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 _____
\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 _____
0 | |
1 | |
-1 | |
-2 |
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
\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
\lambda | |
0.5\lambda | |
0.25\lambda | |
0.75\lambda |
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.
5.55 | |
2.25 | |
4.45 | |
6.25 |
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 circle of radius 1 with centre at (0, 0). | |
a point at the centre of the chart. | |
a line passing through the centre of the chart | |
a circle of radius 0.5 with centre at (0.5, 0). |
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:
\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 ___________
0.85 | |
1.25 | |
1.73 | |
1.92 |
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
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
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
1.5 \times 10^8 m/s | |
2 \times 10^8 m/s | |
3 \times 10^8 m/s | |
4.5 \times 10^8 m/s |
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}
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
\frac{\mu _0 I}{6 \pi r} | |
\frac{6 \mu _0 I}{5 \pi r} | |
\frac{5 \mu _0 I}{6 \pi r} | |
\frac{{\mu _0}^2 I^2}{2 \pi r^2} |
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 _____ %.
0.86 | |
2.01 | |
2.54 | |
2.68 |
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}
\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.
1-P, 2-R, 3-Q, 4-S | |
1-Q, 2-R, 3-P, 4-S | |
1-Q, 2-S, 3-P, 4-R | |
1-R, 2-Q, 3-S, 4-P |
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?
\frac{HQ}{\varepsilon _0} | |
\frac{HQ}{4\varepsilon _0} | |
\frac{H\varepsilon _0}{4Q} | |
\frac{4H}{Q\varepsilon _0} |
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}
\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.