Poynting Theorem Questions and Answers

Q: A laser beam of intensity 50 W/cm\(^{2}\) falls on a perfectly reflecting plane mirror for an hour. The area of the mirror is 5 cm\(^{2}\). The average force acting on the mirror is few :

The force is a few Micro Newtons.The power falling on the mirror is \(P = IA = 50 \times 5 = 250\) W. For perfect reflection the momentum of the light is reversed, so the force is twice P/c:\[ \begin{aligned} F &= \frac{2P}{c} = \frac{2 \times 250}{3 \times 10^{8}} \\ &\approx 1.7 \times 10^{-6}\ \text{N} \end{aligned} \]That is about 1.7 micronewtons. The duration of one hour does not affect the force, only the total momentum transferred.

Q: What is the magnitude of the Poynting vector at the surface of a long cylindrical wire of radius R, length L, carrying current I, when its ends are kept at a potential difference V ?

The Poynting vector has magnitude \(\frac{VI}{2\pi RL}\).Along the wire the electric field is \(E = V/L\), and at its surface the magnetic field is \(H = \frac{I}{2\pi R}\), at right angles to E. The Poynting vector \(\bar{S} = \bar{E} \times \bar{H}\) points radially into the wire, with magnitude\[ \begin{aligned} S = EH &= \frac{V}{L}\cdot\frac{I}{2\pi R} \\ &= \frac{VI}{2\pi RL} \end{aligned} \]Check: multiplying by the curved surface area \(2\pi RL\) gives the power \(VI\), exactly the Joule heat in the wire, so the energy flows in through the sides. The options with \(\pi R^{2}L\) or the end faces \(2\pi R^{2}\) divide by the wrong area, and it is not zero, because both fields are present.

Q: An oscillating magnetic dipole is formed by making sinusoidal current flow through a circular loop. Consider this dipole as a perfect dipole. If peak value of current flowing is doubled, then time averaged power radiated by this dipole becomes ............. times the initial power.

The radiated power becomes 4 times the initial power.The time-averaged power of an oscillating magnetic dipole of moment amplitude \(m_{0}\) at angular frequency \(\omega\) is\[ \begin{aligned} \langle P \rangle &= \frac{\mu_{0}m_{0}^{2}\omega^{4}}{12\pi c^{3}} \\ m_{0} &= I_{0}\pi b^{2} \end{aligned} \]The moment is proportional to the current amplitude, so the power goes as \(I_{0}^{2}\). Doubling \(I_{0}\) at the same frequency and loop size multiplies the power by \(2^{2} = 4\).16 is what one gets by confusing the current with the frequency, which enters as \(\omega^{4}\); the fractions 1/4 and 1/16 would mean the power falls when the current rises.

Q: Let q and a denote the charge and the acceleration of a moving point charge. Larmor formula states that total power radiated by the point charge is proportional to \(q^{\alpha}a^{\beta}\), where ..............

The radiated power goes as \(q^{2}a^{2}\), so \(\alpha = 2\) and \(\beta = 2\).The Larmor formula for a non-relativistic accelerated point charge is\[ \begin{aligned} P &= \frac{\mu_{0}q^{2}a^{2}}{6\pi c} \\ &= \frac{q^{2}a^{2}}{6\pi\varepsilon_{0}c^{3}} \end{aligned} \]The radiation fields are proportional to \(qa\), and the radiated power, coming from the Poynting vector, goes as the square of the fields. That is why both the charge and the acceleration appear squared, and why the sign of the charge does not matter.Any option with a first power of \(q\) or \(a\) would make the power change sign with the charge or the direction of acceleration, which a radiated power cannot do.

Q: A relativistic particle of charge q and rest mass \(m_{0}\) moves with velocity \(\vec{v} = v_{x}\hat{x} + v_{y}\hat{y}\) in a uniform magnetic field, \(\vec{B} = B_{0}\hat{z}\). The relativistic Larmor radius R (the radius of the circular path the charged particle takes) is given by ..............

