QBManager

PHXI15:WAVES

358886 The velocity of electromagnetic wave is along the direction of:

1

\vec{B}

2

\frac{\vec{E} \times \vec{B}}{μ_{0}}

3

\frac{\vec{B} \times \vec{E}}{μ_{0}}

4

E

Explanation:

An electromagnetic wave is the wave composed of the oscillations of electric and magnetic fields in mutually perpendicular planes and these oscllilations are perpendicular to the direction of propagations of wave.
The direction of propagation of electromagnetic wave is given by poynting vector
Along $\vec{S} = \vec{E} \times \vec{H} = \frac{\vec{E} \times \vec{B}}{μ_{0}}$ this is parallel to $\vec{E} \times \vec{B}$

PHXI15:WAVES

358887 In an apparatus, the electric field was found to oscillate with amplitude of $18 V / m$ . The amplitude of the oscillating magnetic field will be

1

11 \times 10^{- 11} T

2

6 \times 10^{- 8} T

3

4 \times 10^{- 6} T

4

9 \times 10^{- 9} T

Explanation:

$B_{0} = \frac{E_{0}}{c} = \frac{18}{3 \times 10^{8}} = 6 \times 10^{- 8} T$

PHXI15:WAVES

358888 The electric field in an electromagnetic wave is given by $\vec{E} = \hat{i} 40 \cos ω (t - \frac{z}{c}) N C^{- 1}$ . The magnetic field induction of this wave is (in $S I$ unit)

1

\vec{B} = \hat{k} \frac{40}{c} \cos ω (t - \frac{z}{c})

2

\vec{B} = \hat{i} \frac{40}{c} \cos ω (t - \frac{z}{c})

3

\vec{B} = \hat{j} 40 \cos ω (t - \frac{z}{c})

4

\vec{B} = \hat{j} \frac{40}{c} \cos ω (t - \frac{z}{c})

Explanation:

$\vec{E} = \hat{i} 40 \cos ω (t - \frac{z}{c}) N / C$
$\vec{E}$ is along $+ X$ axis: $\vec{v}$ is along $+ Z$ axis ;
Direction of $E M$ wave is given by
$\vec{v} = \vec{E} \times \vec{B}$
So, $\vec{B}$ is along $+ Y$ axis, $\frac{E}{B} = c$
$\Rightarrow B = \frac{E}{c} = \frac{40}{3 \times 10^{8}}$
$∴ \vec{B} = \frac{40}{c} \hat{j} \cos ω (t - \frac{z}{c})$

JEE - 2024

PHXI15:WAVES

358889 The amplitude of magnetic field in an electromagnetic wave propagating along $y$ -axis is $6.0 \times 10^{- 7} T$ . The maximum value of electric field in the electromagnetic wave is

1

2 \times 10^{15} V m^{- 1}

2

6.0 \times 10^{- 7} V m^{- 1}

3

5 \times 10^{14} V m^{- 1}

4

180 V m^{- 1}

Explanation:

For a electromagnetic wave, amplitude of electric and magnetic field are related by,
$\frac{E_{m}}{B_{m}} = c$
$∴ E_{m} = B_{m} \times C$
$= 6 \times 10^{- 7} \times 3 \times 10^{8} = 18 \times 10; E = 180 {V m}^{- 1}$

JEE - 2023

PHXI15:WAVES

358890 In a plane electromagnetic wave travelling in free space, the elctric field component oscillates sinusoidally at a frequency of $2.0 \times 10^{10} H z$ and amplitude $48 V m^{- 1}$ . Then the amplitude of oscillating magnetic field is: (Speed of light in free space $= 3 \times 10^{8} m s^{- 1})$

1

1.6 \times 10^{8} T

2

1.6 \times 10^{- 7} T

3

1.6 \times 10^{- 6} T

4

1.6 \times 10^{- 9} T

Explanation:

$c = \frac{E_{0}}{B_{0}}$
$\Rightarrow B_{0} = \frac{E_{0}}{c} = \frac{48}{3 \times 10^{8}} = 1.6 \times 10^{- 7} T$ .
Correct option is (2).

