Uniform Plane Waves Questions and Answers

Q: In an electromagnetic wave, the electric field of amplitude 6.2 V/m oscillates with a frequency of 2.4 × 1010 Hz. Estimate energy density of the wave :(ε = 8.85 × 10-12 C2 / Nm2)

The key gives \(3.4 \times 10^{-10}\) J/m\(^{3}\), which is the peak total energy density, \(\varepsilon_{0}E_{0}^{2}\).In a plane wave the electric and magnetic fields carry equal energy, so the total energy density is \(u = \varepsilon_{0}E^{2}\). At the instant the field is at its amplitude:\[ \begin{aligned} u &= \varepsilon_{0} E_{0}^{2} \\ &= 8.85 \times 10^{-12} \times (6.2)^{2} \\ &= 8.85 \times 10^{-12} \times 38.44 \\ &\approx 3.4 \times 10^{-10}\ \text{J/m}^{3} \end{aligned} \]The frequency does not enter the calculation. The time-averaged value would be half of this, about \(1.7 \times 10^{-10}\) J/m\(^{3}\), which is not among the options. 1.4, 2.4 and 4.4 do not come from either formula.

Q: Calculate the skin depth of a material with attenuation constant of 2 units.

The skin depth is 0.5 units.The skin depth \(\delta\) is the distance over which a wave's amplitude falls to \(1/e\) (about 37%) of its value at the surface. Since the amplitude varies as \(e^{-\alpha z}\), this happens when \(\alpha z = 1\):\[ \delta = \frac{1}{\alpha} = \frac{1}{2} = 0.5 \]The answer 2 would follow from mistaking skin depth for the attenuation constant itself, 4 from multiplying instead of dividing, and 1 from ignoring the attenuation altogether. Skin depth and attenuation constant are reciprocals, so a larger \(\alpha\) always means a thinner skin.

Q: The expression for velocity of a wave in the conductor is :

The velocity of a wave in a good conductor is \(V = \sqrt{2\omega/\mu\sigma}\).In a good conductor (\(\sigma \gg \omega\varepsilon\)) the attenuation and phase constants are equal:\[ \alpha = \beta = \sqrt{\frac{\omega\mu\sigma}{2}} \]The phase velocity is then\[ \begin{aligned} v &= \frac{\omega}{\beta} = \omega\sqrt{\frac{2}{\omega\mu\sigma}} \\ &= \sqrt{\frac{2\omega}{\mu\sigma}} \end{aligned} \]This is very slow compared with light; in copper at 1 MHz it is only about 400 m/s.Option B has the conductivity and permeability in the numerator, which would make the wave faster in better conductors, the wrong way round. Options C and D drop the square root, so they do not even have the units of velocity.

Q: In a good conductor the phase relation between the tangential components of electric E and the magnetic field H is as follows :

In a good conductor E leads H by 45 degrees.The ratio of the tangential electric and magnetic fields is the intrinsic impedance. For a good conductor\[ \begin{aligned} \eta &= \sqrt{\frac{j\omega\mu}{\sigma}} \\ &= \sqrt{\frac{\omega\mu}{\sigma}}\,\angle 45^{\circ} \end{aligned} \]because \(\sqrt{j} = e^{j\pi/4}\). Since \(E = \eta H\), the electric field is 45° ahead of the magnetic field.E and H are in phase only in a lossless dielectric, where \(\eta\) is real. They are never 90° apart in a propagating wave; that limit would need \(\eta\) purely imaginary. 'Out of phase' (180°) does not arise either.

Q: The ratio of electric field vector \(\vec{E}\) and magnetic field vector \(\vec{H}\) has the dimension of :

The ratio has the dimension of Resistance.\(E\) is in V/m and \(H\) is in A/m, so \(\frac{E}{H}\) is in V/A, that is ohms. For a plane wave in free space, \(\frac{E}{H} = \sqrt{\mu_{0}/\varepsilon_{0}} \approx 377\ \Omega\), the impedance of free space.Inductance is V s/A and capacitance A s/V, and their product LC has dimensions of time squared, so none of these matches.

Q: For a good conductor skin depth varies inversely with ......... power of frequency.

