358818
The ratio of average electric energy density and total average energy density of electromagnetic wave is
1 2
2 3
3 \(\dfrac{1}{2}\)
4 1
Explanation:
Average electric energy density = average magnetic energy density \(=\dfrac{1}{2}\) total average energy density So, the ratio of, \(\dfrac{\text { average electric energy density }}{\text { total average energy density }}=\dfrac{1}{2}\)
JEE - 2023
PHXI15:WAVES
358819
A point source of \(100\;W\) emits light with \(5 \%\) efficiency. At a distance of \(5\;m\) from the source, the intensity produced by the electric field component is
1 \(\dfrac{1}{40 \pi} \dfrac{W}{m^{2}}\)
2 \(\dfrac{1}{20 \pi} \dfrac{W}{m^{2}}\)
3 \(\dfrac{1}{2 \pi} \dfrac{W}{m^{2}}\)
4 \(\dfrac{1}{10 \pi} \dfrac{W}{m^{2}}\)
Explanation:
Effective power emitted by the source \(P_{e}=100 \times \dfrac{5}{100}\) Watt \(=5\) Watt. Intensity at a distance of \(5 {~m}\) form the source is given by \(I = \frac{{{P_e}}}{{4\pi {r^2}}} = \frac{5}{{4\pi {{(5)}^2}}}W/{m^2} = \frac{1}{{20\pi }}.\) \(50 \%\) of this energy is carried by electric field and \(50 \%\) is carried by magnetic field. \(\therefore\) Intensity produced by electric field component is given by \({I_E} = \frac{1}{2},{I_e} = \frac{1}{{40\pi }}W/{m^2}.\)
JEE - 2023
PHXI15:WAVES
358820
The energy density associated with electric field \(\vec{E}\) and magnetic field \(\vec{B}\) of an electromagnetic wave in free space is given by \(\left(\varepsilon_{0}\right.\) - permittivity of free space, \(\mu_{0}\) - permeability of free space)
In free space, the energy density of a static field \(E\) is \({u_E} = \frac{1}{2}{\varepsilon _0}{E^2},\) and the energy density of magnetic field is \({u_B} = \frac{1}{{2{\mu _0}}}{B^2}\)
JEE - 2023
PHXI15:WAVES
358821
Predict the correct relation between average energy density in electric field \({\mu _E}\) magnetic field \(u_{B}\)
1 \(u_{E}=u_{B}\)
2 \(u_{E}=\dfrac{1}{2} u_{B}\)
3 \(\dfrac{1}{2} u_{E}=u_{B}\)
4 \(u_{E}=4 u_{B}\)
Explanation:
Average energy density of \(E\) \(u_{E}=\dfrac{1}{4} \varepsilon_{0} E_{0}^{2}\) Average energy density of \(B\) \(\begin{gathered}u_{B}=\dfrac{1}{4} \dfrac{B_{0}^{2}}{\mu_{0}} \\\because \mathrm{c}=\dfrac{\mathrm{E}_{0}}{\mathrm{~B}_{0}} \mathrm{So}, \\u_{E}=\dfrac{1}{4} \varepsilon_{0}\left(c B_{0}\right)^{2}=\dfrac{1}{4} \varepsilon_{0} \dfrac{1}{\left(\mu_{0} \varepsilon_{0}\right)} \times B_{0}^{2}=\dfrac{1}{4} \dfrac{\mathrm{B}_{0}^{2}}{\mu_{0}} \\u_{E}=u_{B}\end{gathered}\)
358818
The ratio of average electric energy density and total average energy density of electromagnetic wave is
1 2
2 3
3 \(\dfrac{1}{2}\)
4 1
Explanation:
Average electric energy density = average magnetic energy density \(=\dfrac{1}{2}\) total average energy density So, the ratio of, \(\dfrac{\text { average electric energy density }}{\text { total average energy density }}=\dfrac{1}{2}\)
JEE - 2023
PHXI15:WAVES
358819
A point source of \(100\;W\) emits light with \(5 \%\) efficiency. At a distance of \(5\;m\) from the source, the intensity produced by the electric field component is
1 \(\dfrac{1}{40 \pi} \dfrac{W}{m^{2}}\)
2 \(\dfrac{1}{20 \pi} \dfrac{W}{m^{2}}\)
3 \(\dfrac{1}{2 \pi} \dfrac{W}{m^{2}}\)
4 \(\dfrac{1}{10 \pi} \dfrac{W}{m^{2}}\)
Explanation:
