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PHXI15:WAVES

358831 An electromagnetic wave of frequency $1 \times 10^{14}$ hertz is propagating along $z$ - axis. The amplitude of electric field is $4 V / m$ .
If $ε_{0} = 8.8 \times 10^{- 12} C^{2} / N - m^{2}$ , then average energy density of electric field will be :-

1

35.2 \times 10^{- 12} J / m^{3}

2

35.2 \times 10^{- 10} J / m^{3}

3

35.2 \times 10^{- 13} J / m^{3}

4

35.2 \times 10^{- 13} J / m^{3}

Explanation:

Only due to electric field, ave. energy density
$= \frac{1}{4} ε_{o} E_{o}^{2} = \frac{1}{4} \times 8.8 \times 10^{- 12} \times 16 = 35.2 \times 10^{- 12} (\frac{J}{m^{3}})$

JEE - 2014

PHXI15:WAVES

358832 A radio station on the surface of the earth radiates $50 k W$ . If transmitter radiates equally in all directions above the surface of the earth find the amplitude of electric field detected $100 k m$ away. [In the downward direction assume no radiation]

1

2.45 \times 10^{- 1} V m^{- 1}

2

2.45 \times 10^{- 3} V m^{- 1}

3

2.45 \times 10^{- 2} V m^{- 1}

4

2.45 V m^{- 1}

Explanation:

$I = \frac{P}{2 π r^{2}}$
$I = \frac{ε_{0} E_{0}^{2} c}{2} = \frac{P}{2 π r^{2}}$
$E_{0} = \sqrt{\frac{P}{ε_{0} c π r^{2}}} = \sqrt{\frac{50 \times 10^{3}}{8.85 \times 10^{- 12} \times 3 \times 10^{8} \times 10^{10} \times 3.14}}$
$= 2.45 \times 10^{- 2} V m^{- 1}$

PHXI15:WAVES

358833 The sun delivers $10^{3} W / m^{2}$ of electromagnetic flux to the earth's surface. The total power that is incident on a roof of dimensions $8 m \times 20 m$ , will be

1

3.4 \times 10^{4} W

2

6.4 \times 10^{3} W

3

1.6 \times 10^{5} W

4 None of these

Explanation:

$P = 10^{3} \times 160 = 1.6 \times 10^{5} W$

PHXI15:WAVES

358834 The average magnetic energy density of an
electromagnetic wave of wave length $λ$ travelling in free space is given by

1

\frac{B^{2}}{2 λ}

2

\frac{B^{2}}{2 μ_{0}}

3

\frac{2 B^{2}}{μ_{0} λ}

4

\frac{B}{μ_{0} λ}

Explanation:

Energy density of an electromagnetic wave,
$U = \frac{1}{2} ε_{0} E^{2} + \frac{1}{2} \frac{B^{2}}{μ_{0}}$
So, the magnetic energy density is
$\frac{B^{2}}{2 μ_{0}} .$

PHXI15:WAVES

358831 An electromagnetic wave of frequency $1 \times 10^{14}$ hertz is propagating along $z$ - axis. The amplitude of electric field is $4 V / m$ .
If $ε_{0} = 8.8 \times 10^{- 12} C^{2} / N - m^{2}$ , then average energy density of electric field will be :-

1

35.2 \times 10^{- 12} J / m^{3}

2

35.2 \times 10^{- 10} J / m^{3}

3

35.2 \times 10^{- 13} J / m^{3}

4

35.2 \times 10^{- 13} J / m^{3}

Explanation:

Only due to electric field, ave. energy density
$= \frac{1}{4} ε_{o} E_{o}^{2} = \frac{1}{4} \times 8.8 \times 10^{- 12} \times 16 = 35.2 \times 10^{- 12} (\frac{J}{m^{3}})$

