NEET Test Series from KOTA - 10 Papers In MS WORD
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PHXI15:WAVES
358891
In an electromagnetic wave in free space the root mean square value of the electric field is \({E_{rms}} = 6\;V/m\). The peak value of the magnetic field is :-
358892
The ratio of the magnitudes of electric field to the magnetic field of an electromagnetic wave is of the order of
1 \({10^{ - 5}}\;m{s^{ - 1}}\)
2 \({10^8}\;m{s^{ - 1}}\)
3 \({10^{ - 8}}\;m{s^{ - 1}}\)
4 \({10^5}\;m{s^{ - 1}}\)
Explanation:
\(\frac{{{E_{peak{\rm{ }}}}}}{{{B_{peak{\rm{ }}}}}} = c = 3 \times {10^8}\;m/s\) (in free space). Correct option is (2).
KCET - 2023
PHXI15:WAVES
358893
A plane \(E M\) wave is propagating along \(x\) direction. It has a wavelength of \(4\,mm.\) If electric field is in \(y\) direction with the maximum magnitude of \(60\,V{m^{ - 1}},\) the equation for magnetic field is
Given that, \(\lambda = 4\;mm,{E_0} = 60\;V/m\) Amplitude magnetic field is given by \(B_{0}=\dfrac{E_{0}}{c}=\dfrac{60}{3 \times 10^{8}}=2 \times 10^{-7}\) Since wave is travelling in \(+x\) direction and electric field is along \(+y\) axis. So, \(B\) is along \(z\) - axis. General equation of \(\vec{B}\) is given by \(\vec{B}=B_{0} \sin [k(x-v t] \hat{k} T\) \(f=\dfrac{c}{\lambda}=\dfrac{3 \times 10^{8}}{4 \times 10^{-3}}=\dfrac{3}{4} \times 10^{11}\) \(\omega = 2\pi f = 2\pi \times \frac{3}{4} \times {10^{11}} = \frac{\pi }{2} \times {10^3}c\) and \(k=\dfrac{2 \pi}{\lambda}=\dfrac{2 \pi}{4 \times 10^{-3}}=\dfrac{\pi}{2} \times 10^{3}\) So, \(\overrightarrow{B_{z}}=2 \times 10^{-7} \sin \left(\dfrac{\pi}{2} \times 10^{3}\left(x-3 \times 10^{8} t\right)\right) \hat{k} T\)
JEE - 2024
PHXI15:WAVES
358894
The electric field of an electromagnetic wave in free space is represented as \(\vec{E}=E_{0} \cos (\omega t-k z) \hat{i}\). The corresponding magnetic induction vector will be
Given: \(\vec{E}=E_{0} \cos (\omega t-k z) \hat{i}\) From the equation of \(\vec{E}\), we can see the direction of \(EM\) wave is \(+Z\) - direction \(\dfrac{E_{0}}{B_{0}}=c \Rightarrow B_{0}=\dfrac{E_{0}}{c}\) \(\vec{E} \times \vec{B}=\hat{k} \Rightarrow \hat{i} \times \vec{B}=\hat{k} \Rightarrow \vec{B}\) is along \(+\hat{j}\) \(\therefore \quad \vec{B}=\dfrac{E_{0}}{c} \cos (\omega t-k z) \hat{j}\)
358891
In an electromagnetic wave in free space the root mean square value of the electric field is \({E_{rms}} = 6\;V/m\). The peak value of the magnetic field is :-
358892
The ratio of the magnitudes of electric field to the magnetic field of an electromagnetic wave is of the order of
1 \({10^{ - 5}}\;m{s^{ - 1}}\)
2 \({10^8}\;m{s^{ - 1}}\)
3 \({10^{ - 8}}\;m{s^{ - 1}}\)
4 \({10^5}\;m{s^{ - 1}}\)
Explanation:
\(\frac{{{E_{peak{\rm{ }}}}}}{{{B_{peak{\rm{ }}}}}} = c = 3 \times {10^8}\;m/s\) (in free space). Correct option is (2).