The radius is \(R = \frac{\gamma m_{0}\sqrt{v_{x}^{2} + v_{y}^{2}}}{qB_{0}}\).The velocity lies in the plane perpendicular to \(\vec{B}\), so the particle moves on a circle. The magnetic force does no work, so the speed and \(\gamma\) stay constant, and the relativistic equation of motion is \(\gamma m_{0}\,d\vec{v}/dt = q\vec{v}\times\vec{B}\). Equating the force to the centripetal requirement:\[ \begin{aligned} \frac{\gamma m_{0}v^{2}}{R} &= qvB_{0} \\ R &= \frac{\gamma m_{0}v}{qB_{0}} = \frac{p}{qB_{0}} \end{aligned} \]with \(v = \sqrt{v_{x}^{2} + v_{y}^{2}}\). The radius is the relativistic momentum divided by \(qB_{0}\). Leaving out \(\gamma\) gives the non-relativistic radius; dividing by \(\gamma\) or using \(\gamma^{2}\) has no basis in the equation of motion.

Q: For a localized time varying current distribution, if the Poynting vector, \(\vec{S}\) varies as \(|\vec{S}| \propto \frac{1}{r^{n}}\), where r is the distance from the center of the current distribution, then ..........

The answer is \(n \le 2\), because the surface integral of the Poynting vector must be non-zero for the radiating fields.A localized time-varying current distribution radiates. The power crossing a large sphere of radius \(r\) is\[ \begin{aligned} P &= \oint \vec{S}\cdot d\vec{a} \\ &\sim |\vec{S}|\, 4\pi r^{2} \propto r^{2-n} \end{aligned} \]Radiation means energy that escapes to infinity, so \(P\) must stay finite and non-zero as \(r \to \infty\). That requires the radiation part of \(\vec{S}\) to fall exactly as \(1/r^{2}\), which happens because the radiation fields fall as \(1/r\). A faster fall-off (\(n \gt 2\)) would carry no energy to infinity.The option claiming \(n \gt 2\) for radiating fields contradicts this, the option with \(n \gt 2\) and non-radiating fields denies that a time-varying localized source radiates, and the general argument above holds without further detail of the source.

61.

A laser beam of intensity 50 W/cm\(^{2}\) falls on a perfectly reflecting plane mirror for an hour. The area of the mirror is 5 cm\(^{2}\). The average force acting on the mirror is few :

A.Newtons
B.Milli Newtons
C.Micro Newtons
D.Nano Newtons

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

Answer: C

The force is a few Micro Newtons.

The power falling on the mirror is \(P = IA = 50 \times 5 = 250\) W. For perfect reflection the momentum of the light is reversed, so the force is twice P/c:

\[ \begin{aligned} F &= \frac{2P}{c} = \frac{2 \times 250}{3 \times 10^{8}} \\ &\approx 1.7 \times 10^{-6}\ \text{N} \end{aligned} \]

That is about 1.7 micronewtons. The duration of one hour does not affect the force, only the total momentum transferred.

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62.

What is the magnitude of the Poynting vector at the surface of a long cylindrical wire of radius R, length L, carrying current I, when its ends are kept at a potential difference V ?

A.zero
B.\(\frac{VI}{2\pi R^{2} + 2\pi RL}\)
C.\(\frac{VI}{\pi R^{2}L}\)
D.\(\frac{VI}{2\pi RL}\)

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: D

The Poynting vector has magnitude \(\frac{VI}{2\pi RL}\).

Along the wire the electric field is \(E = V/L\), and at its surface the magnetic field is \(H = \frac{I}{2\pi R}\), at right angles to E. The Poynting vector \(\bar{S} = \bar{E} \times \bar{H}\) points radially into the wire, with magnitude

\[ \begin{aligned} S = EH &= \frac{V}{L}\cdot\frac{I}{2\pi R} \\ &= \frac{VI}{2\pi RL} \end{aligned} \]

Check: multiplying by the curved surface area \(2\pi RL\) gives the power \(VI\), exactly the Joule heat in the wire, so the energy flows in through the sides. The options with \(\pi R^{2}L\) or the end faces \(2\pi R^{2}\) divide by the wrong area, and it is not zero, because both fields are present.

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63.

The electric field of an electromagnetic wave is \(\vec{E} = i\sqrt{2}\sin(kz - \omega t)\) V/m. The average flow energy per unit area per unit time, due to this wave is :

A.\(27 \times 10^{4}\) W/m\(^{2}\)
B.\(27 \times 10^{-4}\) W/m\(^{2}\)
C.\(27 \times 10^{-2}\) W/m\(^{2}\)
D.\(27 \times 10^{2}\) W/m\(^{2}\)

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

Answer: B

The average energy flow is \(27 \times 10^{-4}\) W/m\(^{2}\).