JEE - 2014

PHXI15:WAVES

358832 A radio station on the surface of the earth radiates $50 k W$ . If transmitter radiates equally in all directions above the surface of the earth find the amplitude of electric field detected $100 k m$ away. [In the downward direction assume no radiation]

1

2.45 \times 10^{- 1} V m^{- 1}

2

2.45 \times 10^{- 3} V m^{- 1}

3

2.45 \times 10^{- 2} V m^{- 1}

4

2.45 V m^{- 1}

Explanation:

$I = \frac{P}{2 π r^{2}}$
$I = \frac{ε_{0} E_{0}^{2} c}{2} = \frac{P}{2 π r^{2}}$
$E_{0} = \sqrt{\frac{P}{ε_{0} c π r^{2}}} = \sqrt{\frac{50 \times 10^{3}}{8.85 \times 10^{- 12} \times 3 \times 10^{8} \times 10^{10} \times 3.14}}$
$= 2.45 \times 10^{- 2} V m^{- 1}$

PHXI15:WAVES

358833 The sun delivers $10^{3} W / m^{2}$ of electromagnetic flux to the earth's surface. The total power that is incident on a roof of dimensions $8 m \times 20 m$ , will be

1

3.4 \times 10^{4} W

2

6.4 \times 10^{3} W

3

1.6 \times 10^{5} W

4 None of these

Explanation:

$P = 10^{3} \times 160 = 1.6 \times 10^{5} W$

PHXI15:WAVES

358834 The average magnetic energy density of an
electromagnetic wave of wave length $λ$ travelling in free space is given by

1

\frac{B^{2}}{2 λ}

2

\frac{B^{2}}{2 μ_{0}}

3

\frac{2 B^{2}}{μ_{0} λ}

4

\frac{B}{μ_{0} λ}

Explanation:

Energy density of an electromagnetic wave,
$U = \frac{1}{2} ε_{0} E^{2} + \frac{1}{2} \frac{B^{2}}{μ_{0}}$
So, the magnetic energy density is
$\frac{B^{2}}{2 μ_{0}} .$

PHXI15:WAVES

358831 An electromagnetic wave of frequency $1 \times 10^{14}$ hertz is propagating along $z$ - axis. The amplitude of electric field is $4 V / m$ .
If $ε_{0} = 8.8 \times 10^{- 12} C^{2} / N - m^{2}$ , then average energy density of electric field will be :-

1

35.2 \times 10^{- 12} J / m^{3}

2

35.2 \times 10^{- 10} J / m^{3}

3

35.2 \times 10^{- 13} J / m^{3}

4

35.2 \times 10^{- 13} J / m^{3}

Explanation:

Only due to electric field, ave. energy density
$= \frac{1}{4} ε_{o} E_{o}^{2} = \frac{1}{4} \times 8.8 \times 10^{- 12} \times 16 = 35.2 \times 10^{- 12} (\frac{J}{m^{3}})$

JEE - 2014

PHXI15:WAVES

358832 A radio station on the surface of the earth radiates $50 k W$ . If transmitter radiates equally in all directions above the surface of the earth find the amplitude of electric field detected $100 k m$ away. [In the downward direction assume no radiation]

1

2.45 \times 10^{- 1} V m^{- 1}

2

2.45 \times 10^{- 3} V m^{- 1}

3

2.45 \times 10^{- 2} V m^{- 1}

4

2.45 V m^{- 1}

Explanation:

$I = \frac{P}{2 π r^{2}}$
$I = \frac{ε_{0} E_{0}^{2} c}{2} = \frac{P}{2 π r^{2}}$
$E_{0} = \sqrt{\frac{P}{ε_{0} c π r^{2}}} = \sqrt{\frac{50 \times 10^{3}}{8.85 \times 10^{- 12} \times 3 \times 10^{8} \times 10^{10} \times 3.14}}$
$= 2.45 \times 10^{- 2} V m^{- 1}$

