U_{E} = \frac{1}{4} \varepsilon_{0}E_{0}^{2}$
and average energy density of the magnetic field,
$$U_{B} = \frac{B_{0}^2}{4\mu_{0} }$$
Also
$$c= \frac{E_{0}}{B_{0}}$$
$$c=\frac{1}{\sqrt{\mu_{0}\varepsilon_{0}}}$$
$$\frac{E_{0}}{B_{0}} =\frac{1} {\sqrt{\mu_{0}\varepsilon_{0}}}$$
Then
$$\frac{U_{E} }{U_{B}} = \frac{\frac{1}{4} \varepsilon_{0}E_{0}^{2}}{\frac{B_{0}^2}{4\mu_{0} }}$$
$$\frac{U_{E} }{U_{B}} = \frac{\mu_{0}\varepsilon_{0}E_{0}^{2} }{B_{0}^{2}}$$
$$\frac{U_{E} }{U_{B}}=\frac{\mu_{0}\varepsilon_{0}}{\mu_{0}\varepsilon_{0}} =1$$
Hence
$$U_{E} = U_{B} $$