electromagnetics Archives

GATE-2012 ECE Q28 (electromagnetics)

Question 28 on electromagnetics from GATE (Graduate Aptitude Test in Engineering) 2012 Electronics and Communication Engineering paper.

Q28. A transmission line with a characteristic impedance of 100 ${\Omega}$ is used to match a 50 ${\Omega}$ section to a 200 ${\Omega}$ section. If the matching is to be done both at 429MHz and 1GHz, the length of the transmission line can be approximately

(A) 82.5cm

(B) 1.05m

(C) 1.58m

(D) 1.75m

GATE-2012 ECE Q52 (electromagnetics)

Question 52 on electromagnetics from GATE (Graduate Aptitude Test in Engineering) 2012 Electronics and Communication Engineering paper.

An infinitely long uniform solid wire of radius ${a}$ carries a uniform dc current of density ${\vec{j}}$ .

Q52. The magnetic field at a distance ${r}$ from the center of the wire is proportional to

(A) ${r}$ for ${r<a}$ and ${1/r^2}$ for ${r>a}$

(B) ${0}$ for ${r<a}$ and ${1/r}$ for ${r>a}$

(C) ${r}$ for ${r<a}$ and ${1/r}$ for ${r>a}$

(D) ${0}$ for ${r<a}$ and ${1/r^2}$ for ${r>a}$

Solution

To answer this question on magnetic field, we need to determine the magnetic field inside and outside the cable using Ampere’s Law.

Ampere’s Law :

The total current $I_{enc}$ inside a closed curve $C$ is the line integral of the magnetic field $B$ (in Tesla)

$\begin{array}\oint_C{B}.{dl}&=&\mu_oI_{enc}\end{array}$ ,

where

$B$ is the magnetic field (in Tesla)

$dl$ is the vector representing the infinitesimal line on the closed loop $C$ ,

$I_{enc}$ is the net current enclosed by the closed loop and

$\mu_0$ is the permeability of vacuum.

Let us use this result to find the magnetic field in a uniform solid wire of radius ${a}$ .

Magnetic field in the region ${r<a}$ :

Figure : Solid wire showing the imaginary Amperian loop inside the wire (circle with radius $r$ )

Given that the current density is ${\vec{j}}$ , the current through the area in the circle with radius $r$ is

$I = \vec{j}\pi r^2$

Applying Ampere’s law,

$\begin{array}\oint_C{B}.{dl}&=&\mu I&=&\mu\vec{j}\pi r^2\end{array}$ ,

where

$\mu=\mu_r \mu_0$ is the permeability ( $\mu_0$ is the permeability of vacuum. $\mu_r$ is the permeability of the material).

Given that the magnetic field is $B$ is parallel to the line $dl$ and is uniform across the closed loop, it can be moved out of the integral, i.e

$\begin{array}B\oint_C{dl}&=&\mu \vec{j} \pi r^2\end{array}$ .

The term $\begin{array}\oint_C{dl}\end{array}$ is the circumference of the circle with radius $r$ i.e.

$\begin{array}\oint_C{dl}&=&2\pi r\end{array}$ .

Substituting, the magnetic field in the region $\begin{array}r<a\end{array}$ is,

$\begin{array}B 2\pi r &=&{\mu \vec{j} \pi r^2}\\B&=&\frac{\mu \vec{j} r }{2} \end{array}$ .

Magnetic field in the region ${r>a}$ :

Figure : Solid wire showing the imaginary Amperian loop outside the wire (circle with radius $r$ )

Given that the current density is ${\vec{j}}$ , the current through the area in the circle with radius $r$ is determined only by the current flowing through the cable with in the radius ${a}$ , i.e.

$I = \vec{j}\pi a^2$

Applying Ampere’s law,

$\begin{array}\oint_C{B}.{dl}&=&\mu I&=&\mu\vec{j}\pi a^2\end{array}$ ,

Taking $B$ outside the intergral and substituting for the term $\begin{array}\oint_C{dl}&=&2\pi r\end{array}$ ,

The magnetic field in the region $\begin{array}r<a\end{array}$ is,

$\begin{array}B 2\pi r &=&{\mu \vec{j} \pi a^2}\\B&=&\frac{\mu \vec{j} a^2 }{2r} \end{array}$ .

Summarizing,

${\begin{array}{llllr}B&=&\frac{\mu \vec{j} r}{2},&{ r < a}\\&=&\frac{\mu \vec{j} a^2 }{2r},&{ r > a}\end{array}}$

Based on the above, the right choice is (C) i.e. The magnetic field at a distance ${r}$ from the center of the wire is proportional to ${r}$ for ${r<a}$ and ${1/r}$ for ${r>a}$

References

[1] GATE Examination Question Papers [Previous Years] from Indian Institute of Technology, Madras http://gate.iitm.ac.in/gateqps/2012/ec.pdf

[2] Fields and Waves in Communication Electronics, Simon Ramo, John R. Whinnery, Theodore Van Duzer (buy from Amazon.com, buy from Flipkart.com).