NEET - 2023

PHXI15:WAVES

358886 The velocity of electromagnetic wave is along the direction of:

1

\vec{B}

2

\frac{\vec{E} \times \vec{B}}{μ_{0}}

3

\frac{\vec{B} \times \vec{E}}{μ_{0}}

4

E

Explanation:

An electromagnetic wave is the wave composed of the oscillations of electric and magnetic fields in mutually perpendicular planes and these oscllilations are perpendicular to the direction of propagations of wave.
The direction of propagation of electromagnetic wave is given by poynting vector
Along $\vec{S} = \vec{E} \times \vec{H} = \frac{\vec{E} \times \vec{B}}{μ_{0}}$ this is parallel to $\vec{E} \times \vec{B}$

PHXI15:WAVES

358887 In an apparatus, the electric field was found to oscillate with amplitude of $18 V / m$ . The amplitude of the oscillating magnetic field will be

1

11 \times 10^{- 11} T

2

6 \times 10^{- 8} T

3

4 \times 10^{- 6} T

4

9 \times 10^{- 9} T

Explanation:

$B_{0} = \frac{E_{0}}{c} = \frac{18}{3 \times 10^{8}} = 6 \times 10^{- 8} T$

PHXI15:WAVES

358888 The electric field in an electromagnetic wave is given by $\vec{E} = \hat{i} 40 \cos ω (t - \frac{z}{c}) N C^{- 1}$ . The magnetic field induction of this wave is (in $S I$ unit)

1

\vec{B} = \hat{k} \frac{40}{c} \cos ω (t - \frac{z}{c})

2

\vec{B} = \hat{i} \frac{40}{c} \cos ω (t - \frac{z}{c})

3

\vec{B} = \hat{j} 40 \cos ω (t - \frac{z}{c})

4

\vec{B} = \hat{j} \frac{40}{c} \cos ω (t - \frac{z}{c})

Explanation:

$\vec{E} = \hat{i} 40 \cos ω (t - \frac{z}{c}) N / C$
$\vec{E}$ is along $+ X$ axis: $\vec{v}$ is along $+ Z$ axis ;
Direction of $E M$ wave is given by
$\vec{v} = \vec{E} \times \vec{B}$
So, $\vec{B}$ is along $+ Y$ axis, $\frac{E}{B} = c$
$\Rightarrow B = \frac{E}{c} = \frac{40}{3 \times 10^{8}}$
$∴ \vec{B} = \frac{40}{c} \hat{j} \cos ω (t - \frac{z}{c})$

JEE - 2024

PHXI15:WAVES

358889 The amplitude of magnetic field in an electromagnetic wave propagating along $y$ -axis is $6.0 \times 10^{- 7} T$ . The maximum value of electric field in the electromagnetic wave is

1

2 \times 10^{15} V m^{- 1}

2

6.0 \times 10^{- 7} V m^{- 1}

3

5 \times 10^{14} V m^{- 1}

4

180 V m^{- 1}

Explanation:

For a electromagnetic wave, amplitude of electric and magnetic field are related by,
$\frac{E_{m}}{B_{m}} = c$
$∴ E_{m} = B_{m} \times C$
$= 6 \times 10^{- 7} \times 3 \times 10^{8} = 18 \times 10; E = 180 {V m}^{- 1}$

JEE - 2023

PHXI15:WAVES

358890 In a plane electromagnetic wave travelling in free space, the elctric field component oscillates sinusoidally at a frequency of $2.0 \times 10^{10} H z$ and amplitude $48 V m^{- 1}$ . Then the amplitude of oscillating magnetic field is: (Speed of light in free space $= 3 \times 10^{8} m s^{- 1})$

1

1.6 \times 10^{8} T

2

1.6 \times 10^{- 7} T

3

1.6 \times 10^{- 6} T

4

1.6 \times 10^{- 9} T

Explanation:

$c = \frac{E_{0}}{B_{0}}$
$\Rightarrow B_{0} = \frac{E_{0}}{c} = \frac{48}{3 \times 10^{8}} = 1.6 \times 10^{- 7} T$ .
Correct option is (2).

