QBManager

Electromagnetic Wave

155562 A plane electromagnetic wave propagating in $x$ $(-)$ direction as a wave function (in SI units) is given as $ψ (x, t) = 10^{3} \sin π (3 \times 10^{6} x - 9 \times 10^{14} t)$ . The speed of the wave is

1

3 \times 10^{6} m / s

2

3 \times 10^{7} m / s

3

3 \times 10^{8} m / s

4

9 \times 10^{14} m / s

Explanation:

C Given,
$ψ (x, t) = 10^{3} \sin π (3 \times 10^{6} x - 9 \times 10^{14} t)$
General equation,
$E = E_{o} \sin (kx - ω t)$
On comparison equation (i) and (ii) we get-
$k = 3 \times 10^{6} m^{- 1}$
$ω = 9 \times 10^{14} rad / \sec$
$∴ v = \frac{ω}{K} = \frac{9 \times 10^{14}}{3 \times 10^{6}}$
$v = 3 \times 10^{8} m / s$

JCECE-2013

Electromagnetic Wave

155566 An electromagnetic wave of frequency $v = 3.0 MHz$ passes from vacuum into a dielectric medium with relative permittivity $\in= 4.0$ . Then

1 Wavelength is doubled and frequency becomes half

2 Wavelength is halved and frequency remains unchanged

3 Wavelength is frequency both remain unchanged

4 Wavelength is doubled and frequency unchanged.

Explanation:

B Velocity wave in medium
$v_{medium} = \frac{1}{\sqrt{μ_{o} μ_{r} ε_{o} ε_{r}}}$
For air, $μ_{r} = 1$
$v_{medium} = \frac{1}{\sqrt{μ_{o} ε_{o} 4}} = \frac{1}{2} \frac{1}{\sqrt{μ_{o} ε_{o}}}$
We know that,
$c = \frac{1}{\sqrt{μ_{o} ε_{o}}}$
(in vacuum)
$v_{medium} = \frac{c}{2}$
$\frac{λ_{medium}}{λ_{air}} = \frac{v_{medium}}{v_{air}} = \frac{c / 2}{c} = \frac{1}{2}$
$λ_{medium} = \frac{λ_{air}}{2}$
Wavelength is halved and frequency remains unchanged.

COMEDK 2013

Electromagnetic Wave

155568 Assertion: Dipole
electromagnetic waves
oscillations produce
Reason: Accelerated
charge produces
electromagnetic waves.

1 If both Assertion and Reason are correct and the Reason is a correct explanation of the Assertion.

2 If both Assertion and Reason are correct but Reason in not a correct explanation of the Assertion.

3 If the Assertion is correct but Reason is incorrect.

4 If both the Assertion and Reason are incorrect.

5 If the Assertion is incorrect but the Reason is correct.

Explanation:

B Oscillating charge between dipolar electric field produces EM wave.
$∴$ Assertion is correct.
Electromagnetic wave is produce by a accelerated charge particle.
Reason is also correct.
But it is not correct explanation of assertion.

AIIMS-2007

Electromagnetic Wave

155573 The magnetic field amplitude of an electromagnetic wave is $2 \times 10^{- 7} T$ . Its electric field amplitude, if the wave is travelling in free space is :

1

6 {Vm}^{- 1}

2

60 {Vm}^{- 1}

3

\frac{1}{6} {Vm}^{- 1}

4 none of these

Explanation:

B Given,
Magnetic field $B_{o} = 2 \times 10^{- 7} T$
We know that,
$E_{o} = {cB}_{o}$
$E_{o} = 3 \times 10^{8} \times 2 \times 10^{- 7}$
$E_{o} = 6 \times 10 = 60 V / m$

BCECE-2006

Electromagnetic Wave

155577 Find the value of magnetic field between plates of capacitor at a distance $1 m$ from centre, where electric field varies by $10^{10} V / m$ per second.