Effective power emitted by the source \(P_{e}=100 \times \dfrac{5}{100}\) Watt \(=5\) Watt. Intensity at a distance of \(5 {~m}\) form the source is given by \(I = \frac{{{P_e}}}{{4\pi {r^2}}} = \frac{5}{{4\pi {{(5)}^2}}}W/{m^2} = \frac{1}{{20\pi }}.\) \(50 \%\) of this energy is carried by electric field and \(50 \%\) is carried by magnetic field. \(\therefore\) Intensity produced by electric field component is given by \({I_E} = \frac{1}{2},{I_e} = \frac{1}{{40\pi }}W/{m^2}.\)
JEE - 2023
PHXI15:WAVES
358820
The energy density associated with electric field \(\vec{E}\) and magnetic field \(\vec{B}\) of an electromagnetic wave in free space is given by \(\left(\varepsilon_{0}\right.\) - permittivity of free space, \(\mu_{0}\) - permeability of free space)
In free space, the energy density of a static field \(E\) is \({u_E} = \frac{1}{2}{\varepsilon _0}{E^2},\) and the energy density of magnetic field is \({u_B} = \frac{1}{{2{\mu _0}}}{B^2}\)
JEE - 2023
PHXI15:WAVES
358821
Predict the correct relation between average energy density in electric field \({\mu _E}\) magnetic field \(u_{B}\)
1 \(u_{E}=u_{B}\)
2 \(u_{E}=\dfrac{1}{2} u_{B}\)
3 \(\dfrac{1}{2} u_{E}=u_{B}\)
4 \(u_{E}=4 u_{B}\)
Explanation:
Average energy density of \(E\) \(u_{E}=\dfrac{1}{4} \varepsilon_{0} E_{0}^{2}\) Average energy density of \(B\) \(\begin{gathered}u_{B}=\dfrac{1}{4} \dfrac{B_{0}^{2}}{\mu_{0}} \\\because \mathrm{c}=\dfrac{\mathrm{E}_{0}}{\mathrm{~B}_{0}} \mathrm{So}, \\u_{E}=\dfrac{1}{4} \varepsilon_{0}\left(c B_{0}\right)^{2}=\dfrac{1}{4} \varepsilon_{0} \dfrac{1}{\left(\mu_{0} \varepsilon_{0}\right)} \times B_{0}^{2}=\dfrac{1}{4} \dfrac{\mathrm{B}_{0}^{2}}{\mu_{0}} \\u_{E}=u_{B}\end{gathered}\)
358818
The ratio of average electric energy density and total average energy density of electromagnetic wave is
1 2
2 3
3 \(\dfrac{1}{2}\)
4 1
Explanation:
Average electric energy density = average magnetic energy density \(=\dfrac{1}{2}\) total average energy density So, the ratio of, \(\dfrac{\text { average electric energy density }}{\text { total average energy density }}=\dfrac{1}{2}\)
JEE - 2023
PHXI15:WAVES
358819
A point source of \(100\;W\) emits light with \(5 \%\) efficiency. At a distance of \(5\;m\) from the source, the intensity produced by the electric field component is
1 \(\dfrac{1}{40 \pi} \dfrac{W}{m^{2}}\)
2 \(\dfrac{1}{20 \pi} \dfrac{W}{m^{2}}\)
3 \(\dfrac{1}{2 \pi} \dfrac{W}{m^{2}}\)
4 \(\dfrac{1}{10 \pi} \dfrac{W}{m^{2}}\)
Explanation:
Effective power emitted by the source \(P_{e}=100 \times \dfrac{5}{100}\) Watt \(=5\) Watt. Intensity at a distance of \(5 {~m}\) form the source is given by \(I = \frac{{{P_e}}}{{4\pi {r^2}}} = \frac{5}{{4\pi {{(5)}^2}}}W/{m^2} = \frac{1}{{20\pi }}.\) \(50 \%\) of this energy is carried by electric field and \(50 \%\) is carried by magnetic field. \(\therefore\) Intensity produced by electric field component is given by \({I_E} = \frac{1}{2},{I_e} = \frac{1}{{40\pi }}W/{m^2}.\)
JEE - 2023
PHXI15:WAVES
358820
The energy density associated with electric field \(\vec{E}\) and magnetic field \(\vec{B}\) of an electromagnetic wave in free space is given by \(\left(\varepsilon_{0}\right.\) - permittivity of free space, \(\mu_{0}\) - permeability of free space)
In free space, the energy density of a static field \(E\) is \({u_E} = \frac{1}{2}{\varepsilon _0}{E^2},\) and the energy density of magnetic field is \({u_B} = \frac{1}{{2{\mu _0}}}{B^2}\)
