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PHXII08:ELECTROMAGNETIC WAVES

368979 In a plane $E M$ wave, the electric field oscillates sinusoidally at a frequency of $5 \times 10^{10} H z$ and have an amplitude of $50 V m^{- 1} .$ The total average energy density of the electromagnetic field of the wave is [Use $ε_{0} = 8.85 \times 10^{- 12} C^{2} / N m^{2} J]$

1

2.212 \times 10^{- 10} J m^{- 3}

2

4.425 \times 10^{- 8} J m^{- 3}

3

1.106 \times 10^{- 8} J m^{- 3}

4

2.212 \times 10^{- 8} J m^{- 3}

Explanation:

Given,
$f = 5 \times 10^{10} H z, E_{0} = 50 V / m$
Total average density of $E M$ wave $u_{T} = \frac{1}{2} ε_{0}$
$E_{0}^{2} = \frac{1}{2} \frac{B_{0}^{2}}{μ_{0}}$
Here, we will use, $u_{T} = \frac{1}{2} ε_{0} E_{0}^{2}$
$= \frac{1}{2} \times 8.85 \times 10^{- 12} \times 2500 = 1.106 \times 10^{- 8} J m^{- 3}$

JEE - 2024

PHXII08:ELECTROMAGNETIC WAVES

368980 A lamp emits monochromatic green light uniformly in all directions. The lamp is 3% efficient in converting electrical power to electromagnetic waves and consumes $100 W$ of power. The amplitude of the electric field associated with the electromagnetic radiation at a distance of $5 m$ from the lamp will be nearly :-

1

4.02 V / m

2

2.68 V / m

3

5.36 V / m

4

1.34 V / m

Explanation:

$\frac{P}{4 π r^{2}} = \frac{1}{2} ε_{o} E_{o}^{2} c [∵ \frac{I}{c} = \frac{1}{2} ε_{o} E_{o}^{2}] E_{0}^{2} = \frac{P}{2 π r^{2} ε_{0} C}$
$E_{0} = \sqrt{\frac{P}{2 π r^{2} ε_{0} c}} = \sqrt{\frac{3}{2 π {(5)}^{2} \times 8.85 \times 10^{- 12} \times 3 \times 10^{8}}}$
$E_{0} = 2.68 v o l t / m e t r e$

JEE - 2014

PHXII08:ELECTROMAGNETIC WAVES

368981 An $E M$ wave radiates outwards from a dipole antenna, with $E_{0}$ as the amplitude of its electric field vector. The electric field $E_{0}$ which transports significant energy from the source falls off as

1

\frac{1}{r^{2}}

2

\frac{1}{r^{3}}

3 Remains constant

4

\frac{1}{r}

Explanation:

From a diode antenna, the electromagnetic waves are radiated outwards. Intensity is directly proportional to square of amplitude $I α E_{o}^{2} α \frac{1}{r^{2}}$ .
The amplitude of electric field vector $(E_{0})$ which transports significant energy from the source falls off inversely as the distance $(r)$ from the antenna, i.e., $E_{0} \propto \frac{1}{r}$ .

NCERT Exempler

PHXII08:ELECTROMAGNETIC WAVES

368982 A point source of electromagnetic radiation has an average power output of $800 W$ . The maximum value of electric field at a distance $3.5 m$ from the source will be $62.6 V / m,$ the energy density at a distance $3.5 m$ from the source will be -(in joule $/ m^{3}$ )

1

1.73 \times 10^{- 6}

2

1.73 \times 10^{- 7}

3

1.73 \times 10^{- 5}

4

1.73 \times 10^{- 8}

Explanation:

Energy density
$u = \frac{1}{2} ε_{0} E_{m}^{2} = 1.73 \times 10^{- 8} J / m^{3}$

PHXII08:ELECTROMAGNETIC WAVES

368979 In a plane $E M$ wave, the electric field oscillates sinusoidally at a frequency of $5 \times 10^{10} H z$ and have an amplitude of $50 V m^{- 1} .$ The total average energy density of the electromagnetic field of the wave is [Use $ε_{0} = 8.85 \times 10^{- 12} C^{2} / N m^{2} J]$

1

2.212 \times 10^{- 10} J m^{- 3}

2

4.425 \times 10^{- 8} J m^{- 3}

3

1.106 \times 10^{- 8} J m^{- 3}

4

2.212 \times 10^{- 8} J m^{- 3}

Explanation:

Given,
$f = 5 \times 10^{10} H z, E_{0} = 50 V / m$
Total average density of $E M$ wave $u_{T} = \frac{1}{2} ε_{0}$
$E_{0}^{2} = \frac{1}{2} \frac{B_{0}^{2}}{μ_{0}}$
Here, we will use, $u_{T} = \frac{1}{2} ε_{0} E_{0}^{2}$
$= \frac{1}{2} \times 8.85 \times 10^{- 12} \times 2500 = 1.106 \times 10^{- 8} J m^{- 3}$