1

5.56 \times 10^{- 8} T

2

5.56 \times 10^{- 3} T

3

5.56 μ T

4

5.55 T

Explanation:

A Given,
$\frac{dE}{dt} = 10^{10} V / m$
Magnetic field between plates of capacitor
$B = \frac{μ_{o} ε_{o}}{2} \cdot \frac{dE}{dt}$
$∴ B = \frac{4 π \times 10^{- 7} \times 8.85 \times 10^{- 12} \times 10^{10}}{2}$
$B = 5.56 \times 10^{- 8} T$

VITEEE-2010

Electromagnetic Wave

155562 A plane electromagnetic wave propagating in $x$ $(-)$ direction as a wave function (in SI units) is given as $ψ (x, t) = 10^{3} \sin π (3 \times 10^{6} x - 9 \times 10^{14} t)$ . The speed of the wave is

1

3 \times 10^{6} m / s

2

3 \times 10^{7} m / s

3

3 \times 10^{8} m / s

4

9 \times 10^{14} m / s

Explanation:

C Given,
$ψ (x, t) = 10^{3} \sin π (3 \times 10^{6} x - 9 \times 10^{14} t)$
General equation,
$E = E_{o} \sin (kx - ω t)$
On comparison equation (i) and (ii) we get-
$k = 3 \times 10^{6} m^{- 1}$
$ω = 9 \times 10^{14} rad / \sec$
$∴ v = \frac{ω}{K} = \frac{9 \times 10^{14}}{3 \times 10^{6}}$
$v = 3 \times 10^{8} m / s$

JCECE-2013

Electromagnetic Wave

155566 An electromagnetic wave of frequency $v = 3.0 MHz$ passes from vacuum into a dielectric medium with relative permittivity $\in= 4.0$ . Then

1 Wavelength is doubled and frequency becomes half

2 Wavelength is halved and frequency remains unchanged

3 Wavelength is frequency both remain unchanged

4 Wavelength is doubled and frequency unchanged.

Explanation:

B Velocity wave in medium
$v_{medium} = \frac{1}{\sqrt{μ_{o} μ_{r} ε_{o} ε_{r}}}$
For air, $μ_{r} = 1$
$v_{medium} = \frac{1}{\sqrt{μ_{o} ε_{o} 4}} = \frac{1}{2} \frac{1}{\sqrt{μ_{o} ε_{o}}}$
We know that,
$c = \frac{1}{\sqrt{μ_{o} ε_{o}}}$
(in vacuum)
$v_{medium} = \frac{c}{2}$
$\frac{λ_{medium}}{λ_{air}} = \frac{v_{medium}}{v_{air}} = \frac{c / 2}{c} = \frac{1}{2}$
$λ_{medium} = \frac{λ_{air}}{2}$
Wavelength is halved and frequency remains unchanged.

COMEDK 2013

Electromagnetic Wave

155568 Assertion: Dipole
electromagnetic waves
oscillations produce
Reason: Accelerated
charge produces
electromagnetic waves.

1 If both Assertion and Reason are correct and the Reason is a correct explanation of the Assertion.

2 If both Assertion and Reason are correct but Reason in not a correct explanation of the Assertion.

3 If the Assertion is correct but Reason is incorrect.

4 If both the Assertion and Reason are incorrect.

5 If the Assertion is incorrect but the Reason is correct.

Explanation:

B Oscillating charge between dipolar electric field produces EM wave.
$∴$ Assertion is correct.
Electromagnetic wave is produce by a accelerated charge particle.
Reason is also correct.
But it is not correct explanation of assertion.

AIIMS-2007

Electromagnetic Wave

155573 The magnetic field amplitude of an electromagnetic wave is $2 \times 10^{- 7} T$ . Its electric field amplitude, if the wave is travelling in free space is :

1

6 {Vm}^{- 1}

2

60 {Vm}^{- 1}

3

\frac{1}{6} {Vm}^{- 1}

4 none of these

Explanation:

B Given,
Magnetic field $B_{o} = 2 \times 10^{- 7} T$
We know that,
$E_{o} = {cB}_{o}$
$E_{o} = 3 \times 10^{8} \times 2 \times 10^{- 7}$
$E_{o} = 6 \times 10 = 60 V / m$

BCECE-2006

Electromagnetic Wave

155577 Find the value of magnetic field between plates of capacitor at a distance $1 m$ from centre, where electric field varies by $10^{10} V / m$ per second.