NEET - 2023

PHXI15:WAVES

358886 The velocity of electromagnetic wave is along the direction of:

1

\vec{B}

2

\frac{\vec{E} \times \vec{B}}{μ_{0}}

3

\frac{\vec{B} \times \vec{E}}{μ_{0}}

4

E

Explanation:

An electromagnetic wave is the wave composed of the oscillations of electric and magnetic fields in mutually perpendicular planes and these oscllilations are perpendicular to the direction of propagations of wave.
The direction of propagation of electromagnetic wave is given by poynting vector
Along $\vec{S} = \vec{E} \times \vec{H} = \frac{\vec{E} \times \vec{B}}{μ_{0}}$ this is parallel to $\vec{E} \times \vec{B}$

PHXI15:WAVES

358887 In an apparatus, the electric field was found to oscillate with amplitude of $18 V / m$ . The amplitude of the oscillating magnetic field will be

1

11 \times 10^{- 11} T

2

6 \times 10^{- 8} T

3

4 \times 10^{- 6} T

4

9 \times 10^{- 9} T

Explanation:

$B_{0} = \frac{E_{0}}{c} = \frac{18}{3 \times 10^{8}} = 6 \times 10^{- 8} T$

PHXI15:WAVES

358888 The electric field in an electromagnetic wave is given by $\vec{E} = \hat{i} 40 \cos ω (t - \frac{z}{c}) N C^{- 1}$ . The magnetic field induction of this wave is (in $S I$ unit)

1

\vec{B} = \hat{k} \frac{40}{c} \cos ω (t - \frac{z}{c})

2

\vec{B} = \hat{i} \frac{40}{c} \cos ω (t - \frac{z}{c})

3

\vec{B} = \hat{j} 40 \cos ω (t - \frac{z}{c})

4

\vec{B} = \hat{j} \frac{40}{c} \cos ω (t - \frac{z}{c})

Explanation:

$\vec{E} = \hat{i} 40 \cos ω (t - \frac{z}{c}) N / C$
$\vec{E}$ is along $+ X$ axis: $\vec{v}$ is along $+ Z$ axis ;
Direction of $E M$ wave is given by
$\vec{v} = \vec{E} \times \vec{B}$
So, $\vec{B}$ is along $+ Y$ axis, $\frac{E}{B} = c$
$\Rightarrow B = \frac{E}{c} = \frac{40}{3 \times 10^{8}}$
$∴ \vec{B} = \frac{40}{c} \hat{j} \cos ω (t - \frac{z}{c})$

JEE - 2024

PHXI15:WAVES

358889 The amplitude of magnetic field in an electromagnetic wave propagating along $y$ -axis is $6.0 \times 10^{- 7} T$ . The maximum value of electric field in the electromagnetic wave is

1

2 \times 10^{15} V m^{- 1}

2

6.0 \times 10^{- 7} V m^{- 1}

3

5 \times 10^{14} V m^{- 1}

4

180 V m^{- 1}

Explanation:

For a electromagnetic wave, amplitude of electric and magnetic field are related by,
$\frac{E_{m}}{B_{m}} = c$
$∴ E_{m} = B_{m} \times C$
$= 6 \times 10^{- 7} \times 3 \times 10^{8} = 18 \times 10; E = 180 {V m}^{- 1}$

JEE - 2023

PHXI15:WAVES

358890 In a plane electromagnetic wave travelling in free space, the elctric field component oscillates sinusoidally at a frequency of $2.0 \times 10^{10} H z$ and amplitude $48 V m^{- 1}$ . Then the amplitude of oscillating magnetic field is: (Speed of light in free space $= 3 \times 10^{8} m s^{- 1})$

1

1.6 \times 10^{8} T

2

1.6 \times 10^{- 7} T

3

1.6 \times 10^{- 6} T

4

1.6 \times 10^{- 9} T

Explanation:

$c = \frac{E_{0}}{B_{0}}$
$\Rightarrow B_{0} = \frac{E_{0}}{c} = \frac{48}{3 \times 10^{8}} = 1.6 \times 10^{- 7} T$ .
Correct option is (2).

