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PHXI15:WAVES

358818 The ratio of average electric energy density and total average energy density of electromagnetic wave is

1 2

2 3

3

\frac{1}{2}

4 1

Explanation:

Average electric energy density = average magnetic energy density $= \frac{1}{2}$ total average energy density
So, the ratio of,
$\frac{average electric energy density}{total average energy density} = \frac{1}{2}$

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

\frac{1}{40 π} \frac{W}{m^{2}}

2

\frac{1}{20 π} \frac{W}{m^{2}}

3

\frac{1}{2 π} \frac{W}{m^{2}}

4

\frac{1}{10 π} \frac{W}{m^{2}}

Explanation:

Effective power emitted by the source
$P_{e} = 100 \times \frac{5}{100}$ Watt $= 5$ Watt.
Intensity at a distance of $5 m$ form the source is given by
$I = \frac{P_{e}}{4 π r^{2}} = \frac{5}{4 π {(5)}^{2}} W / m^{2} = \frac{1}{20 π} .$
$50 %$ of this energy is carried by electric field and $50 %$ is carried by magnetic field.
$∴$ Intensity produced by electric field component is given by $I_{E} = \frac{1}{2}, I_{e} = \frac{1}{40 π} W / m^{2} .$

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 $(ε_{0}$ - permittivity of free space, $μ_{0}$ - permeability of free space)

1

u_{E} = \frac{E^{2}}{2 ε_{0}}, u_{B} = \frac{μ_{0} B^{2}}{2}

2

u_{E} = \frac{ε_{0} E^{2}}{2}, u_{B} = \frac{B^{2}}{2 μ_{0}}

3

u_{E} = \frac{ε_{0} E^{2}}{2}, u_{B} = \frac{μ_{0} B^{2}}{2}

4

u_{E} = \frac{E^{2}}{2 ε_{0}}, u_{B} = \frac{B^{2}}{2 μ_{0}}

Explanation:

In free space, the energy density of a static field $E$ is $u_{E} = \frac{1}{2} ε_{0} E^{2},$ and the energy density of magnetic field is $u_{B} = \frac{1}{2 μ_{0}} B^{2}$

JEE - 2023

PHXI15:WAVES

358821 Predict the correct relation between average energy density in electric field $μ_{E}$ magnetic field $u_{B}$

1

u_{E} = u_{B}

2

u_{E} = \frac{1}{2} u_{B}

3

\frac{1}{2} u_{E} = u_{B}

4

u_{E} = 4 u_{B}

Explanation:

Average energy density of $E$
$u_{E} = \frac{1}{4} ε_{0} E_{0}^{2}$
Average energy density of $B$
$\begin{matrix} u_{B} = \frac{1}{4} \frac{B_{0}^{2}}{μ_{0}} \\ ∵ c = \frac{E_{0}}{B_{0}} So, \\ u_{E} = \frac{1}{4} ε_{0} {(c B_{0})}^{2} = \frac{1}{4} ε_{0} \frac{1}{(μ_{0} ε_{0})} \times B_{0}^{2} = \frac{1}{4} \frac{B_{0}^{2}}{μ_{0}} \\ u_{E} = u_{B} \end{matrix}$

PHXI15:WAVES

358818 The ratio of average electric energy density and total average energy density of electromagnetic wave is

1 2

2 3

3

\frac{1}{2}

4 1

Explanation:

Average electric energy density = average magnetic energy density $= \frac{1}{2}$ total average energy density
So, the ratio of,
$\frac{average electric energy density}{total average energy density} = \frac{1}{2}$

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

\frac{1}{40 π} \frac{W}{m^{2}}

2

\frac{1}{20 π} \frac{W}{m^{2}}

3

\frac{1}{2 π} \frac{W}{m^{2}}

4

\frac{1}{10 π} \frac{W}{m^{2}}

Explanation:

Effective power emitted by the source
$P_{e} = 100 \times \frac{5}{100}$ Watt $= 5$ Watt.
Intensity at a distance of $5 m$ form the source is given by
$I = \frac{P_{e}}{4 π r^{2}} = \frac{5}{4 π {(5)}^{2}} W / m^{2} = \frac{1}{20 π} .$
$50 %$ of this energy is carried by electric field and $50 %$ is carried by magnetic field.
$∴$ Intensity produced by electric field component is given by $I_{E} = \frac{1}{2}, I_{e} = \frac{1}{40 π} W / m^{2} .$