The time-averaged intensity of a plane wave with amplitude \(E_{0}\) is

\[ \begin{aligned} \langle S\rangle &= \tfrac{1}{2}c\varepsilon_{0}E_{0}^{2} \\ &= \tfrac{1}{2}(3 \times 10^{8})(8.85 \times 10^{-12})(\sqrt{2})^{2} \\ &= 2.66 \times 10^{-3}\ \text{W/m}^{2} \end{aligned} \]

which is about \(27 \times 10^{-4}\) W/m\(^{2}\). Equivalently, \(E_{0}^{2}/(2Z_{0}) = 2/754\). The other options differ by powers of ten: they correspond to errors in the exponent of \(\varepsilon_{0}\) or c, not to a different physical result.

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64.

The value of Poynting vector at the surface of a long cylindrical wire of radius R and length L carrying a current I when its ends are kept at a potential difference of V is :

A.Zero
B.(VI)/(2πR\(^{2}\) + 2πRL)
C.(VI)/(πR\(^{2}\)L)
D.(VI)/(2πRL)

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: D

The Poynting vector at the surface is \(\frac{VI}{2\pi RL}\).

At the surface, \(E = \frac{V}{L}\) along the wire and \(H = \frac{I}{2\pi R}\) round it, so the Poynting vector points radially inward with magnitude:

\[ \begin{aligned} S &= EH \\ &= \frac{V}{L}\cdot\frac{I}{2\pi R} \\ &= \frac{VI}{2\pi RL} \end{aligned} \]

Multiplying by the curved area \(2\pi RL\) gives \(VI\), exactly the Joule heat produced in the wire, so the energy flows in through the side. The end faces carry no flux, so the total area \(2\pi R^{2} + 2\pi RL\) is wrong, as is dividing by the volume \(\pi R^{2}L\). Zero would mean no energy reaches the wire.

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65.

An oscillating magnetic dipole is formed by making sinusoidal current flow through a circular loop. Consider this dipole as a perfect dipole. If peak value of current flowing is doubled, then time averaged power radiated by this dipole becomes ............. times the initial power.

A.4
B.\(\frac{1}{4}\)
C.16
D.\(\frac{1}{16}\)

Exam Year(s)

MH SET - 2025, SET Exam - 2025,

Answer: A

The radiated power becomes 4 times the initial power.

The time-averaged power of an oscillating magnetic dipole of moment amplitude \(m_{0}\) at angular frequency \(\omega\) is

\[ \begin{aligned} \langle P \rangle &= \frac{\mu_{0}m_{0}^{2}\omega^{4}}{12\pi c^{3}} \\ m_{0} &= I_{0}\pi b^{2} \end{aligned} \]

The moment is proportional to the current amplitude, so the power goes as \(I_{0}^{2}\). Doubling \(I_{0}\) at the same frequency and loop size multiplies the power by \(2^{2} = 4\).

16 is what one gets by confusing the current with the frequency, which enters as \(\omega^{4}\); the fractions 1/4 and 1/16 would mean the power falls when the current rises.

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66.

Let q and a denote the charge and the acceleration of a moving point charge. Larmor formula states that total power radiated by the point charge is proportional to \(q^{\alpha}a^{\beta}\), where ..............

A.\(\alpha = 1\) and \(\beta = 1\)
B.\(\alpha = 1\) and \(\beta = 2\)
C.\(\alpha = 2\) and \(\beta = 1\)
D.\(\alpha = 2\) and \(\beta = 2\)

Exam Year(s)

MH SET - 2025, SET Exam - 2025,

Answer: D

The radiated power goes as \(q^{2}a^{2}\), so \(\alpha = 2\) and \(\beta = 2\).

The Larmor formula for a non-relativistic accelerated point charge is

\[ \begin{aligned} P &= \frac{\mu_{0}q^{2}a^{2}}{6\pi c} \\ &= \frac{q^{2}a^{2}}{6\pi\varepsilon_{0}c^{3}} \end{aligned} \]

The radiation fields are proportional to \(qa\), and the radiated power, coming from the Poynting vector, goes as the square of the fields. That is why both the charge and the acceleration appear squared, and why the sign of the charge does not matter.

Any option with a first power of \(q\) or \(a\) would make the power change sign with the charge or the direction of acceleration, which a radiated power cannot do.

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67.

A relativistic particle of charge q and rest mass \(m_{0}\) moves with velocity \(\vec{v} = v_{x}\hat{x} + v_{y}\hat{y}\) in a uniform magnetic field, \(\vec{B} = B_{0}\hat{z}\). The relativistic Larmor radius R (the radius of the circular path the charged particle takes) is given by ..............

