QBManager

Electromagnetic Wave

155579 If the speed of a wave doubles as it passes from shallow water into deeper water, its wavelength will be

1 unchanged

2 halved

3 doubled

4 quadrupled

Explanation:

C Given,
$Initial velocity, v_{1} = v$
$Final velocity, v_{2} = 2 v$
$∵ v = f λ$
Frequency is always constant.
$\frac{v_{1}}{v_{2}} = \frac{λ_{1}}{λ_{2}}$
$λ_{2} = \frac{v_{2} λ_{1}}{v_{1}}$
$λ_{2} = \frac{2 v}{v} λ_{1}$
$λ_{2} = 2 λ_{1}$

VITEEE-2008

Electromagnetic Wave

155580 A new system of units is evolved in which the values of $μ_{0}$ and $ε_{0}$ are 2 and 8 respectively. Then the speed of light in this system will be

1 0.25

2 0.5

3 0.75

4 1

Explanation:

A Given,
$μ_{0} = 2$
$ε_{0} = 8$
Speed of the light
$v = \frac{1}{\sqrt{μ_{0} ε_{0}}} = \frac{1}{\sqrt{2 \times 8}}$
$v = \frac{1}{\sqrt{16}}$
$v = \frac{1}{4} = 0.25$
$v = 0.25$

VITEEE-2006

Electromagnetic Wave

155584 A beam of light travelling along $X$ -axis is described by the electric field $E Y = 600 \frac{V}{m}$ . $\sin ω (t - \frac{x}{c})$ . The maximum magnetic force on a charge $q = 2 e$ , moving along $Y$ -axis with the speed of $3 \times 10^{8} m / s$ is

1

19.2 \times 10^{- 17} N

2

1.92 \times 10^{- 17} N

3

0.192 N

4 None of these

Explanation:

A Given,
$E_{y} = 600 \frac{V}{m} \sin ω (t - \frac{x}{c})$
On comparing with standard equation, we get -
$E = E_{0} \sin ω (t - \frac{x}{c})$
$E_{0} = 600 V / m$
We know that,
$c = \frac{E_{0}}{B_{0}}$
$B_{0} = \frac{E_{0}}{c} = \frac{600}{3 \times 10^{8}}$
$B_{0} = 2 \times 10^{- 6} T$
Maximum force,
$F_{m} = qv B_{0}$
$= 2 e v B_{0} (∵ q = 2 e)$
$= 2 \times 1.6 \times 10^{- 19} \times 3 \times 10^{8} \times 2 \times 10^{- 6}$
$F_{m} = 19.2 \times 10^{- 17} N$

JIPMER-2016

Electromagnetic Wave

155585 The electric field for a electromagnetic wave in free space is $E = i 30 \cos (k z - 5 \times 10^{8} t)$ , magnitude of wave vector $k$ is (velocity of em wave in free space $= 3 \times 10^{8} m / s$ )

1

0.46 rad m^{- 1}

2

3 rad m^{- 1}

3

1.66 rad m^{- 1}

4

0.83 rad m^{- 1}

Explanation:

C Given,
$E = 30 \cos (kz - 5 \times 10^{8} t) \hat{i}$
From the above equation, we get -
$ω = 5 \times 10^{8} rad / s$
Velocity of EM wave $(v) = 3 \times 10^{8} m / s$
$∵ v = \frac{ω}{k}$
$k = \frac{ω}{v} = \frac{5 \times 10^{8}}{3 \times 10^{8}}$
$k = 1.66 rad / m$

AP EAMCET -2014

Electromagnetic Wave

155579 If the speed of a wave doubles as it passes from shallow water into deeper water, its wavelength will be

1 unchanged

2 halved

3 doubled

4 quadrupled

Explanation:

C Given,
$Initial velocity, v_{1} = v$
$Final velocity, v_{2} = 2 v$
$∵ v = f λ$
Frequency is always constant.
$\frac{v_{1}}{v_{2}} = \frac{λ_{1}}{λ_{2}}$
$λ_{2} = \frac{v_{2} λ_{1}}{v_{1}}$
$λ_{2} = \frac{2 v}{v} λ_{1}$
$λ_{2} = 2 λ_{1}$

VITEEE-2008

Electromagnetic Wave

155580 A new system of units is evolved in which the values of $μ_{0}$ and $ε_{0}$ are 2 and 8 respectively. Then the speed of light in this system will be

1 0.25

2 0.5

3 0.75

4 1

Explanation:

A Given,
$μ_{0} = 2$
$ε_{0} = 8$
Speed of the light
$v = \frac{1}{\sqrt{μ_{0} ε_{0}}} = \frac{1}{\sqrt{2 \times 8}}$
$v = \frac{1}{\sqrt{16}}$
$v = \frac{1}{4} = 0.25$
$v = 0.25$