[3] The youtube videos uploaded by user lasseviren1 aided me in understanding the integrals in electric field and magnetic field.

[4] Ampere’s Law

[5] Permeability

GATE-2012 ECE Q16 (electromagnetics)

Question 16 on electromagnetics from GATE (Graduate Aptitude Test in Engineering) 2012 Electronics and Communication Engineering paper.

Q16. A coaxial cable with an inner diameter of 1mm and outer diameter of 2.4mm is filled with a dielectric of relative permittivity 10.89. Given ${\mu_0=4\pi\mbox{x}10^{-7}\mbox{ H/m, }\epsilon_0={10^{-9}}/{36\pi}\mbox{ F/m}}$ , the characteristic impedance of the cable is

(A) ${330\mbox{\Omega}}$

(B) ${100\mbox{\Omega}}$

(C) ${143.3\mbox{\Omega}}$

(D) ${43.4\mbox{\Omega}}$

Solution

To answer this question, am referring to the discussion on capacitance per unit length (Section 1.9) , inductance per unit length (Section 2.4) and characteristic impedance (Section 5.2) of a coaxial cable from Fields and Waves in Communication Electronics, Simon Ramo, John R. Whinnery, Theodore Van Duzer (buy from Amazon.com, buy from Flipkart.com).

Finding the capacitance per unit length:

Using Gauss law :

For a closed Gaussian surface $S$ , the electrical flux is given by :

$\begin{array}\Phi_{\mbox{E}}&=&\oint_S{E}.{dA}&=&\frac{Q_s}{\Large{\epsilon_0}}\end{array}$ , where

$\begin{array}\Phi_{\mbox{E}}\end{array}$ is the electrical flux,

$E$ is the electric field,

$dA$ is the vector representing the infinitesimal area on the surface $S$ ,

$Q_s$ is the net charge enclosed by the surface and

$\epsilon_0$ is the permittivity of vacuum.

Let us use this result to find the electrical field in a coaxial cable formed by two conducting cylinders of radii $a$ and $b$ respectively with a dielectric $\epsilon$ between them (shown in the figure below).

Figure : Coaxial cable showing the imaginary Gaussian surface (cylinder with radius $r$ and length $l$ )

For finding the electric field $E$ in region $\begin{array}a&<&r&<&b\end{array}$ , assume a cylindrical Gaussian surface of radius $r$ and length $l$ (as shown by the dashed line).

Applying Gauss law,

$\begin{array}\oint_S{E}.{dA}&=&\frac{Q_s}{\Large{\epsilon}}&=&\frac{q_ll}{\epsilon}\end{array}$ ,

where

$q_l$ is the charge per unit length,

total charge enclosed is the charge per unit length multiplied by the length i.e. $Q_s = q_ll$ and

$\epsilon=\epsilon_r \epsilon_0$ is the permittivity ( $\epsilon_0$ is the permittivity of vacuum, $\epsilon_r$ is the permittivity of the material).

Given that the electrical field is $E$ is parallel to the surface $dA$ and is uniform across the Gaussian surface, it can be moved out of the integral, i.e

$\begin{array}E\oint_S{dA}&=&\frac{q_ll}{\epsilon}\end{array}$ .

The term $\begin{array}\oint_S{dA}\end{array}$ is the surface area of the cylinder with radius $r$ of length $l$ and is,

$\begin{array}\oint_S{dA}&=&2\pi rl\end{array}$ .

Substituting, the electric field in the region $\begin{array}a&<&r&<&b\end{array}$ is,

$\begin{array}E&=&\frac{q_l}{2\pi\epsilon r}\end{array}$ .

With the above electric field, the potential difference between the coaxial cylinders is computed as,

$\begin{array}\phi_a - \phi_b &=&-\int_a^b\frac{q_ldr}{2\pi \epsilon r}\\&=&\frac{q_l}{2\pi\epsilon}\ln$\frac{b}{a}$\end{array}$ .

The capacitance between two electrodes is defined as the charge $Q$ on each electrode per volt of potential difference $\phi_a - \phi_b$ between then,

$C = \frac{Q}{\phi_a - \phi_b}$ .

Assuming that the field is only radial and the total charge per unit length is $q_l$ , the capacitance per unit length is

${\begin{array}C&=&\frac{2\pi\epsilon}{\ln$b/a$},&\mbox{ F/m}\end{aray}}$ .