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 $(ε_{0}$ - permittivity of free space, $μ_{0}$ - permeability of free space)

1

u_{E} = \frac{E^{2}}{2 ε_{0}}, u_{B} = \frac{μ_{0} B^{2}}{2}

2

u_{E} = \frac{ε_{0} E^{2}}{2}, u_{B} = \frac{B^{2}}{2 μ_{0}}

3

u_{E} = \frac{ε_{0} E^{2}}{2}, u_{B} = \frac{μ_{0} B^{2}}{2}

4

u_{E} = \frac{E^{2}}{2 ε_{0}}, u_{B} = \frac{B^{2}}{2 μ_{0}}

Explanation:

In free space, the energy density of a static field $E$ is $u_{E} = \frac{1}{2} ε_{0} E^{2},$ and the energy density of magnetic field is $u_{B} = \frac{1}{2 μ_{0}} B^{2}$

JEE - 2023

PHXI15:WAVES

358821 Predict the correct relation between average energy density in electric field $μ_{E}$ magnetic field $u_{B}$

1

u_{E} = u_{B}

2

u_{E} = \frac{1}{2} u_{B}

3

\frac{1}{2} u_{E} = u_{B}

4

u_{E} = 4 u_{B}

Explanation:

Average energy density of $E$
$u_{E} = \frac{1}{4} ε_{0} E_{0}^{2}$
Average energy density of $B$
$\begin{matrix} u_{B} = \frac{1}{4} \frac{B_{0}^{2}}{μ_{0}} \\ ∵ c = \frac{E_{0}}{B_{0}} So, \\ u_{E} = \frac{1}{4} ε_{0} {(c B_{0})}^{2} = \frac{1}{4} ε_{0} \frac{1}{(μ_{0} ε_{0})} \times B_{0}^{2} = \frac{1}{4} \frac{B_{0}^{2}}{μ_{0}} \\ u_{E} = u_{B} \end{matrix}$

PHXI15:WAVES

358818 The ratio of average electric energy density and total average energy density of electromagnetic wave is

1 2

2 3

3

\frac{1}{2}

4 1

Explanation:

Average electric energy density = average magnetic energy density $= \frac{1}{2}$ total average energy density
So, the ratio of,
$\frac{average electric energy density}{total average energy density} = \frac{1}{2}$

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

\frac{1}{40 π} \frac{W}{m^{2}}

2

\frac{1}{20 π} \frac{W}{m^{2}}

3

\frac{1}{2 π} \frac{W}{m^{2}}

4

\frac{1}{10 π} \frac{W}{m^{2}}

Explanation:

Effective power emitted by the source
$P_{e} = 100 \times \frac{5}{100}$ Watt $= 5$ Watt.
Intensity at a distance of $5 m$ form the source is given by
$I = \frac{P_{e}}{4 π r^{2}} = \frac{5}{4 π {(5)}^{2}} W / m^{2} = \frac{1}{20 π} .$
$50 %$ of this energy is carried by electric field and $50 %$ is carried by magnetic field.
$∴$ Intensity produced by electric field component is given by $I_{E} = \frac{1}{2}, I_{e} = \frac{1}{40 π} W / m^{2} .$

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 $(ε_{0}$ - permittivity of free space, $μ_{0}$ - permeability of free space)

1

u_{E} = \frac{E^{2}}{2 ε_{0}}, u_{B} = \frac{μ_{0} B^{2}}{2}

2

u_{E} = \frac{ε_{0} E^{2}}{2}, u_{B} = \frac{B^{2}}{2 μ_{0}}

3

u_{E} = \frac{ε_{0} E^{2}}{2}, u_{B} = \frac{μ_{0} B^{2}}{2}

4

u_{E} = \frac{E^{2}}{2 ε_{0}}, u_{B} = \frac{B^{2}}{2 μ_{0}}

Explanation:

In free space, the energy density of a static field $E$ is $u_{E} = \frac{1}{2} ε_{0} E^{2},$ and the energy density of magnetic field is $u_{B} = \frac{1}{2 μ_{0}} B^{2}$