VITEEE-2006

Electromagnetic Wave

155584 A beam of light travelling along $X$ -axis is described by the electric field $E Y = 600 \frac{V}{m}$ . $\sin ω (t - \frac{x}{c})$ . The maximum magnetic force on a charge $q = 2 e$ , moving along $Y$ -axis with the speed of $3 \times 10^{8} m / s$ is

1

19.2 \times 10^{- 17} N

2

1.92 \times 10^{- 17} N

3

0.192 N

4 None of these

Explanation:

A Given,
$E_{y} = 600 \frac{V}{m} \sin ω (t - \frac{x}{c})$
On comparing with standard equation, we get -
$E = E_{0} \sin ω (t - \frac{x}{c})$
$E_{0} = 600 V / m$
We know that,
$c = \frac{E_{0}}{B_{0}}$
$B_{0} = \frac{E_{0}}{c} = \frac{600}{3 \times 10^{8}}$
$B_{0} = 2 \times 10^{- 6} T$
Maximum force,
$F_{m} = qv B_{0}$
$= 2 e v B_{0} (∵ q = 2 e)$
$= 2 \times 1.6 \times 10^{- 19} \times 3 \times 10^{8} \times 2 \times 10^{- 6}$
$F_{m} = 19.2 \times 10^{- 17} N$

JIPMER-2016

Electromagnetic Wave

155585 The electric field for a electromagnetic wave in free space is $E = i 30 \cos (k z - 5 \times 10^{8} t)$ , magnitude of wave vector $k$ is (velocity of em wave in free space $= 3 \times 10^{8} m / s$ )

1

0.46 rad m^{- 1}

2

3 rad m^{- 1}

3

1.66 rad m^{- 1}

4

0.83 rad m^{- 1}

Explanation:

C Given,
$E = 30 \cos (kz - 5 \times 10^{8} t) \hat{i}$
From the above equation, we get -
$ω = 5 \times 10^{8} rad / s$
Velocity of EM wave $(v) = 3 \times 10^{8} m / s$
$∵ v = \frac{ω}{k}$
$k = \frac{ω}{v} = \frac{5 \times 10^{8}}{3 \times 10^{8}}$
$k = 1.66 rad / m$

AP EAMCET -2014

Electromagnetic Wave

155579 If the speed of a wave doubles as it passes from shallow water into deeper water, its wavelength will be

1 unchanged

2 halved

3 doubled

4 quadrupled

Explanation:

C Given,
$Initial velocity, v_{1} = v$
$Final velocity, v_{2} = 2 v$
$∵ v = f λ$
Frequency is always constant.
$\frac{v_{1}}{v_{2}} = \frac{λ_{1}}{λ_{2}}$
$λ_{2} = \frac{v_{2} λ_{1}}{v_{1}}$
$λ_{2} = \frac{2 v}{v} λ_{1}$
$λ_{2} = 2 λ_{1}$

VITEEE-2008

Electromagnetic Wave

155580 A new system of units is evolved in which the values of $μ_{0}$ and $ε_{0}$ are 2 and 8 respectively. Then the speed of light in this system will be

1 0.25

2 0.5

3 0.75

4 1

Explanation:

A Given,
$μ_{0} = 2$
$ε_{0} = 8$
Speed of the light
$v = \frac{1}{\sqrt{μ_{0} ε_{0}}} = \frac{1}{\sqrt{2 \times 8}}$
$v = \frac{1}{\sqrt{16}}$
$v = \frac{1}{4} = 0.25$
$v = 0.25$

VITEEE-2006

Electromagnetic Wave

155584 A beam of light travelling along $X$ -axis is described by the electric field $E Y = 600 \frac{V}{m}$ . $\sin ω (t - \frac{x}{c})$ . The maximum magnetic force on a charge $q = 2 e$ , moving along $Y$ -axis with the speed of $3 \times 10^{8} m / s$ is

1

19.2 \times 10^{- 17} N

2

1.92 \times 10^{- 17} N

3

0.192 N

4 None of these

Explanation:

A Given,
$E_{y} = 600 \frac{V}{m} \sin ω (t - \frac{x}{c})$
On comparing with standard equation, we get -
$E = E_{0} \sin ω (t - \frac{x}{c})$
$E_{0} = 600 V / m$
We know that,
$c = \frac{E_{0}}{B_{0}}$
$B_{0} = \frac{E_{0}}{c} = \frac{600}{3 \times 10^{8}}$
$B_{0} = 2 \times 10^{- 6} T$
Maximum force,
$F_{m} = qv B_{0}$
$= 2 e v B_{0} (∵ q = 2 e)$
$= 2 \times 1.6 \times 10^{- 19} \times 3 \times 10^{8} \times 2 \times 10^{- 6}$
$F_{m} = 19.2 \times 10^{- 17} N$

