A transformer transfers power from primary to secondary. In the field picture the transfer takes place
Answer: C
The Poynting vector in the window between the windings carries the power across, guided by the changing flux in the core. No charge crosses between the windings, and the circuit description in terms of mutual inductance gives the same total power.
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Two antennas of equal gain are separated by a distance that is then doubled. The received power falls to
Answer: D
The power density at the receiver falls as the inverse square of the distance and the effective aperture is unchanged, so the received power falls by four, which is six decibels. This is the free space path loss and it dominates every link budget.
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The average power carried by a wave whose peak electric field is 100 V/m in free space is about
Answer: A
Correct answer: 13.3 W/m²
With a *peak* field a factor of one half is needed to obtain the average: \[ S_{avg} = \frac{E_{peak}^2}{2\eta_0} = \frac{100^2}{2\times377} = \frac{10{,}000}{754} = 13.3\ \mathrm{W/m^2}. \] Omitting that factor gives 26.5, which is the instantaneous peak power density rather than the average, and is offered here as the trap.
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The radiation resistance of a very short dipole is small, with the consequence that
Answer: B
A few ohms of radiation resistance sits alongside a loss resistance of comparable size, so much of the input power is dissipated rather than radiated, and the large reactance makes matching narrow and lossy. This is the fundamental difficulty of electrically small antennas.
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An antenna's efficiency is the ratio of
Answer: C
The input power divides between the two resistances in proportion to their values, so the fraction radiated is that ratio. The ratio of gain to directivity is numerically the same quantity, which is why the two definitions are sometimes confused.
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A wave of power density 100 W/m² strikes a perfect reflector normally. The radiation pressure is about
Answer: D
Correct answer: 0.67 \( \mu \)Pa
A perfect reflector reverses the wave’s momentum rather than absorbing it, so the pressure is twice that on a black surface: \[ p = \frac{2S}{c} = \frac{2\times100}{3\times10^8} = 6.7\times10^{-7}\ \mathrm{Pa}. \] The 0.33 microPa offered omits the factor of two and is the absorbing case. The magnitudes explain why solar sails must be enormous - even full sunlight gives only a few micropascals.
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Comparing the energy stored and the energy transported in a length of transmission line carrying power to a load, both are described by
Answer: A
The theorem's three terms are exactly the power crossing the boundary, the rate of change of the stored energy and the dissipation, so a line delivering steady power has the first balanced against the third with the second zero. It is the complete energy account for the line.
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A long straight wire, having radius \(a\) and resistance per unit length \(r\), carries a current \(I\). The magnitude and direction of the Poynting vector on the surface of the wire is
GATE Physics - 2018,
Answer: A
The Poynting vector has magnitude \(I^{2}r/2\pi a\), perpendicular to the axis and pointing inwards.
At the surface, the electric field driving the current is along the wire, \(E = Ir\) (the voltage drop per unit length), and the magnetic field circles the wire with \(H = I/(2\pi a)\). They are perpendicular, so
\[ S = EH = Ir \cdot \frac{I}{2\pi a} = \frac{I^{2}r}{2\pi a} \]
With \(\vec{E}\) along the current and \(\vec{H}\) circling it by the right-hand rule, \(\vec{E} \times \vec{H}\) points radially inward. This is the energy flowing in from the field to be dissipated as heat: multiplying by the surface area per unit length, \(2\pi a\), gives \(I^{2}r\), the Joule heating per unit length. An outward flux would have the wire supplying energy, and \(I^{2}r/\pi a\) would deliver twice the heat produced.
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An infinitely long solenoid, with its axis along \(\hat{k}\), carries a current I. In addition there is a uniform line charge density \(\lambda\) along the axis. If \(\vec{S}\) is the energy flux, in cylindrical coordinates \((\hat{\rho}, \hat{\phi}, \hat{k})\), then
IIT JAM - 2018,
Answer: D
The energy flux \(\vec{S}\) is along \(\hat{\rho} \times \hat{k}\).
Inside the solenoid the magnetic field is axial, \(\vec{B} = \mu_{0} n I\hat{k}\). The line charge gives a radial electric field \(\vec{E} = \frac{\lambda}{2\pi\varepsilon_{0}\rho}\hat{\rho}\). The Poynting vector is
\[ \begin{aligned} \vec{S} &= \frac{1}{\mu_{0}}\vec{E} \times \vec{B} \\ &\propto \hat{\rho} \times \hat{k} = -\hat{\phi} \end{aligned} \]
so the energy flux circulates azimuthally. It has no \(\hat{\rho}\) or \(\hat{k}\) component, which rules out the other three options.
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A constant current I is flowing through a cylindrical conductor. The direction of Poynting vector on the curved surface is :
MH SET - 2019, SET Exam - 2019,
Answer: D
The Poynting vector points radially inward, \(-\hat{r}\).
Inside the conductor the electric field drives the current along the axis, \(\overline{E} \propto \hat{z}\), and the current's magnetic field circles the axis, \(\overline{B} \propto \hat{\phi}\). The Poynting vector \(\overline{S} = \frac{1}{\mu_{0}}\overline{E} \times \overline{B}\) is along \(\hat{z} \times \hat{\phi} = -\hat{r}\).
So energy flows into the wire through its curved surface, supplying the Joule heat \(I^{2}R\). It is not along the current (\(\hat{z}\)), around the wire (\(\hat{\phi}\)) or outward (\(\hat{r}\)).
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