155513
What is the amplitude of the electric field in a parallel beam of light intensity $\left(\frac{15}{\pi}\right) \frac{\mathrm{W}}{\mathrm{m}^{2}}$ ? $\left[\text { Assume, } \frac{1}{4 \pi \varepsilon_{0}}=9 \times 10^{9} \frac{\mathrm{Nm}^{2}}{\mathrm{C}^{2}}\right]$
1 $60 \mathrm{~N} / \mathrm{C}$
2 $50 \mathrm{~N} / \mathrm{C}$
3 $40 \mathrm{~N} / \mathrm{C}$
4 $30 \mathrm{~N} / \mathrm{C}$
Explanation:
A Given, $\mathrm{c}=3 \times 10^{8} \mathrm{~m} / \mathrm{s}$ Assume, $\frac{1}{4 \pi \varepsilon_{0}}=9 \times 10^{9} \frac{\mathrm{Nm}^{2}}{\mathrm{C}^{2}}$ Intensity of light $(\mathrm{I})=\frac{15}{\pi} \mathrm{W} / \mathrm{m}^{2}$ Amplitude of electric field, $\mathrm{E}=\sqrt{\frac{2 \mathrm{I}}{\varepsilon_{0} \mathrm{c}}}$ Putting the value of $\mathrm{I}$ in equation (i), $\mathrm{E} =\sqrt{\frac{2 \times 15}{\varepsilon_{0} \pi \mathrm{c}}}=\sqrt{\frac{8 \times 15}{4 \pi \varepsilon_{0} \mathrm{c}}}$ $\mathrm{E} =\sqrt{\frac{8 \times 15 \times 9 \times 10^{9}}{3 \times 10^{8}}}$ $=\sqrt{8 \times 15 \times 3 \times 10}$ $\mathrm{E} =60 \mathrm{~N} / \mathrm{C}$
TS- EAMCET-10.09.2020
Electromagnetic Wave
155514
At an instant, a plane electromagnetic wave has its magnetic field in the direction of the vector $\hat{\mathbf{i}}-\hat{\mathbf{j}}$ and its electric field is in the direction of $\hat{\mathbf{i}}+\hat{\mathbf{j}}$. The wave is travelling along which direction?
1 $+x$-direction
2 - x-direction
3 + z-direction
4 - z-direction
Explanation:
D Direction of magnetic field, $\vec{B}=\hat{i}-\hat{j}$ Direction of electric field, $\vec{E}=\hat{i}+\hat{j}$ Direction of propagation of EM wave is given according to pointing vector as $\hat{n}=\vec{E} \times \vec{B} =(\hat{i}+\hat{j}) \times(\hat{i}-\hat{j})$ $=\hat{i} \times \hat{i}+\hat{j} \times \hat{i}-\hat{i} \times \hat{j}-\hat{j} \times \hat{j}$ $=0+(-\hat{k})-\hat{k}-0=-2 \hat{k}$ Hence, electromagnetic wave will travel along z-direction.
TS- EAMCET-11.09.2020
Electromagnetic Wave
155515
An electromagnetic wave of frequency $3.0 \mathrm{MHz}$ passes from vacuum into a non-magnetic medium with permittivity, $\varepsilon=16 \varepsilon_{0}$. Where, $\epsilon_{0}$ is the free space permittivity. The change in wavelength is
1 $-75 \mathrm{~m}$
2 $+75 \mathrm{~m}$
3 $-50 \mathrm{~m}$
4 $+50 \mathrm{~m}$
Explanation:
A Frequency EM wave $=3 \mathrm{M} \mathrm{Hz}=\mathrm{n}$ Velocity of light in a medium of permittivity $\mathrm{v}=\frac{\mathrm{c}}{\sqrt{\varepsilon}}$ Initial wave length $\left(\lambda_{1}\right)=\frac{\mathrm{v}}{\mathrm{n}}=\frac{\mathrm{c}}{\mathrm{n}}$ Wave length in medium $\left(\lambda_{2}\right)=\frac{\mathrm{c} / \sqrt{\varepsilon}}{\mathrm{n}}$. Change in wavelength, $\lambda_{2}-\lambda_{1}=\frac{\mathrm{c} / \sqrt{\varepsilon}}{\mathrm{n}}-\frac{\mathrm{c}}{\mathrm{n}}=\frac{\mathrm{c}}{\mathrm{n}}\left[\frac{1}{\sqrt{\varepsilon}}-1\right]$ $\Delta \lambda=\frac{3 \times 10^{8}}{3 \times 10^{6}}\left[\frac{1}{\sqrt{16 \varepsilon_{0}}}-1\right] \quad\left[\varepsilon_{0}=1\right]$ $=10^{2}\left[\frac{1}{4}-1\right] \quad=100 \times \frac{3}{4}$ $=10^{2}\left[\frac{-3}{4}\right]=-100$ $\Delta \lambda=-75 \mathrm{~m}$
TS- EAMCET-03.05.2019
Electromagnetic Wave
155516
The electric field of an electromagnetic wave in free space is given by $\vec{E}=5 \sin \left(\frac{2 \pi}{3} z-\omega t\right) \hat{y} V / m$. which of the following statements is correct?