JEE - 2024

PHXII08:ELECTROMAGNETIC WAVES

368980 A lamp emits monochromatic green light uniformly in all directions. The lamp is 3% efficient in converting electrical power to electromagnetic waves and consumes $100 W$ of power. The amplitude of the electric field associated with the electromagnetic radiation at a distance of $5 m$ from the lamp will be nearly :-

1

4.02 V / m

2

2.68 V / m

3

5.36 V / m

4

1.34 V / m

Explanation:

$\frac{P}{4 π r^{2}} = \frac{1}{2} ε_{o} E_{o}^{2} c [∵ \frac{I}{c} = \frac{1}{2} ε_{o} E_{o}^{2}] E_{0}^{2} = \frac{P}{2 π r^{2} ε_{0} C}$
$E_{0} = \sqrt{\frac{P}{2 π r^{2} ε_{0} c}} = \sqrt{\frac{3}{2 π {(5)}^{2} \times 8.85 \times 10^{- 12} \times 3 \times 10^{8}}}$
$E_{0} = 2.68 v o l t / m e t r e$

JEE - 2014

PHXII08:ELECTROMAGNETIC WAVES

368981 An $E M$ wave radiates outwards from a dipole antenna, with $E_{0}$ as the amplitude of its electric field vector. The electric field $E_{0}$ which transports significant energy from the source falls off as

1

\frac{1}{r^{2}}

2

\frac{1}{r^{3}}

3 Remains constant

4

\frac{1}{r}

Explanation:

From a diode antenna, the electromagnetic waves are radiated outwards. Intensity is directly proportional to square of amplitude $I α E_{o}^{2} α \frac{1}{r^{2}}$ .
The amplitude of electric field vector $(E_{0})$ which transports significant energy from the source falls off inversely as the distance $(r)$ from the antenna, i.e., $E_{0} \propto \frac{1}{r}$ .

NCERT Exempler

PHXII08:ELECTROMAGNETIC WAVES

368982 A point source of electromagnetic radiation has an average power output of $800 W$ . The maximum value of electric field at a distance $3.5 m$ from the source will be $62.6 V / m,$ the energy density at a distance $3.5 m$ from the source will be -(in joule $/ m^{3}$ )

1

1.73 \times 10^{- 6}

2

1.73 \times 10^{- 7}

3

1.73 \times 10^{- 5}

4

1.73 \times 10^{- 8}

Explanation:

Energy density
$u = \frac{1}{2} ε_{0} E_{m}^{2} = 1.73 \times 10^{- 8} J / m^{3}$

PHXII08:ELECTROMAGNETIC WAVES

368979 In a plane $E M$ wave, the electric field oscillates sinusoidally at a frequency of $5 \times 10^{10} H z$ and have an amplitude of $50 V m^{- 1} .$ The total average energy density of the electromagnetic field of the wave is [Use $ε_{0} = 8.85 \times 10^{- 12} C^{2} / N m^{2} J]$

1

2.212 \times 10^{- 10} J m^{- 3}

2

4.425 \times 10^{- 8} J m^{- 3}

3

1.106 \times 10^{- 8} J m^{- 3}

4

2.212 \times 10^{- 8} J m^{- 3}

Explanation:

Given,
$f = 5 \times 10^{10} H z, E_{0} = 50 V / m$
Total average density of $E M$ wave $u_{T} = \frac{1}{2} ε_{0}$
$E_{0}^{2} = \frac{1}{2} \frac{B_{0}^{2}}{μ_{0}}$
Here, we will use, $u_{T} = \frac{1}{2} ε_{0} E_{0}^{2}$
$= \frac{1}{2} \times 8.85 \times 10^{- 12} \times 2500 = 1.106 \times 10^{- 8} J m^{- 3}$

JEE - 2024

PHXII08:ELECTROMAGNETIC WAVES

368980 A lamp emits monochromatic green light uniformly in all directions. The lamp is 3% efficient in converting electrical power to electromagnetic waves and consumes $100 W$ of power. The amplitude of the electric field associated with the electromagnetic radiation at a distance of $5 m$ from the lamp will be nearly :-

1

4.02 V / m

2

2.68 V / m

3

5.36 V / m

4

1.34 V / m

Explanation:

$\frac{P}{4 π r^{2}} = \frac{1}{2} ε_{o} E_{o}^{2} c [∵ \frac{I}{c} = \frac{1}{2} ε_{o} E_{o}^{2}] E_{0}^{2} = \frac{P}{2 π r^{2} ε_{0} C}$
$E_{0} = \sqrt{\frac{P}{2 π r^{2} ε_{0} c}} = \sqrt{\frac{3}{2 π {(5)}^{2} \times 8.85 \times 10^{- 12} \times 3 \times 10^{8}}}$
$E_{0} = 2.68 v o l t / m e t r e$

JEE - 2014

PHXII08:ELECTROMAGNETIC WAVES

368981 An $E M$ wave radiates outwards from a dipole antenna, with $E_{0}$ as the amplitude of its electric field vector. The electric field $E_{0}$ which transports significant energy from the source falls off as

1

\frac{1}{r^{2}}

2

\frac{1}{r^{3}}

3 Remains constant

4

\frac{1}{r}

Explanation:

From a diode antenna, the electromagnetic waves are radiated outwards. Intensity is directly proportional to square of amplitude $I α E_{o}^{2} α \frac{1}{r^{2}}$ .
The amplitude of electric field vector $(E_{0})$ which transports significant energy from the source falls off inversely as the distance $(r)$ from the antenna, i.e., $E_{0} \propto \frac{1}{r}$ .