1

5.56 \times 10^{- 8} T

2

5.56 \times 10^{- 3} T

3

5.56 μ T

4

5.55 T

Explanation:

A Given,
$\frac{dE}{dt} = 10^{10} V / m$
Magnetic field between plates of capacitor
$B = \frac{μ_{o} ε_{o}}{2} \cdot \frac{dE}{dt}$
$∴ B = \frac{4 π \times 10^{- 7} \times 8.85 \times 10^{- 12} \times 10^{10}}{2}$
$B = 5.56 \times 10^{- 8} T$

VITEEE-2010

Electromagnetic Wave

155562 A plane electromagnetic wave propagating in $x$ $(-)$ direction as a wave function (in SI units) is given as $ψ (x, t) = 10^{3} \sin π (3 \times 10^{6} x - 9 \times 10^{14} t)$ . The speed of the wave is

1

3 \times 10^{6} m / s

2

3 \times 10^{7} m / s

3

3 \times 10^{8} m / s

4

9 \times 10^{14} m / s

Explanation:

C Given,
$ψ (x, t) = 10^{3} \sin π (3 \times 10^{6} x - 9 \times 10^{14} t)$
General equation,
$E = E_{o} \sin (kx - ω t)$
On comparison equation (i) and (ii) we get-
$k = 3 \times 10^{6} m^{- 1}$
$ω = 9 \times 10^{14} rad / \sec$
$∴ v = \frac{ω}{K} = \frac{9 \times 10^{14}}{3 \times 10^{6}}$
$v = 3 \times 10^{8} m / s$

JCECE-2013

Electromagnetic Wave

155566 An electromagnetic wave of frequency $v = 3.0 MHz$ passes from vacuum into a dielectric medium with relative permittivity $\in= 4.0$ . Then

1 Wavelength is doubled and frequency becomes half

2 Wavelength is halved and frequency remains unchanged

3 Wavelength is frequency both remain unchanged

4 Wavelength is doubled and frequency unchanged.

Explanation:

B Velocity wave in medium
$v_{medium} = \frac{1}{\sqrt{μ_{o} μ_{r} ε_{o} ε_{r}}}$
For air, $μ_{r} = 1$
$v_{medium} = \frac{1}{\sqrt{μ_{o} ε_{o} 4}} = \frac{1}{2} \frac{1}{\sqrt{μ_{o} ε_{o}}}$
We know that,
$c = \frac{1}{\sqrt{μ_{o} ε_{o}}}$
(in vacuum)
$v_{medium} = \frac{c}{2}$
$\frac{λ_{medium}}{λ_{air}} = \frac{v_{medium}}{v_{air}} = \frac{c / 2}{c} = \frac{1}{2}$
$λ_{medium} = \frac{λ_{air}}{2}$
Wavelength is halved and frequency remains unchanged.

COMEDK 2013

Electromagnetic Wave

155568 Assertion: Dipole
electromagnetic waves
oscillations produce
Reason: Accelerated
charge produces
electromagnetic waves.

1 If both Assertion and Reason are correct and the Reason is a correct explanation of the Assertion.

2 If both Assertion and Reason are correct but Reason in not a correct explanation of the Assertion.

3 If the Assertion is correct but Reason is incorrect.

4 If both the Assertion and Reason are incorrect.

5 If the Assertion is incorrect but the Reason is correct.

Explanation:

B Oscillating charge between dipolar electric field produces EM wave.
$∴$ Assertion is correct.
Electromagnetic wave is produce by a accelerated charge particle.
Reason is also correct.
But it is not correct explanation of assertion.

AIIMS-2007

Electromagnetic Wave

155573 The magnetic field amplitude of an electromagnetic wave is $2 \times 10^{- 7} T$ . Its electric field amplitude, if the wave is travelling in free space is :

1

6 {Vm}^{- 1}

2

60 {Vm}^{- 1}

3

\frac{1}{6} {Vm}^{- 1}

4 none of these

Explanation:

B Given,
Magnetic field $B_{o} = 2 \times 10^{- 7} T$
We know that,
$E_{o} = {cB}_{o}$
$E_{o} = 3 \times 10^{8} \times 2 \times 10^{- 7}$
$E_{o} = 6 \times 10 = 60 V / m$

BCECE-2006

Electromagnetic Wave

155577 Find the value of magnetic field between plates of capacitor at a distance $1 m$ from centre, where electric field varies by $10^{10} V / m$ per second.