1 The wave propagates along $\hat{y}$
2 The wave vector is given $\overrightarrow{\mathrm{K}}=\frac{2 \pi}{3} \hat{\mathrm{Z}}$
3 The wavelength of the electromagnetic wave is $\frac{1}{3} \mathrm{~m}$
4 The corresponding magnetic field is $\overrightarrow{\mathrm{B}}=\frac{5}{\mathrm{c}} \cos \left(\frac{2 \pi}{3} \mathrm{z}-\omega \mathrm{t}\right) \hat{\mathrm{x} T}$
5 The frequency of the wave is approximately $10^{6} \mathrm{~Hz}$
Explanation:
B Given, $E=5 \sin \left(\frac{2 \pi}{3} z-\omega t\right) \hat{y} V / m$ General $\mathrm{eq}^{\mathrm{n}}, \mathrm{E}=\mathrm{E}_{0} \sin (\mathrm{kz}-\omega \mathrm{t})$ (b) $\mathrm{k}=\frac{2 \pi}{3} \hat{\mathrm{z}}$ (c) $\quad \lambda=\frac{2 \pi}{\mathrm{k}}=\frac{2 \pi}{\frac{2 \pi}{3}}=3 \mathrm{~m}$ (a) Direction of propagation $=\hat{z}$ direction (e) frequency $\mathrm{f}=\frac{2 \pi}{\omega} \mathrm{Hz}$ (d) $\quad \mathrm{B}_{0}=\frac{\mathrm{E}_{0}}{\mathrm{c}}=\frac{5}{\mathrm{c}}$ $\mathrm{B}=\mathrm{B}_{0} \sin \left(\frac{2 \pi}{3} \mathrm{z}-\omega \mathrm{t}\right)$ $\mathrm{B}=\frac{5}{\mathrm{c}} \sin \left(\frac{2 \pi}{3} \mathrm{z}-\omega \mathrm{t}\right)(-\hat{\mathrm{x}})$
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Electromagnetic Wave
155513
What is the amplitude of the electric field in a parallel beam of light intensity $\left(\frac{15}{\pi}\right) \frac{\mathrm{W}}{\mathrm{m}^{2}}$ ? $\left[\text { Assume, } \frac{1}{4 \pi \varepsilon_{0}}=9 \times 10^{9} \frac{\mathrm{Nm}^{2}}{\mathrm{C}^{2}}\right]$
1 $60 \mathrm{~N} / \mathrm{C}$
2 $50 \mathrm{~N} / \mathrm{C}$
3 $40 \mathrm{~N} / \mathrm{C}$
4 $30 \mathrm{~N} / \mathrm{C}$
Explanation:
A Given, $\mathrm{c}=3 \times 10^{8} \mathrm{~m} / \mathrm{s}$ Assume, $\frac{1}{4 \pi \varepsilon_{0}}=9 \times 10^{9} \frac{\mathrm{Nm}^{2}}{\mathrm{C}^{2}}$ Intensity of light $(\mathrm{I})=\frac{15}{\pi} \mathrm{W} / \mathrm{m}^{2}$ Amplitude of electric field, $\mathrm{E}=\sqrt{\frac{2 \mathrm{I}}{\varepsilon_{0} \mathrm{c}}}$ Putting the value of $\mathrm{I}$ in equation (i), $\mathrm{E} =\sqrt{\frac{2 \times 15}{\varepsilon_{0} \pi \mathrm{c}}}=\sqrt{\frac{8 \times 15}{4 \pi \varepsilon_{0} \mathrm{c}}}$ $\mathrm{E} =\sqrt{\frac{8 \times 15 \times 9 \times 10^{9}}{3 \times 10^{8}}}$ $=\sqrt{8 \times 15 \times 3 \times 10}$ $\mathrm{E} =60 \mathrm{~N} / \mathrm{C}$
TS- EAMCET-10.09.2020
Electromagnetic Wave
155514
At an instant, a plane electromagnetic wave has its magnetic field in the direction of the vector $\hat{\mathbf{i}}-\hat{\mathbf{j}}$ and its electric field is in the direction of $\hat{\mathbf{i}}+\hat{\mathbf{j}}$. The wave is travelling along which direction?
