# Electromagnetics

 Question 1
The magnetic field of a uniform plane wave in vacuum is given by

$\vec{H}(x,y,z,t)=(\hat{a}_x+2\hat{a}_y+b\hat{a}_z) \cos (\omega t+3x-y-z)$

The value of b is _____
 A 0 B 1 C -1 D -2
GATE EC 2020      Uniform Plane Waves
Question 1 Explanation:
For uniform plane wave
$\hat{a}_{H}\cdot \hat{a}_{\rho }=0$
$\hat{a}_{H}$is unit vector in magnetic field direction $\hat{a}_{\rho }$ is unit vector in power flow direction
$\hat{a}_{H }=\frac{1\hat{a}_{x}+2\hat{a}_{y}+b\hat{a}_{z}}{\sqrt{1^{2}+2^{2}+b^{2}}}$
$\hat{a}_{\rho }=\frac{-3\hat{a}_{x}+\hat{a}_{y}+\hat{a}_{z}}{\sqrt{3^{2}+1^{2}+1^{2}}}$
$\hat{a}_{H}\cdot \hat{a}_{\rho }=0 (\hat{a}_{x}+2\hat{a}_{y}+b\hat{a}_{z})\cdot (-3\hat{a}_{x}+\hat{a}_{y}+\hat{a}_{z})=0$
$-3+2+b=0$
$b=1$
 Question 2
For an infinitesimally small dipole in free space, the electric field $E_\theta$ in the far field is proportional to $(e^{-jkr}/r) \sin \theta$, where $k = 2 \pi /\lambda$. A vertical infinitesimally small electric dipole ($\delta l \lt \lt \lambda$) is placed at a distance $h(h \gt 0)$ above an infinite ideal conducting plane, as shown in the figure. The minimum value of h, for which one of the maxima in the far field radiation pattern occurs at $\theta =60^{\circ}$, is
 A $\lambda$ B 0.5$\lambda$ C 0.25$\lambda$ D 0.75$\lambda$
GATE EC 2020      Antennas
Question 2 Explanation:

\begin{aligned}\left | \text{Total E }\right |&=\left | (E_{\text{single element}}) \right | \left | (A.F.) \right | \\ \left | (A.F.) \right |&=\frac{\sin\left ( N\frac{\psi }{2} \right )}{\sin (\frac{\psi }{2})}=\frac{\sin \left ( 2\frac{\psi }{2} \right )}{\sin \left ( \frac{\psi }{2} \right )} \\ & =\frac{2\sin \left ( \frac{\psi }{2} \right )\cos (\frac{\psi }{2})}{\sin \frac{\psi }{2}} \\ & =2\cos (\frac{\psi }{2})\\ \left | A.F_{N} \right | &=\frac{A.F}{A.F_{max}}=\frac{2\cos (\frac{\psi }{2})}{2} \\ &=\left | \cos \left ( \frac{\psi }{2} \right ) \right | \\ \text{where, }\psi &=\beta d\cos \theta =\frac{2\pi }{\lambda }(2h)\cos \theta \\ \left.\begin{matrix} \left | A.F_{N} \right |_{\theta =60^{\circ}} \end{matrix}\right|& =\left | \cos (\frac{2\pi }{\lambda }h\cos 60^{\circ}) \right |\\ &=\left | \cos \left ( \frac{\pi h}{\lambda } \right ) \right |\\ \cos \theta \text{ is maximum ,where } \theta &=n\pi \;\; n=0,1,2..\\ \frac{\pi h}{\lambda }&=n\pi \Rightarrow h=n\lambda \\ \Rightarrow \; \text{For } n=1, \;\; h_{min}&=\lambda \end{aligned}
 Question 3
A transmission line of length $3\lambda /4$ and having a characteristic impedance of 50$\Omega$ is terminated with a load of 400$\Omega$. The impedance (rounded off to two decimal places) seen at the input end of the transmission line is __________ $\Omega$.
 A 5.55 B 2.25 C 4.45 D 6.25
GATE EC 2020      Transmission Lines
Question 3 Explanation:

$Z_{in}\,\, for \, (l=\lambda /4)=\frac{Z_{0}^{2}}{Z_{L}}=\frac{50^{2}}{400}=\frac{25}{4}=6.25\Omega$
 Question 4
The impedances Z=jX, for all X in the range ($-\infty ,\infty$), map to the Smith chart as
 A a circle of radius 1 with centre at (0, 0). B a point at the centre of the chart. C a line passing through the centre of the chart D a circle of radius 0.5 with centre at (0.5, 0).
GATE EC 2020      Transmission Lines
Question 4 Explanation:
For given impedance Normalized impedance is
\begin{aligned} \frac{Z}{Z_{0}}&=\frac{jX}{Z_{0}}\\ Z&=jX \\ \Rightarrow \; Z&=0+jX \end{aligned}
Normalized Resistance $=0 \; \Rightarrow \; r=0$
$X=-\infty \text{ to } \infty$
$r=0$ and X from $-\infty \, to\, \infty$ is a unit circle (radius 1) and centre (0,0) on a complex reflection coeffient plane:
 Question 5
A rectangular waveguide of width w and height h has cut-off frequencies for $TE_{10} \; and \;TE_{11}$ modes in the ratio 1:2. The aspect ratio w/h, rounded off to two decimal places, is ___________
 A 0.85 B 1.25 C 1.73 D 1.92
GATE EC 2019      Waveguides
Question 5 Explanation:
$f_{c m n}=\frac{c}{2} \sqrt{\left(\frac{m}{a}\right)^{2}+\left(\frac{n}{b}\right)^{2}}$
For TE mode
$f_{c10}=\frac{c}{2 w} \quad\ldots(i)$
and For $TE_{11}$ mode.
\begin{aligned} f_{c11} &=\frac{c}{2} \sqrt{\frac{1}{w}^{2}+\left(\frac{1}{h}\right)^{2}} &\ldots(i)\\ &=\frac{c}{2 w} \sqrt{1+\left(\frac{w}{h}\right)^{2}} &\ldots(ii)\\ \text{given}\quad\frac{f_{c10}}{f_{11}} &=\frac{1}{2} &\ldots(iii) \end{aligned}
put (i). (in) in (ii)
$\Rightarrow \frac{\frac{c}{2 w}}{\frac{c}{2 w} \sqrt{1+\left(\frac{w}{h}\right)^{2}}}=\frac{1}{2} \Rightarrow \sqrt{1+\left(\frac{w}{h}\right)^{2}}=2$
On solving above equation, we get,
$\frac{w}{h}=\sqrt{3}=1.732$
 Question 6
The dispersion equation of a waveguide,which relates the wavenumber kto the frequency $\omega$, is

$k(\omega) =(1/c)\sqrt{\omega ^2 -{\omega _0}^2}$

where the speed of light $c=3 \times 10^8$ m/s, and $\omega_0$ is a constant. If the group velocity is $2 \times 10^8$ m/s, then the phase velocity is
 A $1.5 \times 10^8$ m/s B $2 \times 10^8$ m/s C $3 \times 10^8$ m/s D $4.5 \times 10^8$ m/s
GATE EC 2019      Waveguides
Question 6 Explanation:
By definition $v_{p}=\frac{\omega}{\beta}=\frac{\omega}{k}$
where, $k(\omega)=\left(\frac{1}{c}\right) \sqrt{\omega^{2}-\omega_{0}^{2}} \quad$(given)
$\therefore v_{p}=\frac{c}{\sqrt{1-\left(\frac{\omega_{0}}{\omega}\right)^{2}}}$
by definition,
\begin{aligned} v_{g} &=\frac{d \omega}{d \beta}=\frac{d \omega}{d k} \\ &=\frac{d k}{d \omega}=\frac{1}{c} \frac{1}{2 \sqrt{\omega^{2}-\omega_{0}^{2}}} \times 2 \omega \\ \text{or}\quad v_{g} &=c \sqrt{1-\left(\frac{\omega_{0}}{\omega}\right)^{2}}\\ \because \quad v_{p} \cdot v_{g} &=c^{2} \\ \therefore \quad v_{p} &=\frac{c^{2}}{v_{g}}=\frac{\left(3 \times 10^{8}\right)^{2}}{2 \times 10^{8}} \\ &=4.5 \times 10^{8} \mathrm{m} / \mathrm{sec} \end{aligned}
 Question 7
Two identical copper wires W1 and W2, placed in parallel as shown in the figure, carry currents I and 2I, respectively, in opposite directions. If the two wires are separated by a distance of 4r, then the magnitude of the magnetic field $\vec{B}$ between the wires at a distance r from W1 is
 A $\frac{\mu _0 I}{6 \pi r}$ B $\frac{6 \mu _0 I}{5 \pi r}$ C $\frac{5 \mu _0 I}{6 \pi r}$ D $\frac{{\mu _0}^2 I^2}{2 \pi r^2}$
GATE EC 2019      Basics of Electromagnetics
Question 7 Explanation:

Magnetic flux density $(\vec{B})$ at r distance due to infinite line carrying current I is $|\vec{B}|=\frac{\mu_{0} I}{2 \pi \rho} .$
$\vec{B}$ at r distance due to $W_{1}$ wire
$=\left|\vec{B}_{1}\right|=\frac{\mu_{0} I}{2 \pi r}\qquad \ldots(i)$
$\vec{B}$ at 3r distance due to $W_{2}$ wire
$=\left|\vec{B}_{2}\right|=\frac{\mu_{0}(2 I)}{2 \pi(3 r)}\qquad \ldots(ii)$
From right hand thumb rule, $\vec{B}$ due to both lines add in between conductors.
\begin{aligned} {So,}\qquad |\vec{B}|&=\left|\vec{B}_{1}\right|+\left|\vec{B}_{2}\right|\\ \therefore \qquad |\vec{B}|&=\frac{\mu_{0} I}{2 \pi r}+\frac{2 \mu_{0} I}{6 \pi r}=\frac{5 \mu_{0} I}{6 \pi r} \end{aligned}
 Question 8
Radiation resistance of a small dipole current element of length l at a frequency of 3 GHz is 3 ohms. If the length is changed by 1%, then the percentage change in the radiation resistance, rounded off to two decimal places, is _____ %.
 A 0.86 B 2.01 C 2.54 D 2.68
GATE EC 2019      Antennas
Question 8 Explanation:
Radiation resistance of a small dipole current element of length 'I' is
$\begin{array}{l} R_{\mathrm{rad}}=80 \pi^{2}\left(\frac{l}{\lambda}\right)^{2} \Rightarrow R \propto l^{2} \\ \frac{R_{2}}{R_{1}}=\left(\frac{l_{2}}{l_{1}}\right)^{2} \end{array}$
If length is changed by 1% then percentage change in the radiation resistance.
$\frac{R_{2}}{R_{1}}=\left(\frac{1.01 l}{l}\right)^{2}=1.0201$
$\text { Percentage change in radiation resistance }$
$\begin{array}{l} =\frac{R_{2}-R_{1}}{R_{1}} \times 100 \\ =0.0201 \times 100=2.01 \% \end{array}$
 Question 9
In the table shown, List I and List II, respectively, contain terms appearing on the left-hand side and the right-hand side of Maxwell's equations (in their standard form). Match the left-hand side with the corresponding right-hand side.
 A 1-P, 2-R, 3-Q, 4-S B 1-Q, 2-R, 3-P, 4-S C 1-Q, 2-S, 3-P, 4-R D 1-R, 2-Q, 3-S, 4-P
GATE EC 2019      Basics of Electromagnetics
Question 9 Explanation:
\begin{aligned} \nabla \cdot \vec{D} &=\rho_{v} \\ \nabla \times \vec{E} &=-\frac{\partial \vec{B}}{\partial t} \\ \nabla \cdot \vec{B} &=0 \\ \nabla \times \vec{H} &=\vec{J}+\frac{\partial \vec{D}}{\partial t} \end{aligned}
 Question 10
What is the electric flux ($\int \vec{E}\cdot d\hat{a}$) through a quarter-cylinder of height H (as shown in the figure) due to an infinitely long line charge along the axis of the cylinder with a charge density of Q?
 A $\frac{HQ}{\varepsilon _0}$ B $\frac{HQ}{4\varepsilon _0}$ C $\frac{H\varepsilon _0}{4Q}$ D $\frac{4H}{Q\varepsilon _0}$
GATE EC 2019      Basics of Electromagnetics
Question 10 Explanation:
Electric field intensity $(\vec{E})$ at '$\rho$' distance due to infinite long line having line charge density Q is
\begin{aligned} \vec{E} &=\frac{Q}{2 \pi \varepsilon_{0} \rho} \hat{a}_{\rho} \\ \int \vec{E} \cdot \vec{da} &=\iint \frac{Q}{2 \pi \varepsilon_{0} \rho} \hat{a}_{p} \cdot \rho d \phi d z \hat{a}_{\rho} \\ &=\frac{Q}{2 \pi \varepsilon_{0}} \int_{<\pi / 2>} d \phi \int_{z=0}^{H} d z \\ &=\frac{Q}{2 \pi \varepsilon_{0}}\left(\frac{\pi}{2}\right) H=\frac{H Q}{4 \varepsilon_{0}} \end{aligned}

There are 10 questions to complete.