1

5.56 \times 10^{- 8} T

2

5.56 \times 10^{- 3} T

3

5.56 μ T

4

5.55 T

Explanation:

A Given,
$\frac{dE}{dt} = 10^{10} V / m$
Magnetic field between plates of capacitor
$B = \frac{μ_{o} ε_{o}}{2} \cdot \frac{dE}{dt}$
$∴ B = \frac{4 π \times 10^{- 7} \times 8.85 \times 10^{- 12} \times 10^{10}}{2}$
$B = 5.56 \times 10^{- 8} T$

VITEEE-2010

Electromagnetic Wave

155562 A plane electromagnetic wave propagating in $x$ $(-)$ direction as a wave function (in SI units) is given as $ψ (x, t) = 10^{3} \sin π (3 \times 10^{6} x - 9 \times 10^{14} t)$ . The speed of the wave is

1

3 \times 10^{6} m / s

2

3 \times 10^{7} m / s

3

3 \times 10^{8} m / s

4

9 \times 10^{14} m / s

Explanation:

C Given,
$ψ (x, t) = 10^{3} \sin π (3 \times 10^{6} x - 9 \times 10^{14} t)$
General equation,
$E = E_{o} \sin (kx - ω t)$
On comparison equation (i) and (ii) we get-
$k = 3 \times 10^{6} m^{- 1}$
$ω = 9 \times 10^{14} rad / \sec$
$∴ v = \frac{ω}{K} = \frac{9 \times 10^{14}}{3 \times 10^{6}}$
$v = 3 \times 10^{8} m / s$

JCECE-2013

Electromagnetic Wave

155566 An electromagnetic wave of frequency $v = 3.0 MHz$ passes from vacuum into a dielectric medium with relative permittivity $\in= 4.0$ . Then

1 Wavelength is doubled and frequency becomes half

2 Wavelength is halved and frequency remains unchanged

3 Wavelength is frequency both remain unchanged

4 Wavelength is doubled and frequency unchanged.

Explanation:

B Velocity wave in medium
$v_{medium} = \frac{1}{\sqrt{μ_{o} μ_{r} ε_{o} ε_{r}}}$
For air, $μ_{r} = 1$
$v_{medium} = \frac{1}{\sqrt{μ_{o} ε_{o} 4}} = \frac{1}{2} \frac{1}{\sqrt{μ_{o} ε_{o}}}$
We know that,
$c = \frac{1}{\sqrt{μ_{o} ε_{o}}}$
(in vacuum)
$v_{medium} = \frac{c}{2}$
$\frac{λ_{medium}}{λ_{air}} = \frac{v_{medium}}{v_{air}} = \frac{c / 2}{c} = \frac{1}{2}$
$λ_{medium} = \frac{λ_{air}}{2}$
Wavelength is halved and frequency remains unchanged.

COMEDK 2013

Electromagnetic Wave

155568 Assertion: Dipole
electromagnetic waves
oscillations produce
Reason: Accelerated
charge produces
electromagnetic waves.

1 If both Assertion and Reason are correct and the Reason is a correct explanation of the Assertion.

2 If both Assertion and Reason are correct but Reason in not a correct explanation of the Assertion.

3 If the Assertion is correct but Reason is incorrect.

4 If both the Assertion and Reason are incorrect.

5 If the Assertion is incorrect but the Reason is correct.

Explanation:

B Oscillating charge between dipolar electric field produces EM wave.
$∴$ Assertion is correct.
Electromagnetic wave is produce by a accelerated charge particle.
Reason is also correct.
But it is not correct explanation of assertion.

AIIMS-2007

Electromagnetic Wave

155573 The magnetic field amplitude of an electromagnetic wave is $2 \times 10^{- 7} T$ . Its electric field amplitude, if the wave is travelling in free space is :

1

6 {Vm}^{- 1}

2

60 {Vm}^{- 1}

3

\frac{1}{6} {Vm}^{- 1}

4 none of these

Explanation:

B Given,
Magnetic field $B_{o} = 2 \times 10^{- 7} T$
We know that,
$E_{o} = {cB}_{o}$
$E_{o} = 3 \times 10^{8} \times 2 \times 10^{- 7}$
$E_{o} = 6 \times 10 = 60 V / m$

BCECE-2006

Electromagnetic Wave

155577 Find the value of magnetic field between plates of capacitor at a distance $1 m$ from centre, where electric field varies by $10^{10} V / m$ per second.