1 $+x$-direction
2 - x-direction
3 + z-direction
4 - z-direction
Explanation:
D Direction of magnetic field, $\vec{B}=\hat{i}-\hat{j}$ Direction of electric field, $\vec{E}=\hat{i}+\hat{j}$ Direction of propagation of EM wave is given according to pointing vector as $\hat{n}=\vec{E} \times \vec{B} =(\hat{i}+\hat{j}) \times(\hat{i}-\hat{j})$ $=\hat{i} \times \hat{i}+\hat{j} \times \hat{i}-\hat{i} \times \hat{j}-\hat{j} \times \hat{j}$ $=0+(-\hat{k})-\hat{k}-0=-2 \hat{k}$ Hence, electromagnetic wave will travel along z-direction.
TS- EAMCET-11.09.2020
Electromagnetic Wave
155515
An electromagnetic wave of frequency $3.0 \mathrm{MHz}$ passes from vacuum into a non-magnetic medium with permittivity, $\varepsilon=16 \varepsilon_{0}$. Where, $\epsilon_{0}$ is the free space permittivity. The change in wavelength is
1 $-75 \mathrm{~m}$
2 $+75 \mathrm{~m}$
3 $-50 \mathrm{~m}$
4 $+50 \mathrm{~m}$
Explanation:
A Frequency EM wave $=3 \mathrm{M} \mathrm{Hz}=\mathrm{n}$ Velocity of light in a medium of permittivity $\mathrm{v}=\frac{\mathrm{c}}{\sqrt{\varepsilon}}$ Initial wave length $\left(\lambda_{1}\right)=\frac{\mathrm{v}}{\mathrm{n}}=\frac{\mathrm{c}}{\mathrm{n}}$ Wave length in medium $\left(\lambda_{2}\right)=\frac{\mathrm{c} / \sqrt{\varepsilon}}{\mathrm{n}}$. Change in wavelength, $\lambda_{2}-\lambda_{1}=\frac{\mathrm{c} / \sqrt{\varepsilon}}{\mathrm{n}}-\frac{\mathrm{c}}{\mathrm{n}}=\frac{\mathrm{c}}{\mathrm{n}}\left[\frac{1}{\sqrt{\varepsilon}}-1\right]$ $\Delta \lambda=\frac{3 \times 10^{8}}{3 \times 10^{6}}\left[\frac{1}{\sqrt{16 \varepsilon_{0}}}-1\right] \quad\left[\varepsilon_{0}=1\right]$ $=10^{2}\left[\frac{1}{4}-1\right] \quad=100 \times \frac{3}{4}$ $=10^{2}\left[\frac{-3}{4}\right]=-100$ $\Delta \lambda=-75 \mathrm{~m}$
TS- EAMCET-03.05.2019
Electromagnetic Wave
155516
The electric field of an electromagnetic wave in free space is given by $\vec{E}=5 \sin \left(\frac{2 \pi}{3} z-\omega t\right) \hat{y} V / m$. which of the following statements is correct?