Skin depth varies inversely with the Half power of frequency.For a good conductor, the skin depth is:\[ \begin{aligned} \delta &= \sqrt{\frac{2}{\omega\mu\sigma}} \\ &\propto \frac{1}{\sqrt{f}} \end{aligned} \]So quadrupling the frequency halves the depth to which the field penetrates. That is why high-frequency currents crowd near the surface of conductors.First, second or third powers do not follow from this expression.

Q: The SI unit of the quantity \(\frac{|E|}{|B|}\) is :

The SI unit of \(\frac{|E|}{|B|}\) is m/s.From the Lorentz force, \(qE\) and \(qvB\) are both forces, so \(E\) and \(vB\) have the same units and \(\frac{E}{B}\) has the units of velocity. For an electromagnetic wave in vacuum \(\frac{E}{B} = c\). Explicitly, (V/m)/T = (V/m)/(V s/m2) = m/s.s/m is the inverse, and the options with \(c^{2}\) put a speed squared where none belongs.

Q: The dispersion relation for a low density plasma is \(\omega^{2} = \omega_{0}^{2} + c^{2}k^{2}\), where \(\omega_{0}\) is the plasma frequency and c is the speed of light in free space. The relationship between the group velocity (Vg) and phase velocity (Vp) is :

The relation is \(V_{p}V_{g} = c^{2}\).From \(\omega^{2} = \omega_{0}^{2} + c^{2}k^{2}\):\[ \begin{aligned} V_{p} &= \frac{\omega}{k} \\ 2\omega\,d\omega &= 2c^{2}k\,dk, \quad V_{g} \\ &= \frac{d\omega}{dk} \\ &= \frac{c^{2}k}{\omega} \\ V_{p}V_{g} &= \frac{\omega}{k}\cdot\frac{c^{2}k}{\omega} \\ &= c^{2} \end{aligned} \]The phase velocity exceeds \(c\) while the group velocity, which carries energy and information, stays below \(c\). They are equal only without the plasma term, and the square-root relations are dimensionally inconsistent.

Q: In free space, which of the following will be zero ?

In free space, conductivity is zero.Free space (vacuum) contains no free charges to carry a conduction current, so its conductivity \(\sigma = 0\). This is why electromagnetic waves travel through free space without loss.Permittivity and permeability of free space are not zero: \(\varepsilon_{0} \approx 8.854 \times 10^{-12}\) F/m and \(\mu_{0} = 4\pi \times 10^{-7}\) H/m, and together they set the speed of light, \(c = 1/\sqrt{\mu_{0}\varepsilon_{0}}\). Resistivity is the reciprocal of conductivity, so in free space it is infinite, not zero.

101.

The reflection coefficient \(k_{r}\), characteristic impedance \(Z_{0}\), and load impedance \(Z_{L}\) of a transmission line are connected together by the relation :

A.\(k_{r} = \frac{Z_{L} + Z_{0}}{Z_{0} - Z_{L}}\)
B.\(k_{r} = \frac{Z_{L}Z_{0}}{Z_{0} - Z_{L}}\)
C.\(k_{r} = \frac{Z_{L} - Z_{0}}{Z_{0} + Z_{L}}\)
D.\(k_{r} = \frac{Z_{L} - Z_{0}}{Z_{0}Z_{L}}\)

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

Answer: C

The relation is \(k_{r} = \frac{Z_{L} - Z_{0}}{Z_{0} + Z_{L}}\).

At the load, the voltage and current must satisfy \(V/I = Z_{L}\), while each travelling wave on the line has \(V/I = \pm Z_{0}\). With incident wave \(V^{+}\) and reflected wave \(V^{-}\):

\[ \begin{aligned} Z_{L} &= Z_{0}\frac{V^{+} + V^{-}}{V^{+} - V^{-}} \\ k_{r} &= \frac{V^{-}}{V^{+}} = \frac{Z_{L} - Z_{0}}{Z_{L} + Z_{0}} \end{aligned} \]

Checks: a matched load (\(Z_{L} = Z_{0}\)) gives no reflection, a short circuit gives -1 and an open circuit +1. The first option becomes infinite for a matched load, and the others do not even have the dimensionless form a reflection coefficient needs.