JEE - 2023
PHXI15:WAVES
358821
Predict the correct relation between average energy density in electric field \({\mu _E}\) magnetic field \(u_{B}\)
1 \(u_{E}=u_{B}\)
2 \(u_{E}=\dfrac{1}{2} u_{B}\)
3 \(\dfrac{1}{2} u_{E}=u_{B}\)
4 \(u_{E}=4 u_{B}\)
Explanation:
Average energy density of \(E\) \(u_{E}=\dfrac{1}{4} \varepsilon_{0} E_{0}^{2}\) Average energy density of \(B\) \(\begin{gathered}u_{B}=\dfrac{1}{4} \dfrac{B_{0}^{2}}{\mu_{0}} \\\because \mathrm{c}=\dfrac{\mathrm{E}_{0}}{\mathrm{~B}_{0}} \mathrm{So}, \\u_{E}=\dfrac{1}{4} \varepsilon_{0}\left(c B_{0}\right)^{2}=\dfrac{1}{4} \varepsilon_{0} \dfrac{1}{\left(\mu_{0} \varepsilon_{0}\right)} \times B_{0}^{2}=\dfrac{1}{4} \dfrac{\mathrm{B}_{0}^{2}}{\mu_{0}} \\u_{E}=u_{B}\end{gathered}\)
358818
The ratio of average electric energy density and total average energy density of electromagnetic wave is
1 2
2 3
3 \(\dfrac{1}{2}\)
4 1
Explanation:
Average electric energy density = average magnetic energy density \(=\dfrac{1}{2}\) total average energy density So, the ratio of, \(\dfrac{\text { average electric energy density }}{\text { total average energy density }}=\dfrac{1}{2}\)
JEE - 2023
PHXI15:WAVES
358819
A point source of \(100\;W\) emits light with \(5 \%\) efficiency. At a distance of \(5\;m\) from the source, the intensity produced by the electric field component is
1 \(\dfrac{1}{40 \pi} \dfrac{W}{m^{2}}\)
2 \(\dfrac{1}{20 \pi} \dfrac{W}{m^{2}}\)
3 \(\dfrac{1}{2 \pi} \dfrac{W}{m^{2}}\)
4 \(\dfrac{1}{10 \pi} \dfrac{W}{m^{2}}\)
Explanation:
Effective power emitted by the source \(P_{e}=100 \times \dfrac{5}{100}\) Watt \(=5\) Watt. Intensity at a distance of \(5 {~m}\) form the source is given by \(I = \frac{{{P_e}}}{{4\pi {r^2}}} = \frac{5}{{4\pi {{(5)}^2}}}W/{m^2} = \frac{1}{{20\pi }}.\) \(50 \%\) of this energy is carried by electric field and \(50 \%\) is carried by magnetic field. \(\therefore\) Intensity produced by electric field component is given by \({I_E} = \frac{1}{2},{I_e} = \frac{1}{{40\pi }}W/{m^2}.\)
JEE - 2023
PHXI15:WAVES
358820
The energy density associated with electric field \(\vec{E}\) and magnetic field \(\vec{B}\) of an electromagnetic wave in free space is given by \(\left(\varepsilon_{0}\right.\) - permittivity of free space, \(\mu_{0}\) - permeability of free space)
In free space, the energy density of a static field \(E\) is \({u_E} = \frac{1}{2}{\varepsilon _0}{E^2},\) and the energy density of magnetic field is \({u_B} = \frac{1}{{2{\mu _0}}}{B^2}\)
JEE - 2023
PHXI15:WAVES
358821
Predict the correct relation between average energy density in electric field \({\mu _E}\) magnetic field \(u_{B}\)
1 \(u_{E}=u_{B}\)
2 \(u_{E}=\dfrac{1}{2} u_{B}\)
3 \(\dfrac{1}{2} u_{E}=u_{B}\)
4 \(u_{E}=4 u_{B}\)
Explanation:
Average energy density of \(E\) \(u_{E}=\dfrac{1}{4} \varepsilon_{0} E_{0}^{2}\) Average energy density of \(B\) \(\begin{gathered}u_{B}=\dfrac{1}{4} \dfrac{B_{0}^{2}}{\mu_{0}} \\\because \mathrm{c}=\dfrac{\mathrm{E}_{0}}{\mathrm{~B}_{0}} \mathrm{So}, \\u_{E}=\dfrac{1}{4} \varepsilon_{0}\left(c B_{0}\right)^{2}=\dfrac{1}{4} \varepsilon_{0} \dfrac{1}{\left(\mu_{0} \varepsilon_{0}\right)} \times B_{0}^{2}=\dfrac{1}{4} \dfrac{\mathrm{B}_{0}^{2}}{\mu_{0}} \\u_{E}=u_{B}\end{gathered}\)