1 The wave propagates along $\hat{y}$
2 The wave vector is given $\overrightarrow{\mathrm{K}}=\frac{2 \pi}{3} \hat{\mathrm{Z}}$
3 The wavelength of the electromagnetic wave is $\frac{1}{3} \mathrm{~m}$
4 The corresponding magnetic field is $\overrightarrow{\mathrm{B}}=\frac{5}{\mathrm{c}} \cos \left(\frac{2 \pi}{3} \mathrm{z}-\omega \mathrm{t}\right) \hat{\mathrm{x} T}$
5 The frequency of the wave is approximately $10^{6} \mathrm{~Hz}$
Explanation:
B Given, $E=5 \sin \left(\frac{2 \pi}{3} z-\omega t\right) \hat{y} V / m$ General $\mathrm{eq}^{\mathrm{n}}, \mathrm{E}=\mathrm{E}_{0} \sin (\mathrm{kz}-\omega \mathrm{t})$ (b) $\mathrm{k}=\frac{2 \pi}{3} \hat{\mathrm{z}}$ (c) $\quad \lambda=\frac{2 \pi}{\mathrm{k}}=\frac{2 \pi}{\frac{2 \pi}{3}}=3 \mathrm{~m}$ (a) Direction of propagation $=\hat{z}$ direction (e) frequency $\mathrm{f}=\frac{2 \pi}{\omega} \mathrm{Hz}$ (d) $\quad \mathrm{B}_{0}=\frac{\mathrm{E}_{0}}{\mathrm{c}}=\frac{5}{\mathrm{c}}$ $\mathrm{B}=\mathrm{B}_{0} \sin \left(\frac{2 \pi}{3} \mathrm{z}-\omega \mathrm{t}\right)$ $\mathrm{B}=\frac{5}{\mathrm{c}} \sin \left(\frac{2 \pi}{3} \mathrm{z}-\omega \mathrm{t}\right)(-\hat{\mathrm{x}})$
155513
What is the amplitude of the electric field in a parallel beam of light intensity $\left(\frac{15}{\pi}\right) \frac{\mathrm{W}}{\mathrm{m}^{2}}$ ? $\left[\text { Assume, } \frac{1}{4 \pi \varepsilon_{0}}=9 \times 10^{9} \frac{\mathrm{Nm}^{2}}{\mathrm{C}^{2}}\right]$
1 $60 \mathrm{~N} / \mathrm{C}$
2 $50 \mathrm{~N} / \mathrm{C}$
3 $40 \mathrm{~N} / \mathrm{C}$
4 $30 \mathrm{~N} / \mathrm{C}$
Explanation:
A Given, $\mathrm{c}=3 \times 10^{8} \mathrm{~m} / \mathrm{s}$ Assume, $\frac{1}{4 \pi \varepsilon_{0}}=9 \times 10^{9} \frac{\mathrm{Nm}^{2}}{\mathrm{C}^{2}}$ Intensity of light $(\mathrm{I})=\frac{15}{\pi} \mathrm{W} / \mathrm{m}^{2}$ Amplitude of electric field, $\mathrm{E}=\sqrt{\frac{2 \mathrm{I}}{\varepsilon_{0} \mathrm{c}}}$ Putting the value of $\mathrm{I}$ in equation (i), $\mathrm{E} =\sqrt{\frac{2 \times 15}{\varepsilon_{0} \pi \mathrm{c}}}=\sqrt{\frac{8 \times 15}{4 \pi \varepsilon_{0} \mathrm{c}}}$ $\mathrm{E} =\sqrt{\frac{8 \times 15 \times 9 \times 10^{9}}{3 \times 10^{8}}}$ $=\sqrt{8 \times 15 \times 3 \times 10}$ $\mathrm{E} =60 \mathrm{~N} / \mathrm{C}$
TS- EAMCET-10.09.2020
Electromagnetic Wave
155514
At an instant, a plane electromagnetic wave has its magnetic field in the direction of the vector $\hat{\mathbf{i}}-\hat{\mathbf{j}}$ and its electric field is in the direction of $\hat{\mathbf{i}}+\hat{\mathbf{j}}$. The wave is travelling along which direction?