JEE - 2023

PHXI15:WAVES

358821 Predict the correct relation between average energy density in electric field $μ_{E}$ magnetic field $u_{B}$

1

u_{E} = u_{B}

2

u_{E} = \frac{1}{2} u_{B}

3

\frac{1}{2} u_{E} = u_{B}

4

u_{E} = 4 u_{B}

Explanation:

Average energy density of $E$
$u_{E} = \frac{1}{4} ε_{0} E_{0}^{2}$
Average energy density of $B$
$\begin{matrix} u_{B} = \frac{1}{4} \frac{B_{0}^{2}}{μ_{0}} \\ ∵ c = \frac{E_{0}}{B_{0}} So, \\ u_{E} = \frac{1}{4} ε_{0} {(c B_{0})}^{2} = \frac{1}{4} ε_{0} \frac{1}{(μ_{0} ε_{0})} \times B_{0}^{2} = \frac{1}{4} \frac{B_{0}^{2}}{μ_{0}} \\ u_{E} = u_{B} \end{matrix}$

PHXI15:WAVES

358818 The ratio of average electric energy density and total average energy density of electromagnetic wave is

1 2

2 3

3

\frac{1}{2}

4 1

Explanation:

Average electric energy density = average magnetic energy density $= \frac{1}{2}$ total average energy density
So, the ratio of,
$\frac{average electric energy density}{total average energy density} = \frac{1}{2}$

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

\frac{1}{40 π} \frac{W}{m^{2}}

2

\frac{1}{20 π} \frac{W}{m^{2}}

3

\frac{1}{2 π} \frac{W}{m^{2}}

4

\frac{1}{10 π} \frac{W}{m^{2}}

Explanation:

Effective power emitted by the source
$P_{e} = 100 \times \frac{5}{100}$ Watt $= 5$ Watt.
Intensity at a distance of $5 m$ form the source is given by
$I = \frac{P_{e}}{4 π r^{2}} = \frac{5}{4 π {(5)}^{2}} W / m^{2} = \frac{1}{20 π} .$
$50 %$ of this energy is carried by electric field and $50 %$ is carried by magnetic field.
$∴$ Intensity produced by electric field component is given by $I_{E} = \frac{1}{2}, I_{e} = \frac{1}{40 π} W / m^{2} .$

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 $(ε_{0}$ - permittivity of free space, $μ_{0}$ - permeability of free space)

1

u_{E} = \frac{E^{2}}{2 ε_{0}}, u_{B} = \frac{μ_{0} B^{2}}{2}

2

u_{E} = \frac{ε_{0} E^{2}}{2}, u_{B} = \frac{B^{2}}{2 μ_{0}}

3

u_{E} = \frac{ε_{0} E^{2}}{2}, u_{B} = \frac{μ_{0} B^{2}}{2}

4

u_{E} = \frac{E^{2}}{2 ε_{0}}, u_{B} = \frac{B^{2}}{2 μ_{0}}

Explanation:

In free space, the energy density of a static field $E$ is $u_{E} = \frac{1}{2} ε_{0} E^{2},$ and the energy density of magnetic field is $u_{B} = \frac{1}{2 μ_{0}} B^{2}$

JEE - 2023

PHXI15:WAVES

358821 Predict the correct relation between average energy density in electric field $μ_{E}$ magnetic field $u_{B}$

1

u_{E} = u_{B}

2

u_{E} = \frac{1}{2} u_{B}

3

\frac{1}{2} u_{E} = u_{B}

4

u_{E} = 4 u_{B}

Explanation:

Average energy density of $E$
$u_{E} = \frac{1}{4} ε_{0} E_{0}^{2}$
Average energy density of $B$
$\begin{matrix} u_{B} = \frac{1}{4} \frac{B_{0}^{2}}{μ_{0}} \\ ∵ c = \frac{E_{0}}{B_{0}} So, \\ u_{E} = \frac{1}{4} ε_{0} {(c B_{0})}^{2} = \frac{1}{4} ε_{0} \frac{1}{(μ_{0} ε_{0})} \times B_{0}^{2} = \frac{1}{4} \frac{B_{0}^{2}}{μ_{0}} \\ u_{E} = u_{B} \end{matrix}$