NCERT Exempler

PHXII08:ELECTROMAGNETIC WAVES

368982 A point source of electromagnetic radiation has an average power output of $800 W$ . The maximum value of electric field at a distance $3.5 m$ from the source will be $62.6 V / m,$ the energy density at a distance $3.5 m$ from the source will be -(in joule $/ m^{3}$ )

1

1.73 \times 10^{- 6}

2

1.73 \times 10^{- 7}

3

1.73 \times 10^{- 5}

4

1.73 \times 10^{- 8}

Explanation:

Energy density
$u = \frac{1}{2} ε_{0} E_{m}^{2} = 1.73 \times 10^{- 8} J / m^{3}$

PHXII08:ELECTROMAGNETIC WAVES

368979 In a plane $E M$ wave, the electric field oscillates sinusoidally at a frequency of $5 \times 10^{10} H z$ and have an amplitude of $50 V m^{- 1} .$ The total average energy density of the electromagnetic field of the wave is [Use $ε_{0} = 8.85 \times 10^{- 12} C^{2} / N m^{2} J]$

1

2.212 \times 10^{- 10} J m^{- 3}

2

4.425 \times 10^{- 8} J m^{- 3}

3

1.106 \times 10^{- 8} J m^{- 3}

4

2.212 \times 10^{- 8} J m^{- 3}

Explanation:

Given,
$f = 5 \times 10^{10} H z, E_{0} = 50 V / m$
Total average density of $E M$ wave $u_{T} = \frac{1}{2} ε_{0}$
$E_{0}^{2} = \frac{1}{2} \frac{B_{0}^{2}}{μ_{0}}$
Here, we will use, $u_{T} = \frac{1}{2} ε_{0} E_{0}^{2}$
$= \frac{1}{2} \times 8.85 \times 10^{- 12} \times 2500 = 1.106 \times 10^{- 8} J m^{- 3}$

JEE - 2024

PHXII08:ELECTROMAGNETIC WAVES

368980 A lamp emits monochromatic green light uniformly in all directions. The lamp is 3% efficient in converting electrical power to electromagnetic waves and consumes $100 W$ of power. The amplitude of the electric field associated with the electromagnetic radiation at a distance of $5 m$ from the lamp will be nearly :-

1

4.02 V / m

2

2.68 V / m

3

5.36 V / m

4

1.34 V / m

Explanation:

$\frac{P}{4 π r^{2}} = \frac{1}{2} ε_{o} E_{o}^{2} c [∵ \frac{I}{c} = \frac{1}{2} ε_{o} E_{o}^{2}] E_{0}^{2} = \frac{P}{2 π r^{2} ε_{0} C}$
$E_{0} = \sqrt{\frac{P}{2 π r^{2} ε_{0} c}} = \sqrt{\frac{3}{2 π {(5)}^{2} \times 8.85 \times 10^{- 12} \times 3 \times 10^{8}}}$
$E_{0} = 2.68 v o l t / m e t r e$

JEE - 2014

PHXII08:ELECTROMAGNETIC WAVES

368981 An $E M$ wave radiates outwards from a dipole antenna, with $E_{0}$ as the amplitude of its electric field vector. The electric field $E_{0}$ which transports significant energy from the source falls off as

1

\frac{1}{r^{2}}

2

\frac{1}{r^{3}}

3 Remains constant

4

\frac{1}{r}

Explanation:

From a diode antenna, the electromagnetic waves are radiated outwards. Intensity is directly proportional to square of amplitude $I α E_{o}^{2} α \frac{1}{r^{2}}$ .
The amplitude of electric field vector $(E_{0})$ which transports significant energy from the source falls off inversely as the distance $(r)$ from the antenna, i.e., $E_{0} \propto \frac{1}{r}$ .

NCERT Exempler

PHXII08:ELECTROMAGNETIC WAVES

368982 A point source of electromagnetic radiation has an average power output of $800 W$ . The maximum value of electric field at a distance $3.5 m$ from the source will be $62.6 V / m,$ the energy density at a distance $3.5 m$ from the source will be -(in joule $/ m^{3}$ )

1

1.73 \times 10^{- 6}

2

1.73 \times 10^{- 7}

3

1.73 \times 10^{- 5}

4

1.73 \times 10^{- 8}

Explanation:

Energy density
$u = \frac{1}{2} ε_{0} E_{m}^{2} = 1.73 \times 10^{- 8} J / m^{3}$