1 $+x$-direction
2 - x-direction
3 + z-direction
4 - z-direction
Explanation:
D Direction of magnetic field, $\vec{B}=\hat{i}-\hat{j}$ Direction of electric field, $\vec{E}=\hat{i}+\hat{j}$ Direction of propagation of EM wave is given according to pointing vector as $\hat{n}=\vec{E} \times \vec{B} =(\hat{i}+\hat{j}) \times(\hat{i}-\hat{j})$ $=\hat{i} \times \hat{i}+\hat{j} \times \hat{i}-\hat{i} \times \hat{j}-\hat{j} \times \hat{j}$ $=0+(-\hat{k})-\hat{k}-0=-2 \hat{k}$ Hence, electromagnetic wave will travel along z-direction.
TS- EAMCET-11.09.2020
Electromagnetic Wave
155515
An electromagnetic wave of frequency $3.0 \mathrm{MHz}$ passes from vacuum into a non-magnetic medium with permittivity, $\varepsilon=16 \varepsilon_{0}$. Where, $\epsilon_{0}$ is the free space permittivity. The change in wavelength is
1 $-75 \mathrm{~m}$
2 $+75 \mathrm{~m}$
3 $-50 \mathrm{~m}$
4 $+50 \mathrm{~m}$
Explanation:
A Frequency EM wave $=3 \mathrm{M} \mathrm{Hz}=\mathrm{n}$ Velocity of light in a medium of permittivity $\mathrm{v}=\frac{\mathrm{c}}{\sqrt{\varepsilon}}$ Initial wave length $\left(\lambda_{1}\right)=\frac{\mathrm{v}}{\mathrm{n}}=\frac{\mathrm{c}}{\mathrm{n}}$ Wave length in medium $\left(\lambda_{2}\right)=\frac{\mathrm{c} / \sqrt{\varepsilon}}{\mathrm{n}}$. Change in wavelength, $\lambda_{2}-\lambda_{1}=\frac{\mathrm{c} / \sqrt{\varepsilon}}{\mathrm{n}}-\frac{\mathrm{c}}{\mathrm{n}}=\frac{\mathrm{c}}{\mathrm{n}}\left[\frac{1}{\sqrt{\varepsilon}}-1\right]$ $\Delta \lambda=\frac{3 \times 10^{8}}{3 \times 10^{6}}\left[\frac{1}{\sqrt{16 \varepsilon_{0}}}-1\right] \quad\left[\varepsilon_{0}=1\right]$ $=10^{2}\left[\frac{1}{4}-1\right] \quad=100 \times \frac{3}{4}$ $=10^{2}\left[\frac{-3}{4}\right]=-100$ $\Delta \lambda=-75 \mathrm{~m}$
TS- EAMCET-03.05.2019
Electromagnetic Wave
155516
The electric field of an electromagnetic wave in free space is given by $\vec{E}=5 \sin \left(\frac{2 \pi}{3} z-\omega t\right) \hat{y} V / m$. which of the following statements is correct?