JIPMER-2016

Electromagnetic Wave

155585 The electric field for a electromagnetic wave in free space is $E = i 30 \cos (k z - 5 \times 10^{8} t)$ , magnitude of wave vector $k$ is (velocity of em wave in free space $= 3 \times 10^{8} m / s$ )

1

0.46 rad m^{- 1}

2

3 rad m^{- 1}

3

1.66 rad m^{- 1}

4

0.83 rad m^{- 1}

Explanation:

C Given,
$E = 30 \cos (kz - 5 \times 10^{8} t) \hat{i}$
From the above equation, we get -
$ω = 5 \times 10^{8} rad / s$
Velocity of EM wave $(v) = 3 \times 10^{8} m / s$
$∵ v = \frac{ω}{k}$
$k = \frac{ω}{v} = \frac{5 \times 10^{8}}{3 \times 10^{8}}$
$k = 1.66 rad / m$

AP EAMCET -2014

Electromagnetic Wave

155579 If the speed of a wave doubles as it passes from shallow water into deeper water, its wavelength will be

1 unchanged

2 halved

3 doubled

4 quadrupled

Explanation:

C Given,
$Initial velocity, v_{1} = v$
$Final velocity, v_{2} = 2 v$
$∵ v = f λ$
Frequency is always constant.
$\frac{v_{1}}{v_{2}} = \frac{λ_{1}}{λ_{2}}$
$λ_{2} = \frac{v_{2} λ_{1}}{v_{1}}$
$λ_{2} = \frac{2 v}{v} λ_{1}$
$λ_{2} = 2 λ_{1}$

VITEEE-2008

Electromagnetic Wave

155580 A new system of units is evolved in which the values of $μ_{0}$ and $ε_{0}$ are 2 and 8 respectively. Then the speed of light in this system will be

1 0.25

2 0.5

3 0.75

4 1

Explanation:

A Given,
$μ_{0} = 2$
$ε_{0} = 8$
Speed of the light
$v = \frac{1}{\sqrt{μ_{0} ε_{0}}} = \frac{1}{\sqrt{2 \times 8}}$
$v = \frac{1}{\sqrt{16}}$
$v = \frac{1}{4} = 0.25$
$v = 0.25$

VITEEE-2006

Electromagnetic Wave

155584 A beam of light travelling along $X$ -axis is described by the electric field $E Y = 600 \frac{V}{m}$ . $\sin ω (t - \frac{x}{c})$ . The maximum magnetic force on a charge $q = 2 e$ , moving along $Y$ -axis with the speed of $3 \times 10^{8} m / s$ is

1

19.2 \times 10^{- 17} N

2

1.92 \times 10^{- 17} N

3

0.192 N

4 None of these

Explanation:

A Given,
$E_{y} = 600 \frac{V}{m} \sin ω (t - \frac{x}{c})$
On comparing with standard equation, we get -
$E = E_{0} \sin ω (t - \frac{x}{c})$
$E_{0} = 600 V / m$
We know that,
$c = \frac{E_{0}}{B_{0}}$
$B_{0} = \frac{E_{0}}{c} = \frac{600}{3 \times 10^{8}}$
$B_{0} = 2 \times 10^{- 6} T$
Maximum force,
$F_{m} = qv B_{0}$
$= 2 e v B_{0} (∵ q = 2 e)$
$= 2 \times 1.6 \times 10^{- 19} \times 3 \times 10^{8} \times 2 \times 10^{- 6}$
$F_{m} = 19.2 \times 10^{- 17} N$

JIPMER-2016

Electromagnetic Wave

155585 The electric field for a electromagnetic wave in free space is $E = i 30 \cos (k z - 5 \times 10^{8} t)$ , magnitude of wave vector $k$ is (velocity of em wave in free space $= 3 \times 10^{8} m / s$ )

1

0.46 rad m^{- 1}

2

3 rad m^{- 1}

3

1.66 rad m^{- 1}

4

0.83 rad m^{- 1}

Explanation:

C Given,
$E = 30 \cos (kz - 5 \times 10^{8} t) \hat{i}$
From the above equation, we get -
$ω = 5 \times 10^{8} rad / s$
Velocity of EM wave $(v) = 3 \times 10^{8} m / s$
$∵ v = \frac{ω}{k}$
$k = \frac{ω}{v} = \frac{5 \times 10^{8}}{3 \times 10^{8}}$
$k = 1.66 rad / m$

AP EAMCET -2014

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