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102.

In an electromagnetic wave, the electric field of amplitude 6.2 V/m oscillates with a frequency of 2.4 × 10¹⁰ Hz. Estimate energy density of the wave :

(ε = 8.85 × 10^-12 C² / Nm²)

A.\(1.4 \times 10^{-10}\ \text{J/m}^{3}\)
B.\(2.4 \times 10^{-10}\ \text{J/m}^{3}\)
C.\(3.4 \times 10^{-10}\ \text{J/m}^{3}\)
D.\(4.4 \times 10^{-10}\ \text{J/m}^{3}\)

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: C

The key gives \(3.4 \times 10^{-10}\) J/m\(^{3}\), which is the peak total energy density, \(\varepsilon_{0}E_{0}^{2}\).

In a plane wave the electric and magnetic fields carry equal energy, so the total energy density is \(u = \varepsilon_{0}E^{2}\). At the instant the field is at its amplitude:

\[ \begin{aligned} u &= \varepsilon_{0} E_{0}^{2} \\ &= 8.85 \times 10^{-12} \times (6.2)^{2} \\ &= 8.85 \times 10^{-12} \times 38.44 \\ &\approx 3.4 \times 10^{-10}\ \text{J/m}^{3} \end{aligned} \]

The frequency does not enter the calculation. The time-averaged value would be half of this, about \(1.7 \times 10^{-10}\) J/m\(^{3}\), which is not among the options. 1.4, 2.4 and 4.4 do not come from either formula.

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103.

Calculate the skin depth of a material with attenuation constant of 2 units.

A.2
B.1
C.0.5
D.4

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: C

The skin depth is 0.5 units.

The skin depth \(\delta\) is the distance over which a wave's amplitude falls to \(1/e\) (about 37%) of its value at the surface. Since the amplitude varies as \(e^{-\alpha z}\), this happens when \(\alpha z = 1\):

\[ \delta = \frac{1}{\alpha} = \frac{1}{2} = 0.5 \]

The answer 2 would follow from mistaking skin depth for the attenuation constant itself, 4 from multiplying instead of dividing, and 1 from ignoring the attenuation altogether. Skin depth and attenuation constant are reciprocals, so a larger \(\alpha\) always means a thinner skin.

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104.

The expression for velocity of a wave in the conductor is :

A.\(V = \sqrt{(2\omega / \mu\sigma)}\)
B.\(V = \sqrt{(2\omega\mu\sigma)}\)
C.\(V = (2\omega / \mu\sigma)\)
D.\(V = (2\omega\mu\sigma)\)

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: A

The velocity of a wave in a good conductor is \(V = \sqrt{2\omega/\mu\sigma}\).

In a good conductor (\(\sigma \gg \omega\varepsilon\)) the attenuation and phase constants are equal:

\[ \alpha = \beta = \sqrt{\frac{\omega\mu\sigma}{2}} \]

The phase velocity is then

\[ \begin{aligned} v &= \frac{\omega}{\beta} = \omega\sqrt{\frac{2}{\omega\mu\sigma}} \\ &= \sqrt{\frac{2\omega}{\mu\sigma}} \end{aligned} \]

This is very slow compared with light; in copper at 1 MHz it is only about 400 m/s.

Option B has the conductivity and permeability in the numerator, which would make the wave faster in better conductors, the wrong way round. Options C and D drop the square root, so they do not even have the units of velocity.

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105.

In a good conductor the phase relation between the tangential components of electric E and the magnetic field H is as follows :

A.E and H are in phase
B.E and H are out of phase
C.H leads E by 90
D.E leads H by 45

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: D

In a good conductor E leads H by 45 degrees.

The ratio of the tangential electric and magnetic fields is the intrinsic impedance. For a good conductor

\[ \begin{aligned} \eta &= \sqrt{\frac{j\omega\mu}{\sigma}} \\ &= \sqrt{\frac{\omega\mu}{\sigma}}\,\angle 45^{\circ} \end{aligned} \]

because \(\sqrt{j} = e^{j\pi/4}\). Since \(E = \eta H\), the electric field is 45° ahead of the magnetic field.