1 The wave propagates along $\hat{y}$
2 The wave vector is given $\overrightarrow{\mathrm{K}}=\frac{2 \pi}{3} \hat{\mathrm{Z}}$
3 The wavelength of the electromagnetic wave is $\frac{1}{3} \mathrm{~m}$
4 The corresponding magnetic field is $\overrightarrow{\mathrm{B}}=\frac{5}{\mathrm{c}} \cos \left(\frac{2 \pi}{3} \mathrm{z}-\omega \mathrm{t}\right) \hat{\mathrm{x} T}$
5 The frequency of the wave is approximately $10^{6} \mathrm{~Hz}$
Explanation:
B Given, $E=5 \sin \left(\frac{2 \pi}{3} z-\omega t\right) \hat{y} V / m$ General $\mathrm{eq}^{\mathrm{n}}, \mathrm{E}=\mathrm{E}_{0} \sin (\mathrm{kz}-\omega \mathrm{t})$ (b) $\mathrm{k}=\frac{2 \pi}{3} \hat{\mathrm{z}}$ (c) $\quad \lambda=\frac{2 \pi}{\mathrm{k}}=\frac{2 \pi}{\frac{2 \pi}{3}}=3 \mathrm{~m}$ (a) Direction of propagation $=\hat{z}$ direction (e) frequency $\mathrm{f}=\frac{2 \pi}{\omega} \mathrm{Hz}$ (d) $\quad \mathrm{B}_{0}=\frac{\mathrm{E}_{0}}{\mathrm{c}}=\frac{5}{\mathrm{c}}$ $\mathrm{B}=\mathrm{B}_{0} \sin \left(\frac{2 \pi}{3} \mathrm{z}-\omega \mathrm{t}\right)$ $\mathrm{B}=\frac{5}{\mathrm{c}} \sin \left(\frac{2 \pi}{3} \mathrm{z}-\omega \mathrm{t}\right)(-\hat{\mathrm{x}})$
155513
What is the amplitude of the electric field in a parallel beam of light intensity $\left(\frac{15}{\pi}\right) \frac{\mathrm{W}}{\mathrm{m}^{2}}$ ? $\left[\text { Assume, } \frac{1}{4 \pi \varepsilon_{0}}=9 \times 10^{9} \frac{\mathrm{Nm}^{2}}{\mathrm{C}^{2}}\right]$
1 $60 \mathrm{~N} / \mathrm{C}$
2 $50 \mathrm{~N} / \mathrm{C}$
3 $40 \mathrm{~N} / \mathrm{C}$
4 $30 \mathrm{~N} / \mathrm{C}$
Explanation:
A Given, $\mathrm{c}=3 \times 10^{8} \mathrm{~m} / \mathrm{s}$ Assume, $\frac{1}{4 \pi \varepsilon_{0}}=9 \times 10^{9} \frac{\mathrm{Nm}^{2}}{\mathrm{C}^{2}}$ Intensity of light $(\mathrm{I})=\frac{15}{\pi} \mathrm{W} / \mathrm{m}^{2}$ Amplitude of electric field, $\mathrm{E}=\sqrt{\frac{2 \mathrm{I}}{\varepsilon_{0} \mathrm{c}}}$ Putting the value of $\mathrm{I}$ in equation (i), $\mathrm{E} =\sqrt{\frac{2 \times 15}{\varepsilon_{0} \pi \mathrm{c}}}=\sqrt{\frac{8 \times 15}{4 \pi \varepsilon_{0} \mathrm{c}}}$ $\mathrm{E} =\sqrt{\frac{8 \times 15 \times 9 \times 10^{9}}{3 \times 10^{8}}}$ $=\sqrt{8 \times 15 \times 3 \times 10}$ $\mathrm{E} =60 \mathrm{~N} / \mathrm{C}$
TS- EAMCET-10.09.2020
Electromagnetic Wave
155514
At an instant, a plane electromagnetic wave has its magnetic field in the direction of the vector $\hat{\mathbf{i}}-\hat{\mathbf{j}}$ and its electric field is in the direction of $\hat{\mathbf{i}}+\hat{\mathbf{j}}$. The wave is travelling along which direction?
1 $+x$-direction
2 - x-direction
3 + z-direction
4 - z-direction
Explanation:
D Direction of magnetic field, $\vec{B}=\hat{i}-\hat{j}$ Direction of electric field, $\vec{E}=\hat{i}+\hat{j}$ Direction of propagation of EM wave is given according to pointing vector as $\hat{n}=\vec{E} \times \vec{B} =(\hat{i}+\hat{j}) \times(\hat{i}-\hat{j})$ $=\hat{i} \times \hat{i}+\hat{j} \times \hat{i}-\hat{i} \times \hat{j}-\hat{j} \times \hat{j}$ $=0+(-\hat{k})-\hat{k}-0=-2 \hat{k}$ Hence, electromagnetic wave will travel along z-direction.