KCET - 2023
PHXI15:WAVES
358893
A plane \(E M\) wave is propagating along \(x\) direction. It has a wavelength of \(4\,mm.\) If electric field is in \(y\) direction with the maximum magnitude of \(60\,V{m^{ - 1}},\) the equation for magnetic field is
Given that, \(\lambda = 4\;mm,{E_0} = 60\;V/m\) Amplitude magnetic field is given by \(B_{0}=\dfrac{E_{0}}{c}=\dfrac{60}{3 \times 10^{8}}=2 \times 10^{-7}\) Since wave is travelling in \(+x\) direction and electric field is along \(+y\) axis. So, \(B\) is along \(z\) - axis. General equation of \(\vec{B}\) is given by \(\vec{B}=B_{0} \sin [k(x-v t] \hat{k} T\) \(f=\dfrac{c}{\lambda}=\dfrac{3 \times 10^{8}}{4 \times 10^{-3}}=\dfrac{3}{4} \times 10^{11}\) \(\omega = 2\pi f = 2\pi \times \frac{3}{4} \times {10^{11}} = \frac{\pi }{2} \times {10^3}c\) and \(k=\dfrac{2 \pi}{\lambda}=\dfrac{2 \pi}{4 \times 10^{-3}}=\dfrac{\pi}{2} \times 10^{3}\) So, \(\overrightarrow{B_{z}}=2 \times 10^{-7} \sin \left(\dfrac{\pi}{2} \times 10^{3}\left(x-3 \times 10^{8} t\right)\right) \hat{k} T\)
JEE - 2024
PHXI15:WAVES
358894
The electric field of an electromagnetic wave in free space is represented as \(\vec{E}=E_{0} \cos (\omega t-k z) \hat{i}\). The corresponding magnetic induction vector will be
Given: \(\vec{E}=E_{0} \cos (\omega t-k z) \hat{i}\) From the equation of \(\vec{E}\), we can see the direction of \(EM\) wave is \(+Z\) - direction \(\dfrac{E_{0}}{B_{0}}=c \Rightarrow B_{0}=\dfrac{E_{0}}{c}\) \(\vec{E} \times \vec{B}=\hat{k} \Rightarrow \hat{i} \times \vec{B}=\hat{k} \Rightarrow \vec{B}\) is along \(+\hat{j}\) \(\therefore \quad \vec{B}=\dfrac{E_{0}}{c} \cos (\omega t-k z) \hat{j}\)
358891
In an electromagnetic wave in free space the root mean square value of the electric field is \({E_{rms}} = 6\;V/m\). The peak value of the magnetic field is :-
358892
The ratio of the magnitudes of electric field to the magnetic field of an electromagnetic wave is of the order of
1 \({10^{ - 5}}\;m{s^{ - 1}}\)
2 \({10^8}\;m{s^{ - 1}}\)
3 \({10^{ - 8}}\;m{s^{ - 1}}\)
4 \({10^5}\;m{s^{ - 1}}\)
Explanation:
\(\frac{{{E_{peak{\rm{ }}}}}}{{{B_{peak{\rm{ }}}}}} = c = 3 \times {10^8}\;m/s\) (in free space). Correct option is (2).
KCET - 2023
PHXI15:WAVES
358893
A plane \(E M\) wave is propagating along \(x\) direction. It has a wavelength of \(4\,mm.\) If electric field is in \(y\) direction with the maximum magnitude of \(60\,V{m^{ - 1}},\) the equation for magnetic field is
Given that, \(\lambda = 4\;mm,{E_0} = 60\;V/m\) Amplitude magnetic field is given by \(B_{0}=\dfrac{E_{0}}{c}=\dfrac{60}{3 \times 10^{8}}=2 \times 10^{-7}\) Since wave is travelling in \(+x\) direction and electric field is along \(+y\) axis. So, \(B\) is along \(z\) - axis. General equation of \(\vec{B}\) is given by \(\vec{B}=B_{0} \sin [k(x-v t] \hat{k} T\) \(f=\dfrac{c}{\lambda}=\dfrac{3 \times 10^{8}}{4 \times 10^{-3}}=\dfrac{3}{4} \times 10^{11}\) \(\omega = 2\pi f = 2\pi \times \frac{3}{4} \times {10^{11}} = \frac{\pi }{2} \times {10^3}c\) and \(k=\dfrac{2 \pi}{\lambda}=\dfrac{2 \pi}{4 \times 10^{-3}}=\dfrac{\pi}{2} \times 10^{3}\) So, \(\overrightarrow{B_{z}}=2 \times 10^{-7} \sin \left(\dfrac{\pi}{2} \times 10^{3}\left(x-3 \times 10^{8} t\right)\right) \hat{k} T\)
