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PHXII08:ELECTROMAGNETIC WAVES

368970 An intense light source radiates uniformly in all directions. At a distance $5 m$ from the source, the radiation pressure on absorbing surface is $9 \times 10^{- 6} P a$ . Find the total average power output.

1

8.5 \times 10^{5} W

2

8.5 \times 10^{3} W

3

6.5 \times 10^{3} W

4

6.5 \times 10^{5} W

Explanation:

$p_{rad} = \frac{I}{c}$ and Power $= I 4 π r^{2}$
$= p_{r a d} \times c 4 π r^{2}$
$= 9 \times 10^{- 6} \times 3 \times 10^{8} \times 12.56 \times 25$
$= 8.5 \times 10^{5} W$

PHXII08:ELECTROMAGNETIC WAVES

368971 The amplitude of electric field in a parallel light beam of intensity $4 W m^{- 2}$ is

1

49.5 N C^{- 1}

2

35.5 N C^{- 1}

3

45.5 N C^{- 1}

4

55.5 N C^{- 1}

Explanation:

$I = \frac{1}{2} ε_{0} E_{0}^{2} c$ or $E_{0} = \sqrt{\frac{2 I}{ε_{0} c}}$
$= \sqrt{\frac{2 \times 4}{8.85 \times 10^{- 12} \times 3 \times 10^{8}}} = 55.5 N C^{- 1}$

PHXII08:ELECTROMAGNETIC WAVES

368972 Light with an average flux of $20 W / c m^{2}$ falls on a non-reflectiong surface at normal incidence having surface area $20 c m^{2}$ . The energy received by the surface during time span of 1 minute is:

1

12 \times 10^{3} J

2

24 \times 10^{3} J

3

48 \times 10^{3} J

4

10 \times 10^{3} J

Explanation:

$I = \frac{E}{A t}$
$E = I A t$
$= \frac{20}{10^{- 4}} \times 20 \times 10^{- 4} \times 60$
$= 24 \times 10^{3} J$

NEET - 2020

PHXII08:ELECTROMAGNETIC WAVES

368973 A $27 m W$ laser beam has a cross-sectional area of $10 {m m}^{2}$ . Find the magnitude of the maximum electric field in this electromagnetic wave is given by [Given permittivity of space $ε_{0} = 9 \times 10^{- 12}$ $S I$ units, Speed of light $c = 3 \times 10^{8} m / s]$ :

1

3.1 k V / m

2

1.4 k V / m

3

5.2 k V / m

4

7.4 k V / m

Explanation:

Intensity of $E M$ wave is given by
$I = \frac{P o w e r}{A r e a} = \frac{1}{2} ε_{0} E_{0}^{2} C = \frac{27 \times 10^{- 3}}{10 \times 10^{- 6}}$
$= \frac{1}{2} \times 9 \times 10^{- 12} \times E^{2} \times 3 \times 10^{8}$
$E = \sqrt{2} \times 10^{3} k V / m = 1.4 k V / m$

PHXII08:ELECTROMAGNETIC WAVES

368974 In an electromagnetic wave, the average energy density associated with magnetic field is

1

\frac{L i_{0}^{2}}{2}

2

\frac{B^{2}}{2 μ_{0}}

3

\frac{μ_{0} B^{2}}{2}

4

\frac{μ_{0}}{2 B^{2}}

Explanation:

$U_{B} = \frac{B^{2}}{2 μ_{0}}$

PHXII08:ELECTROMAGNETIC WAVES

368970 An intense light source radiates uniformly in all directions. At a distance $5 m$ from the source, the radiation pressure on absorbing surface is $9 \times 10^{- 6} P a$ . Find the total average power output.