TS- EAMCET-11.09.2020
Electromagnetic Wave
155515
An electromagnetic wave of frequency $3.0 \mathrm{MHz}$ passes from vacuum into a non-magnetic medium with permittivity, $\varepsilon=16 \varepsilon_{0}$. Where, $\epsilon_{0}$ is the free space permittivity. The change in wavelength is
1 $-75 \mathrm{~m}$
2 $+75 \mathrm{~m}$
3 $-50 \mathrm{~m}$
4 $+50 \mathrm{~m}$
Explanation:
A Frequency EM wave $=3 \mathrm{M} \mathrm{Hz}=\mathrm{n}$ Velocity of light in a medium of permittivity $\mathrm{v}=\frac{\mathrm{c}}{\sqrt{\varepsilon}}$ Initial wave length $\left(\lambda_{1}\right)=\frac{\mathrm{v}}{\mathrm{n}}=\frac{\mathrm{c}}{\mathrm{n}}$ Wave length in medium $\left(\lambda_{2}\right)=\frac{\mathrm{c} / \sqrt{\varepsilon}}{\mathrm{n}}$. Change in wavelength, $\lambda_{2}-\lambda_{1}=\frac{\mathrm{c} / \sqrt{\varepsilon}}{\mathrm{n}}-\frac{\mathrm{c}}{\mathrm{n}}=\frac{\mathrm{c}}{\mathrm{n}}\left[\frac{1}{\sqrt{\varepsilon}}-1\right]$ $\Delta \lambda=\frac{3 \times 10^{8}}{3 \times 10^{6}}\left[\frac{1}{\sqrt{16 \varepsilon_{0}}}-1\right] \quad\left[\varepsilon_{0}=1\right]$ $=10^{2}\left[\frac{1}{4}-1\right] \quad=100 \times \frac{3}{4}$ $=10^{2}\left[\frac{-3}{4}\right]=-100$ $\Delta \lambda=-75 \mathrm{~m}$
TS- EAMCET-03.05.2019
Electromagnetic Wave
155516
The electric field of an electromagnetic wave in free space is given by $\vec{E}=5 \sin \left(\frac{2 \pi}{3} z-\omega t\right) \hat{y} V / m$. which of the following statements is correct?
1 The wave propagates along $\hat{y}$
2 The wave vector is given $\overrightarrow{\mathrm{K}}=\frac{2 \pi}{3} \hat{\mathrm{Z}}$
3 The wavelength of the electromagnetic wave is $\frac{1}{3} \mathrm{~m}$
4 The corresponding magnetic field is $\overrightarrow{\mathrm{B}}=\frac{5}{\mathrm{c}} \cos \left(\frac{2 \pi}{3} \mathrm{z}-\omega \mathrm{t}\right) \hat{\mathrm{x} T}$
5 The frequency of the wave is approximately $10^{6} \mathrm{~Hz}$
Explanation:
B Given, $E=5 \sin \left(\frac{2 \pi}{3} z-\omega t\right) \hat{y} V / m$ General $\mathrm{eq}^{\mathrm{n}}, \mathrm{E}=\mathrm{E}_{0} \sin (\mathrm{kz}-\omega \mathrm{t})$ (b) $\mathrm{k}=\frac{2 \pi}{3} \hat{\mathrm{z}}$ (c) $\quad \lambda=\frac{2 \pi}{\mathrm{k}}=\frac{2 \pi}{\frac{2 \pi}{3}}=3 \mathrm{~m}$ (a) Direction of propagation $=\hat{z}$ direction (e) frequency $\mathrm{f}=\frac{2 \pi}{\omega} \mathrm{Hz}$ (d) $\quad \mathrm{B}_{0}=\frac{\mathrm{E}_{0}}{\mathrm{c}}=\frac{5}{\mathrm{c}}$ $\mathrm{B}=\mathrm{B}_{0} \sin \left(\frac{2 \pi}{3} \mathrm{z}-\omega \mathrm{t}\right)$ $\mathrm{B}=\frac{5}{\mathrm{c}} \sin \left(\frac{2 \pi}{3} \mathrm{z}-\omega \mathrm{t}\right)(-\hat{\mathrm{x}})$