E and H are in phase only in a lossless dielectric, where \(\eta\) is real. They are never 90° apart in a propagating wave; that limit would need \(\eta\) purely imaginary. 'Out of phase' (180°) does not arise either.

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106.

The ratio of electric field vector \(\vec{E}\) and magnetic field vector \(\vec{H}\) has the dimension of :

A.Inductance
B.Capacitance
C.Resistance
D.Product of inductance and capacitance

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: C

The ratio has the dimension of Resistance.

\(E\) is in V/m and \(H\) is in A/m, so \(\frac{E}{H}\) is in V/A, that is ohms. For a plane wave in free space, \(\frac{E}{H} = \sqrt{\mu_{0}/\varepsilon_{0}} \approx 377\ \Omega\), the impedance of free space.

Inductance is V s/A and capacitance A s/V, and their product LC has dimensions of time squared, so none of these matches.

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107.

For a good conductor skin depth varies inversely with ......... power of frequency.

A.First
B.Second
C.Half
D.Third

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: C

Skin depth varies inversely with the Half power of frequency.

For a good conductor, the skin depth is:

\[ \begin{aligned} \delta &= \sqrt{\frac{2}{\omega\mu\sigma}} \\ &\propto \frac{1}{\sqrt{f}} \end{aligned} \]

So quadrupling the frequency halves the depth to which the field penetrates. That is why high-frequency currents crowd near the surface of conductors.

First, second or third powers do not follow from this expression.

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108.

The SI unit of the quantity \(\frac{|E|}{|B|}\) is :

A.s/m
B.m/s
C.c\(^{2}\)m/s
D.c\(^{2}\)m\(^{2}\)/s

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: B

The SI unit of \(\frac{|E|}{|B|}\) is m/s.

From the Lorentz force, \(qE\) and \(qvB\) are both forces, so \(E\) and \(vB\) have the same units and \(\frac{E}{B}\) has the units of velocity. For an electromagnetic wave in vacuum \(\frac{E}{B} = c\). Explicitly, (V/m)/T = (V/m)/(V s/m²) = m/s.

s/m is the inverse, and the options with \(c^{2}\) put a speed squared where none belongs.

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109.

The dispersion relation for a low density plasma is \(\omega^{2} = \omega_{0}^{2} + c^{2}k^{2}\), where \(\omega_{0}\) is the plasma frequency and c is the speed of light in free space. The relationship between the group velocity (V_g) and phase velocity (V_p) is :

A.\(V_{p} = V_{g}\)
B.\(V_{p} = (V_{g})^{1/2}\)
C.\(V_{p}V_{g} = c^{2}\)
D.\(V_{g} = (V_{p})^{1/2}\)

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: C

The relation is \(V_{p}V_{g} = c^{2}\).

From \(\omega^{2} = \omega_{0}^{2} + c^{2}k^{2}\):

\[ \begin{aligned} V_{p} &= \frac{\omega}{k} \\ 2\omega\,d\omega &= 2c^{2}k\,dk, \quad V_{g} \\ &= \frac{d\omega}{dk} \\ &= \frac{c^{2}k}{\omega} \\ V_{p}V_{g} &= \frac{\omega}{k}\cdot\frac{c^{2}k}{\omega} \\ &= c^{2} \end{aligned} \]

The phase velocity exceeds \(c\) while the group velocity, which carries energy and information, stays below \(c\). They are equal only without the plasma term, and the square-root relations are dimensionally inconsistent.

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110.

In free space, which of the following will be zero ?

A.Permittivity
B.Permeability
C.Conductivity
D.Resistivity

Exam Year(s)

MH SET - 2025, SET Exam - 2025,

Answer: C

In free space, conductivity is zero.

Free space (vacuum) contains no free charges to carry a conduction current, so its conductivity \(\sigma = 0\). This is why electromagnetic waves travel through free space without loss.

Permittivity and permeability of free space are not zero: \(\varepsilon_{0} \approx 8.854 \times 10^{-12}\) F/m and \(\mu_{0} = 4\pi \times 10^{-7}\) H/m, and together they set the speed of light, \(c = 1/\sqrt{\mu_{0}\varepsilon_{0}}\). Resistivity is the reciprocal of conductivity, so in free space it is infinite, not zero.

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