A.\(R = \frac{\gamma m_{0}\sqrt{(v_{x}^{2} + v_{y}^{2})}}{qB_{0}}\)
B.\(R = \frac{m_{0}\sqrt{(v_{x}^{2} + v_{y}^{2})}}{qB_{0}}\)
C.\(R = \frac{m_{0}\sqrt{(v_{x}^{2} + v_{y}^{2})}}{\gamma qB_{0}}\)
D.\(R = \frac{\gamma^{2}m_{0}\sqrt{(v_{x}^{2} + v_{y}^{2})}}{qB_{0}}\)

Exam Year(s)

MH SET - 2025, SET Exam - 2025,

Answer: A

The radius is \(R = \frac{\gamma m_{0}\sqrt{v_{x}^{2} + v_{y}^{2}}}{qB_{0}}\).

The velocity lies in the plane perpendicular to \(\vec{B}\), so the particle moves on a circle. The magnetic force does no work, so the speed and \(\gamma\) stay constant, and the relativistic equation of motion is \(\gamma m_{0}\,d\vec{v}/dt = q\vec{v}\times\vec{B}\). Equating the force to the centripetal requirement:

\[ \begin{aligned} \frac{\gamma m_{0}v^{2}}{R} &= qvB_{0} \\ R &= \frac{\gamma m_{0}v}{qB_{0}} = \frac{p}{qB_{0}} \end{aligned} \]

with \(v = \sqrt{v_{x}^{2} + v_{y}^{2}}\). The radius is the relativistic momentum divided by \(qB_{0}\). Leaving out \(\gamma\) gives the non-relativistic radius; dividing by \(\gamma\) or using \(\gamma^{2}\) has no basis in the equation of motion.

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68.

For a localized time varying current distribution, if the Poynting vector, \(\vec{S}\) varies as \(|\vec{S}| \propto \frac{1}{r^{n}}\), where r is the distance from the center of the current distribution, then ..........

A.\(n \le 2\) because surface integral of Poynting vector has to be non-zero; electromagnetic fields created by this configuration are radiating fields.
B.\(n \gt 2\) because the localized time varying current distribution creates a radiating electromagnetic fields.
C.\(n \gt 2\) and the electromagnetic fields are not radiating.
D.one cannot say anything about n without more information about the current distribution.

Exam Year(s)

MH SET - 2025, SET Exam - 2025,

Answer: A

The answer is \(n \le 2\), because the surface integral of the Poynting vector must be non-zero for the radiating fields.

A localized time-varying current distribution radiates. The power crossing a large sphere of radius \(r\) is

\[ \begin{aligned} P &= \oint \vec{S}\cdot d\vec{a} \\ &\sim |\vec{S}|\, 4\pi r^{2} \propto r^{2-n} \end{aligned} \]

Radiation means energy that escapes to infinity, so \(P\) must stay finite and non-zero as \(r \to \infty\). That requires the radiation part of \(\vec{S}\) to fall exactly as \(1/r^{2}\), which happens because the radiation fields fall as \(1/r\). A faster fall-off (\(n \gt 2\)) would carry no energy to infinity.

The option claiming \(n \gt 2\) for radiating fields contradicts this, the option with \(n \gt 2\) and non-radiating fields denies that a time-varying localized source radiates, and the general argument above holds without further detail of the source.

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69.

The energy per unit time, per unit area transported by the electromagnetic fields is expressed as

A.\(\vec{S} = \dfrac{1}{\mu_{o}}(\vec{E} \times \vec{B})\)
B.\(\vec{S} = (\vec{E} \times \vec{B})\)
C.\(\vec{S} = \mu_{o}(\vec{E} \times \vec{B})\)
D.\(\vec{S} = \dfrac{1}{\epsilon_{o}}(\vec{E} \times \vec{B})\)

Exam Year(s)

UGC NET - 2012,

Answer: A

The energy per unit time per unit area carried by an electromagnetic field is the Poynting vector, \( \vec{S} = \dfrac{1}{\mu_{o}}(\vec{E} \times \vec{B}) \), in watts per square metre, so option A is correct. It points in the direction of propagation. Option B lacks the factor \( 1/\mu_{o} \) and so is dimensionally wrong, option C multiplies by \( \mu_{o} \) instead of dividing, and option D divides by \( \epsilon_{o} \) where the permeability belongs, which would not give a power flux.

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