1

5.56 \times 10^{- 8} T

2

5.56 \times 10^{- 3} T

3

5.56 μ T

4

5.55 T

Explanation:

A Given,
$\frac{dE}{dt} = 10^{10} V / m$
Magnetic field between plates of capacitor
$B = \frac{μ_{o} ε_{o}}{2} \cdot \frac{dE}{dt}$
$∴ B = \frac{4 π \times 10^{- 7} \times 8.85 \times 10^{- 12} \times 10^{10}}{2}$
$B = 5.56 \times 10^{- 8} T$

VITEEE-2010

Electromagnetic Wave

155562 A plane electromagnetic wave propagating in $x$ $(-)$ direction as a wave function (in SI units) is given as $ψ (x, t) = 10^{3} \sin π (3 \times 10^{6} x - 9 \times 10^{14} t)$ . The speed of the wave is

1

3 \times 10^{6} m / s

2

3 \times 10^{7} m / s

3

3 \times 10^{8} m / s

4

9 \times 10^{14} m / s

Explanation:

C Given,
$ψ (x, t) = 10^{3} \sin π (3 \times 10^{6} x - 9 \times 10^{14} t)$
General equation,
$E = E_{o} \sin (kx - ω t)$
On comparison equation (i) and (ii) we get-
$k = 3 \times 10^{6} m^{- 1}$
$ω = 9 \times 10^{14} rad / \sec$
$∴ v = \frac{ω}{K} = \frac{9 \times 10^{14}}{3 \times 10^{6}}$
$v = 3 \times 10^{8} m / s$

JCECE-2013

Electromagnetic Wave

155566 An electromagnetic wave of frequency $v = 3.0 MHz$ passes from vacuum into a dielectric medium with relative permittivity $\in= 4.0$ . Then

1 Wavelength is doubled and frequency becomes half

2 Wavelength is halved and frequency remains unchanged

3 Wavelength is frequency both remain unchanged

4 Wavelength is doubled and frequency unchanged.

Explanation:

B Velocity wave in medium
$v_{medium} = \frac{1}{\sqrt{μ_{o} μ_{r} ε_{o} ε_{r}}}$
For air, $μ_{r} = 1$
$v_{medium} = \frac{1}{\sqrt{μ_{o} ε_{o} 4}} = \frac{1}{2} \frac{1}{\sqrt{μ_{o} ε_{o}}}$
We know that,
$c = \frac{1}{\sqrt{μ_{o} ε_{o}}}$
(in vacuum)
$v_{medium} = \frac{c}{2}$
$\frac{λ_{medium}}{λ_{air}} = \frac{v_{medium}}{v_{air}} = \frac{c / 2}{c} = \frac{1}{2}$
$λ_{medium} = \frac{λ_{air}}{2}$
Wavelength is halved and frequency remains unchanged.

COMEDK 2013

Electromagnetic Wave

155568 Assertion: Dipole
electromagnetic waves
oscillations produce
Reason: Accelerated
charge produces
electromagnetic waves.

1 If both Assertion and Reason are correct and the Reason is a correct explanation of the Assertion.

2 If both Assertion and Reason are correct but Reason in not a correct explanation of the Assertion.

3 If the Assertion is correct but Reason is incorrect.

4 If both the Assertion and Reason are incorrect.

5 If the Assertion is incorrect but the Reason is correct.

Explanation:

B Oscillating charge between dipolar electric field produces EM wave.
$∴$ Assertion is correct.
Electromagnetic wave is produce by a accelerated charge particle.
Reason is also correct.
But it is not correct explanation of assertion.

AIIMS-2007

Electromagnetic Wave

155573 The magnetic field amplitude of an electromagnetic wave is $2 \times 10^{- 7} T$ . Its electric field amplitude, if the wave is travelling in free space is :

1

6 {Vm}^{- 1}

2

60 {Vm}^{- 1}

3

\frac{1}{6} {Vm}^{- 1}

4 none of these

Explanation:

B Given,
Magnetic field $B_{o} = 2 \times 10^{- 7} T$
We know that,
$E_{o} = {cB}_{o}$
$E_{o} = 3 \times 10^{8} \times 2 \times 10^{- 7}$
$E_{o} = 6 \times 10 = 60 V / m$

BCECE-2006

Electromagnetic Wave

155577 Find the value of magnetic field between plates of capacitor at a distance $1 m$ from centre, where electric field varies by $10^{10} V / m$ per second.

1

5.56 \times 10^{- 8} T

2

5.56 \times 10^{- 3} T

3

5.56 μ T

4

5.55 T

Explanation:

A Given,
$\frac{dE}{dt} = 10^{10} V / m$
Magnetic field between plates of capacitor
$B = \frac{μ_{o} ε_{o}}{2} \cdot \frac{dE}{dt}$
$∴ B = \frac{4 π \times 10^{- 7} \times 8.85 \times 10^{- 12} \times 10^{10}}{2}$
$B = 5.56 \times 10^{- 8} T$

VITEEE-2010