NEET - 2023

PHXI15:WAVES

358886 The velocity of electromagnetic wave is along the direction of:

1

\vec{B}

2

\frac{\vec{E} \times \vec{B}}{μ_{0}}

3

\frac{\vec{B} \times \vec{E}}{μ_{0}}

4

E

Explanation:

An electromagnetic wave is the wave composed of the oscillations of electric and magnetic fields in mutually perpendicular planes and these oscllilations are perpendicular to the direction of propagations of wave.
The direction of propagation of electromagnetic wave is given by poynting vector
Along $\vec{S} = \vec{E} \times \vec{H} = \frac{\vec{E} \times \vec{B}}{μ_{0}}$ this is parallel to $\vec{E} \times \vec{B}$

PHXI15:WAVES

358887 In an apparatus, the electric field was found to oscillate with amplitude of $18 V / m$ . The amplitude of the oscillating magnetic field will be

1

11 \times 10^{- 11} T

2

6 \times 10^{- 8} T

3

4 \times 10^{- 6} T

4

9 \times 10^{- 9} T

Explanation:

$B_{0} = \frac{E_{0}}{c} = \frac{18}{3 \times 10^{8}} = 6 \times 10^{- 8} T$

PHXI15:WAVES

358888 The electric field in an electromagnetic wave is given by $\vec{E} = \hat{i} 40 \cos ω (t - \frac{z}{c}) N C^{- 1}$ . The magnetic field induction of this wave is (in $S I$ unit)

1

\vec{B} = \hat{k} \frac{40}{c} \cos ω (t - \frac{z}{c})

2

\vec{B} = \hat{i} \frac{40}{c} \cos ω (t - \frac{z}{c})

3

\vec{B} = \hat{j} 40 \cos ω (t - \frac{z}{c})

4

\vec{B} = \hat{j} \frac{40}{c} \cos ω (t - \frac{z}{c})

Explanation:

$\vec{E} = \hat{i} 40 \cos ω (t - \frac{z}{c}) N / C$
$\vec{E}$ is along $+ X$ axis: $\vec{v}$ is along $+ Z$ axis ;
Direction of $E M$ wave is given by
$\vec{v} = \vec{E} \times \vec{B}$
So, $\vec{B}$ is along $+ Y$ axis, $\frac{E}{B} = c$
$\Rightarrow B = \frac{E}{c} = \frac{40}{3 \times 10^{8}}$
$∴ \vec{B} = \frac{40}{c} \hat{j} \cos ω (t - \frac{z}{c})$

JEE - 2024

PHXI15:WAVES

358889 The amplitude of magnetic field in an electromagnetic wave propagating along $y$ -axis is $6.0 \times 10^{- 7} T$ . The maximum value of electric field in the electromagnetic wave is

1

2 \times 10^{15} V m^{- 1}

2

6.0 \times 10^{- 7} V m^{- 1}

3

5 \times 10^{14} V m^{- 1}

4

180 V m^{- 1}

Explanation:

For a electromagnetic wave, amplitude of electric and magnetic field are related by,
$\frac{E_{m}}{B_{m}} = c$
$∴ E_{m} = B_{m} \times C$
$= 6 \times 10^{- 7} \times 3 \times 10^{8} = 18 \times 10; E = 180 {V m}^{- 1}$

JEE - 2023

PHXI15:WAVES

358890 In a plane electromagnetic wave travelling in free space, the elctric field component oscillates sinusoidally at a frequency of $2.0 \times 10^{10} H z$ and amplitude $48 V m^{- 1}$ . Then the amplitude of oscillating magnetic field is: (Speed of light in free space $= 3 \times 10^{8} m s^{- 1})$

1

1.6 \times 10^{8} T

2

1.6 \times 10^{- 7} T

3

1.6 \times 10^{- 6} T

4

1.6 \times 10^{- 9} T

Explanation:

$c = \frac{E_{0}}{B_{0}}$
$\Rightarrow B_{0} = \frac{E_{0}}{c} = \frac{48}{3 \times 10^{8}} = 1.6 \times 10^{- 7} T$ .
Correct option is (2).