PHXI15:WAVES

358833 The sun delivers $10^{3} W / m^{2}$ of electromagnetic flux to the earth's surface. The total power that is incident on a roof of dimensions $8 m \times 20 m$ , will be

1

3.4 \times 10^{4} W

2

6.4 \times 10^{3} W

3

1.6 \times 10^{5} W

4 None of these

Explanation:

$P = 10^{3} \times 160 = 1.6 \times 10^{5} W$

PHXI15:WAVES

358834 The average magnetic energy density of an
electromagnetic wave of wave length $λ$ travelling in free space is given by

1

\frac{B^{2}}{2 λ}

2

\frac{B^{2}}{2 μ_{0}}

3

\frac{2 B^{2}}{μ_{0} λ}

4

\frac{B}{μ_{0} λ}

Explanation:

Energy density of an electromagnetic wave,
$U = \frac{1}{2} ε_{0} E^{2} + \frac{1}{2} \frac{B^{2}}{μ_{0}}$
So, the magnetic energy density is
$\frac{B^{2}}{2 μ_{0}} .$

PHXI15:WAVES

358831 An electromagnetic wave of frequency $1 \times 10^{14}$ hertz is propagating along $z$ - axis. The amplitude of electric field is $4 V / m$ .
If $ε_{0} = 8.8 \times 10^{- 12} C^{2} / N - m^{2}$ , then average energy density of electric field will be :-

1

35.2 \times 10^{- 12} J / m^{3}

2

35.2 \times 10^{- 10} J / m^{3}

3

35.2 \times 10^{- 13} J / m^{3}

4

35.2 \times 10^{- 13} J / m^{3}

Explanation:

Only due to electric field, ave. energy density
$= \frac{1}{4} ε_{o} E_{o}^{2} = \frac{1}{4} \times 8.8 \times 10^{- 12} \times 16 = 35.2 \times 10^{- 12} (\frac{J}{m^{3}})$

JEE - 2014

PHXI15:WAVES

358832 A radio station on the surface of the earth radiates $50 k W$ . If transmitter radiates equally in all directions above the surface of the earth find the amplitude of electric field detected $100 k m$ away. [In the downward direction assume no radiation]

1

2.45 \times 10^{- 1} V m^{- 1}

2

2.45 \times 10^{- 3} V m^{- 1}

3

2.45 \times 10^{- 2} V m^{- 1}

4

2.45 V m^{- 1}

Explanation:

$I = \frac{P}{2 π r^{2}}$
$I = \frac{ε_{0} E_{0}^{2} c}{2} = \frac{P}{2 π r^{2}}$
$E_{0} = \sqrt{\frac{P}{ε_{0} c π r^{2}}} = \sqrt{\frac{50 \times 10^{3}}{8.85 \times 10^{- 12} \times 3 \times 10^{8} \times 10^{10} \times 3.14}}$
$= 2.45 \times 10^{- 2} V m^{- 1}$

PHXI15:WAVES

358833 The sun delivers $10^{3} W / m^{2}$ of electromagnetic flux to the earth's surface. The total power that is incident on a roof of dimensions $8 m \times 20 m$ , will be

1

3.4 \times 10^{4} W

2

6.4 \times 10^{3} W

3

1.6 \times 10^{5} W

4 None of these

Explanation:

$P = 10^{3} \times 160 = 1.6 \times 10^{5} W$

PHXI15:WAVES

358834 The average magnetic energy density of an
electromagnetic wave of wave length $λ$ travelling in free space is given by

1

\frac{B^{2}}{2 λ}

2

\frac{B^{2}}{2 μ_{0}}

3

\frac{2 B^{2}}{μ_{0} λ}

4

\frac{B}{μ_{0} λ}

Explanation:

Energy density of an electromagnetic wave,
$U = \frac{1}{2} ε_{0} E^{2} + \frac{1}{2} \frac{B^{2}}{μ_{0}}$
So, the magnetic energy density is
$\frac{B^{2}}{2 μ_{0}} .$

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