1

8.5 \times 10^{5} W

2

8.5 \times 10^{3} W

3

6.5 \times 10^{3} W

4

6.5 \times 10^{5} W

Explanation:

$p_{rad} = \frac{I}{c}$ and Power $= I 4 π r^{2}$
$= p_{r a d} \times c 4 π r^{2}$
$= 9 \times 10^{- 6} \times 3 \times 10^{8} \times 12.56 \times 25$
$= 8.5 \times 10^{5} W$

PHXII08:ELECTROMAGNETIC WAVES

368971 The amplitude of electric field in a parallel light beam of intensity $4 W m^{- 2}$ is

1

49.5 N C^{- 1}

2

35.5 N C^{- 1}

3

45.5 N C^{- 1}

4

55.5 N C^{- 1}

Explanation:

$I = \frac{1}{2} ε_{0} E_{0}^{2} c$ or $E_{0} = \sqrt{\frac{2 I}{ε_{0} c}}$
$= \sqrt{\frac{2 \times 4}{8.85 \times 10^{- 12} \times 3 \times 10^{8}}} = 55.5 N C^{- 1}$

PHXII08:ELECTROMAGNETIC WAVES

368972 Light with an average flux of $20 W / c m^{2}$ falls on a non-reflectiong surface at normal incidence having surface area $20 c m^{2}$ . The energy received by the surface during time span of 1 minute is:

1

12 \times 10^{3} J

2

24 \times 10^{3} J

3

48 \times 10^{3} J

4

10 \times 10^{3} J

Explanation:

$I = \frac{E}{A t}$
$E = I A t$
$= \frac{20}{10^{- 4}} \times 20 \times 10^{- 4} \times 60$
$= 24 \times 10^{3} J$

NEET - 2020

PHXII08:ELECTROMAGNETIC WAVES

368973 A $27 m W$ laser beam has a cross-sectional area of $10 {m m}^{2}$ . Find the magnitude of the maximum electric field in this electromagnetic wave is given by [Given permittivity of space $ε_{0} = 9 \times 10^{- 12}$ $S I$ units, Speed of light $c = 3 \times 10^{8} m / s]$ :

1

3.1 k V / m

2

1.4 k V / m

3

5.2 k V / m

4

7.4 k V / m

Explanation:

Intensity of $E M$ wave is given by
$I = \frac{P o w e r}{A r e a} = \frac{1}{2} ε_{0} E_{0}^{2} C = \frac{27 \times 10^{- 3}}{10 \times 10^{- 6}}$
$= \frac{1}{2} \times 9 \times 10^{- 12} \times E^{2} \times 3 \times 10^{8}$
$E = \sqrt{2} \times 10^{3} k V / m = 1.4 k V / m$

PHXII08:ELECTROMAGNETIC WAVES

368974 In an electromagnetic wave, the average energy density associated with magnetic field is

1

\frac{L i_{0}^{2}}{2}

2

\frac{B^{2}}{2 μ_{0}}

3

\frac{μ_{0} B^{2}}{2}

4

\frac{μ_{0}}{2 B^{2}}

Explanation:

$U_{B} = \frac{B^{2}}{2 μ_{0}}$

PHXII08:ELECTROMAGNETIC WAVES

368970 An intense light source radiates uniformly in all directions. At a distance $5 m$ from the source, the radiation pressure on absorbing surface is $9 \times 10^{- 6} P a$ . Find the total average power output.

1

8.5 \times 10^{5} W

2

8.5 \times 10^{3} W

3

6.5 \times 10^{3} W

4

6.5 \times 10^{5} W

Explanation:

$p_{rad} = \frac{I}{c}$ and Power $= I 4 π r^{2}$
$= p_{r a d} \times c 4 π r^{2}$
$= 9 \times 10^{- 6} \times 3 \times 10^{8} \times 12.56 \times 25$
$= 8.5 \times 10^{5} W$

PHXII08:ELECTROMAGNETIC WAVES

368971 The amplitude of electric field in a parallel light beam of intensity $4 W m^{- 2}$ is

1

49.5 N C^{- 1}

2

35.5 N C^{- 1}

3

45.5 N C^{- 1}

4

55.5 N C^{- 1}

Explanation:

$I = \frac{1}{2} ε_{0} E_{0}^{2} c$ or $E_{0} = \sqrt{\frac{2 I}{ε_{0} c}}$
$= \sqrt{\frac{2 \times 4}{8.85 \times 10^{- 12} \times 3 \times 10^{8}}} = 55.5 N C^{- 1}$

PHXII08:ELECTROMAGNETIC WAVES

368972 Light with an average flux of $20 W / c m^{2}$ falls on a non-reflectiong surface at normal incidence having surface area $20 c m^{2}$ . The energy received by the surface during time span of 1 minute is:

1

12 \times 10^{3} J

2

24 \times 10^{3} J

3

48 \times 10^{3} J

4

10 \times 10^{3} J

Explanation:

$I = \frac{E}{A t}$
$E = I A t$
$= \frac{20}{10^{- 4}} \times 20 \times 10^{- 4} \times 60$
$= 24 \times 10^{3} J$

NEET - 2020

PHXII08:ELECTROMAGNETIC WAVES

368973 A $27 m W$ laser beam has a cross-sectional area of $10 {m m}^{2}$ . Find the magnitude of the maximum electric field in this electromagnetic wave is given by [Given permittivity of space $ε_{0} = 9 \times 10^{- 12}$ $S I$ units, Speed of light $c = 3 \times 10^{8} m / s]$ :

1

3.1 k V / m

2

1.4 k V / m

3

5.2 k V / m

4

7.4 k V / m

Explanation:

Intensity of $E M$ wave is given by
$I = \frac{P o w e r}{A r e a} = \frac{1}{2} ε_{0} E_{0}^{2} C = \frac{27 \times 10^{- 3}}{10 \times 10^{- 6}}$
$= \frac{1}{2} \times 9 \times 10^{- 12} \times E^{2} \times 3 \times 10^{8}$
$E = \sqrt{2} \times 10^{3} k V / m = 1.4 k V / m$

PHXII08:ELECTROMAGNETIC WAVES

368974 In an electromagnetic wave, the average energy density associated with magnetic field is

1

\frac{L i_{0}^{2}}{2}

2

\frac{B^{2}}{2 μ_{0}}

3

\frac{μ_{0} B^{2}}{2}

4

\frac{μ_{0}}{2 B^{2}}

Explanation:

$U_{B} = \frac{B^{2}}{2 μ_{0}}$

PHXII08:ELECTROMAGNETIC WAVES

368970 An intense light source radiates uniformly in all directions. At a distance $5 m$ from the source, the radiation pressure on absorbing surface is $9 \times 10^{- 6} P a$ . Find the total average power output.

1

8.5 \times 10^{5} W

2

8.5 \times 10^{3} W

3

6.5 \times 10^{3} W

4

6.5 \times 10^{5} W

Explanation:

$p_{rad} = \frac{I}{c}$ and Power $= I 4 π r^{2}$
$= p_{r a d} \times c 4 π r^{2}$
$= 9 \times 10^{- 6} \times 3 \times 10^{8} \times 12.56 \times 25$
$= 8.5 \times 10^{5} W$

PHXII08:ELECTROMAGNETIC WAVES

368971 The amplitude of electric field in a parallel light beam of intensity $4 W m^{- 2}$ is

1

49.5 N C^{- 1}

2

35.5 N C^{- 1}

3

45.5 N C^{- 1}

4

55.5 N C^{- 1}

Explanation:

$I = \frac{1}{2} ε_{0} E_{0}^{2} c$ or $E_{0} = \sqrt{\frac{2 I}{ε_{0} c}}$
$= \sqrt{\frac{2 \times 4}{8.85 \times 10^{- 12} \times 3 \times 10^{8}}} = 55.5 N C^{- 1}$

PHXII08:ELECTROMAGNETIC WAVES

368972 Light with an average flux of $20 W / c m^{2}$ falls on a non-reflectiong surface at normal incidence having surface area $20 c m^{2}$ . The energy received by the surface during time span of 1 minute is:

1

12 \times 10^{3} J

2

24 \times 10^{3} J

3

48 \times 10^{3} J

4

10 \times 10^{3} J

Explanation:

$I = \frac{E}{A t}$
$E = I A t$
$= \frac{20}{10^{- 4}} \times 20 \times 10^{- 4} \times 60$
$= 24 \times 10^{3} J$

NEET - 2020

PHXII08:ELECTROMAGNETIC WAVES

368973 A $27 m W$ laser beam has a cross-sectional area of $10 {m m}^{2}$ . Find the magnitude of the maximum electric field in this electromagnetic wave is given by [Given permittivity of space $ε_{0} = 9 \times 10^{- 12}$ $S I$ units, Speed of light $c = 3 \times 10^{8} m / s]$ :

1

3.1 k V / m

2

1.4 k V / m

3

5.2 k V / m

4

7.4 k V / m

Explanation:

Intensity of $E M$ wave is given by
$I = \frac{P o w e r}{A r e a} = \frac{1}{2} ε_{0} E_{0}^{2} C = \frac{27 \times 10^{- 3}}{10 \times 10^{- 6}}$
$= \frac{1}{2} \times 9 \times 10^{- 12} \times E^{2} \times 3 \times 10^{8}$
$E = \sqrt{2} \times 10^{3} k V / m = 1.4 k V / m$

PHXII08:ELECTROMAGNETIC WAVES

368974 In an electromagnetic wave, the average energy density associated with magnetic field is

1

\frac{L i_{0}^{2}}{2}

2

\frac{B^{2}}{2 μ_{0}}

3

\frac{μ_{0} B^{2}}{2}

4

\frac{μ_{0}}{2 B^{2}}

Explanation:

$U_{B} = \frac{B^{2}}{2 μ_{0}}$

PHXII08:ELECTROMAGNETIC WAVES

368970 An intense light source radiates uniformly in all directions. At a distance $5 m$ from the source, the radiation pressure on absorbing surface is $9 \times 10^{- 6} P a$ . Find the total average power output.

1

8.5 \times 10^{5} W

2

8.5 \times 10^{3} W

3

6.5 \times 10^{3} W

4

6.5 \times 10^{5} W

Explanation:

$p_{rad} = \frac{I}{c}$ and Power $= I 4 π r^{2}$
$= p_{r a d} \times c 4 π r^{2}$
$= 9 \times 10^{- 6} \times 3 \times 10^{8} \times 12.56 \times 25$
$= 8.5 \times 10^{5} W$

PHXII08:ELECTROMAGNETIC WAVES

368971 The amplitude of electric field in a parallel light beam of intensity $4 W m^{- 2}$ is

1

49.5 N C^{- 1}

2

35.5 N C^{- 1}

3

45.5 N C^{- 1}

4

55.5 N C^{- 1}

Explanation:

$I = \frac{1}{2} ε_{0} E_{0}^{2} c$ or $E_{0} = \sqrt{\frac{2 I}{ε_{0} c}}$
$= \sqrt{\frac{2 \times 4}{8.85 \times 10^{- 12} \times 3 \times 10^{8}}} = 55.5 N C^{- 1}$

PHXII08:ELECTROMAGNETIC WAVES

368972 Light with an average flux of $20 W / c m^{2}$ falls on a non-reflectiong surface at normal incidence having surface area $20 c m^{2}$ . The energy received by the surface during time span of 1 minute is:

1

12 \times 10^{3} J

2

24 \times 10^{3} J

3

48 \times 10^{3} J

4

10 \times 10^{3} J

Explanation:

$I = \frac{E}{A t}$
$E = I A t$
$= \frac{20}{10^{- 4}} \times 20 \times 10^{- 4} \times 60$
$= 24 \times 10^{3} J$

NEET - 2020

PHXII08:ELECTROMAGNETIC WAVES

368973 A $27 m W$ laser beam has a cross-sectional area of $10 {m m}^{2}$ . Find the magnitude of the maximum electric field in this electromagnetic wave is given by [Given permittivity of space $ε_{0} = 9 \times 10^{- 12}$ $S I$ units, Speed of light $c = 3 \times 10^{8} m / s]$ :

1

3.1 k V / m

2

1.4 k V / m

3

5.2 k V / m

4

7.4 k V / m

Explanation:

Intensity of $E M$ wave is given by
$I = \frac{P o w e r}{A r e a} = \frac{1}{2} ε_{0} E_{0}^{2} C = \frac{27 \times 10^{- 3}}{10 \times 10^{- 6}}$
$= \frac{1}{2} \times 9 \times 10^{- 12} \times E^{2} \times 3 \times 10^{8}$
$E = \sqrt{2} \times 10^{3} k V / m = 1.4 k V / m$

PHXII08:ELECTROMAGNETIC WAVES

368974 In an electromagnetic wave, the average energy density associated with magnetic field is

1

\frac{L i_{0}^{2}}{2}

2

\frac{B^{2}}{2 μ_{0}}

3

\frac{μ_{0} B^{2}}{2}

4

\frac{μ_{0}}{2 B^{2}}

Explanation:

$U_{B} = \frac{B^{2}}{2 μ_{0}}$