JEE - 2024
PHXI15:WAVES
358894
The electric field of an electromagnetic wave in free space is represented as \(\vec{E}=E_{0} \cos (\omega t-k z) \hat{i}\). The corresponding magnetic induction vector will be
Given: \(\vec{E}=E_{0} \cos (\omega t-k z) \hat{i}\) From the equation of \(\vec{E}\), we can see the direction of \(EM\) wave is \(+Z\) - direction \(\dfrac{E_{0}}{B_{0}}=c \Rightarrow B_{0}=\dfrac{E_{0}}{c}\) \(\vec{E} \times \vec{B}=\hat{k} \Rightarrow \hat{i} \times \vec{B}=\hat{k} \Rightarrow \vec{B}\) is along \(+\hat{j}\) \(\therefore \quad \vec{B}=\dfrac{E_{0}}{c} \cos (\omega t-k z) \hat{j}\)
358891
In an electromagnetic wave in free space the root mean square value of the electric field is \({E_{rms}} = 6\;V/m\). The peak value of the magnetic field is :-
358892
The ratio of the magnitudes of electric field to the magnetic field of an electromagnetic wave is of the order of
1 \({10^{ - 5}}\;m{s^{ - 1}}\)
2 \({10^8}\;m{s^{ - 1}}\)
3 \({10^{ - 8}}\;m{s^{ - 1}}\)
4 \({10^5}\;m{s^{ - 1}}\)
Explanation:
\(\frac{{{E_{peak{\rm{ }}}}}}{{{B_{peak{\rm{ }}}}}} = c = 3 \times {10^8}\;m/s\) (in free space). Correct option is (2).
KCET - 2023
PHXI15:WAVES
358893
A plane \(E M\) wave is propagating along \(x\) direction. It has a wavelength of \(4\,mm.\) If electric field is in \(y\) direction with the maximum magnitude of \(60\,V{m^{ - 1}},\) the equation for magnetic field is
Given that, \(\lambda = 4\;mm,{E_0} = 60\;V/m\) Amplitude magnetic field is given by \(B_{0}=\dfrac{E_{0}}{c}=\dfrac{60}{3 \times 10^{8}}=2 \times 10^{-7}\) Since wave is travelling in \(+x\) direction and electric field is along \(+y\) axis. So, \(B\) is along \(z\) - axis. General equation of \(\vec{B}\) is given by \(\vec{B}=B_{0} \sin [k(x-v t] \hat{k} T\) \(f=\dfrac{c}{\lambda}=\dfrac{3 \times 10^{8}}{4 \times 10^{-3}}=\dfrac{3}{4} \times 10^{11}\) \(\omega = 2\pi f = 2\pi \times \frac{3}{4} \times {10^{11}} = \frac{\pi }{2} \times {10^3}c\) and \(k=\dfrac{2 \pi}{\lambda}=\dfrac{2 \pi}{4 \times 10^{-3}}=\dfrac{\pi}{2} \times 10^{3}\) So, \(\overrightarrow{B_{z}}=2 \times 10^{-7} \sin \left(\dfrac{\pi}{2} \times 10^{3}\left(x-3 \times 10^{8} t\right)\right) \hat{k} T\)
JEE - 2024
PHXI15:WAVES
358894
The electric field of an electromagnetic wave in free space is represented as \(\vec{E}=E_{0} \cos (\omega t-k z) \hat{i}\). The corresponding magnetic induction vector will be
Given: \(\vec{E}=E_{0} \cos (\omega t-k z) \hat{i}\) From the equation of \(\vec{E}\), we can see the direction of \(EM\) wave is \(+Z\) - direction \(\dfrac{E_{0}}{B_{0}}=c \Rightarrow B_{0}=\dfrac{E_{0}}{c}\) \(\vec{E} \times \vec{B}=\hat{k} \Rightarrow \hat{i} \times \vec{B}=\hat{k} \Rightarrow \vec{B}\) is along \(+\hat{j}\) \(\therefore \quad \vec{B}=\dfrac{E_{0}}{c} \cos (\omega t-k z) \hat{j}\)