NEET - 2023

PHXI15:WAVES

358886 The velocity of electromagnetic wave is along the direction of:

1

\vec{B}

2

\frac{\vec{E} \times \vec{B}}{μ_{0}}

3

\frac{\vec{B} \times \vec{E}}{μ_{0}}

4

E

Explanation:

An electromagnetic wave is the wave composed of the oscillations of electric and magnetic fields in mutually perpendicular planes and these oscllilations are perpendicular to the direction of propagations of wave.
The direction of propagation of electromagnetic wave is given by poynting vector
Along $\vec{S} = \vec{E} \times \vec{H} = \frac{\vec{E} \times \vec{B}}{μ_{0}}$ this is parallel to $\vec{E} \times \vec{B}$

PHXI15:WAVES

358887 In an apparatus, the electric field was found to oscillate with amplitude of $18 V / m$ . The amplitude of the oscillating magnetic field will be

1

11 \times 10^{- 11} T

2

6 \times 10^{- 8} T

3

4 \times 10^{- 6} T

4

9 \times 10^{- 9} T

Explanation:

$B_{0} = \frac{E_{0}}{c} = \frac{18}{3 \times 10^{8}} = 6 \times 10^{- 8} T$

PHXI15:WAVES

358888 The electric field in an electromagnetic wave is given by $\vec{E} = \hat{i} 40 \cos ω (t - \frac{z}{c}) N C^{- 1}$ . The magnetic field induction of this wave is (in $S I$ unit)

1

\vec{B} = \hat{k} \frac{40}{c} \cos ω (t - \frac{z}{c})

2

\vec{B} = \hat{i} \frac{40}{c} \cos ω (t - \frac{z}{c})

3

\vec{B} = \hat{j} 40 \cos ω (t - \frac{z}{c})

4

\vec{B} = \hat{j} \frac{40}{c} \cos ω (t - \frac{z}{c})

Explanation:

$\vec{E} = \hat{i} 40 \cos ω (t - \frac{z}{c}) N / C$
$\vec{E}$ is along $+ X$ axis: $\vec{v}$ is along $+ Z$ axis ;
Direction of $E M$ wave is given by
$\vec{v} = \vec{E} \times \vec{B}$
So, $\vec{B}$ is along $+ Y$ axis, $\frac{E}{B} = c$
$\Rightarrow B = \frac{E}{c} = \frac{40}{3 \times 10^{8}}$
$∴ \vec{B} = \frac{40}{c} \hat{j} \cos ω (t - \frac{z}{c})$

JEE - 2024

PHXI15:WAVES

358889 The amplitude of magnetic field in an electromagnetic wave propagating along $y$ -axis is $6.0 \times 10^{- 7} T$ . The maximum value of electric field in the electromagnetic wave is

1

2 \times 10^{15} V m^{- 1}

2

6.0 \times 10^{- 7} V m^{- 1}

3

5 \times 10^{14} V m^{- 1}

4

180 V m^{- 1}

Explanation:

For a electromagnetic wave, amplitude of electric and magnetic field are related by,
$\frac{E_{m}}{B_{m}} = c$
$∴ E_{m} = B_{m} \times C$
$= 6 \times 10^{- 7} \times 3 \times 10^{8} = 18 \times 10; E = 180 {V m}^{- 1}$

JEE - 2023

PHXI15:WAVES

358890 In a plane electromagnetic wave travelling in free space, the elctric field component oscillates sinusoidally at a frequency of $2.0 \times 10^{10} H z$ and amplitude $48 V m^{- 1}$ . Then the amplitude of oscillating magnetic field is: (Speed of light in free space $= 3 \times 10^{8} m s^{- 1})$

1

1.6 \times 10^{8} T

2

1.6 \times 10^{- 7} T

3

1.6 \times 10^{- 6} T

4

1.6 \times 10^{- 9} T

Explanation:

$c = \frac{E_{0}}{B_{0}}$
$\Rightarrow B_{0} = \frac{E_{0}}{c} = \frac{48}{3 \times 10^{8}} = 1.6 \times 10^{- 7} T$ .
Correct option is (2).

NEET - 2023

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