Finding the inductance per unit length:

Using Ampere’s Law :

The total current $I_{enc}$ inside a closed curve $C$ is the line integral of the magnetic field $B$ (in Tesla)

$\begin{array}\oint_C{B}.{dl}&=&\mu_oI_{enc}\end{array}$ ,

where

$B$ is the magnetic field (in Tesla)

$dl$ is the vector representing the infinitesimal line on the closed loop $C$ ,

$I_{enc}$ is the net current enclosed by the closed loop and

$\mu_0$ is the permeability of vacuum.

Let us use this result to find the magnetic field in a coaxial cable formed by two conducting cylinders of radii $a$ and $b$ respectively with a dielectric having permeability of $\mu$ between them (shown in the figure below).

Figure : Coaxial cable showing the imaginary Amperian loop (circle with radius $r$ )

Applying Ampere’s law,

$\begin{array}\oint_C{B}.{dl}&=&\mu I\end{array}$ ,

where

$\mu=\mu_r \mu_0$ is the permeability ( $\mu_0$ is the permeability of vacuum. $\mu_r$ is the permeability of the material).

Given that the magnetic field is $B$ is parallel to the line $dl$ and is uniform across the closed loop, it can be moved out of the integral, i.e

$\begin{array}B\oint_C{dl}&=&\mu I\end{array}$ .

The term $\begin{array}\oint_C{dl}\end{array}$ is the circumference of the circle with radius $r$ and is, $\begin{array}\oint_C{dl}&=&2\pi r\end{array}$ .

Substituting, the magnetic field in the region $\begin{array}a&<&r&<&b\end{array}$ is,

${\begin{array}B&=&\frac{\mu I}{2\pi r}\end{array}}$ .

The inductance is defined as ratio of magnetic field over a surface and the the current.

$\begin{array}L&=&\frac{1}{I}\int_SB.dS\end{array}$ .

For a unit length, the integral of magnetic field in the region $\begin{array}a&<&r&<&b\end{array}$ is,

$\begin{array}\int_SB.dS&=&\int_a^b\mu\frac{I}{2\pi r}dr&=&\frac{\mu I}{2\pi}\ln\(b/a)\end{array}$ .

Substituting, the inductance per unit length is,

$\begin{array}L&=&\frac{\mu}{2\pi}\ln$b/a$&\mbox{ H/m}\end{array}$ .

Note :

YouTube videos uploaded by user lasseviren1 aided me in understanding the integrals in electric field and magnetic field. Do checkout the play-lists : Gauss’s Law , Sources of Magnetic Fields

The characteristic impedance :

From wiki entry on characteristic impedance, the general expression for the characteristic impedance of a transmission line is

$\begin{array}Z_0&=\sqrt{\frac{R+j\omega L}{G+j\omega C}}\end{array}$ ,

where,

$R$ is the resistance per unit length,

$L$ is the inductance per unit length,

$C$ is the capacitance per unit length,

$G$ is the conductance per unit length and

$\omega$ is the angular frequency.

Assuming a loss less transmission line, i.e. $R=G=0$ , the equation reduces to,

$\begin{array}Z_0&=\sqrt{\frac{L}{C}}\end{array}$ .

Substituting the expressions for inductance per unit length and capacitance per unit length for a coaxial cable,

${\begin{array}Z_0&=\sqrt{\frac{\mu}{\epsilon}}\frac{\ln$b/a$}{2\pi}\end{array}}$ .

Substituting the numbers from the problem at hand,

$\mu \simeq \mu_0 = 4\pi\mbox{x}10^{-7}$

$\epsilon =\epsilon_0 \epsilon_r = 10^{-9}/{36\pi} * 10.89$

$b = 2.4$

$a = 1$ ,

the characteristic impedance is,

$\begin{array}Z_0&=&15.918\end{array}$ .

The calculated choice is not listed in the options. However if we ignore the $\2\pi$ term, the calculated number comes to around $100$ . That does not help, does it? 🙂

References

[1] GATE Examination Question Papers [Previous Years] from Indian Institute of Technology, Madras http://gate.iitm.ac.in/gateqps/2012/ec.pdf

[2] Fields and Waves in Communication Electronics, Simon Ramo, John R. Whinnery, Theodore Van Duzer (buy from Amazon.com, buy from Flipkart.com).

[3] The youtube videos uploaded by user lasseviren1 aided me in understanding the integrals in electric field and magnetic field. Do checkout the play-lists : Gauss’s Law

Sources of Magnetic Fields

[4] Characteristic impedance

[5] Ampere’s Law

[6] Gauss law

[7] Permeability

[8] Permittivity

[9] Electrical Flux

Q28. A transmission line with a characteristic impedance of 100 is used to match a 50 section to a 200 section. If the matching is to be done both at 429MHz and 1GHz, the length of the transmission line can be approximately

(A) 82.5cm

(B) 1.05m

(C) 1.58m

(D) 1.75m

An infinitely long uniform solid wire of radius carries a uniform dc current of density .

Q52. The magnetic field at a distance from the center of the wire is proportional to

(A) for and for

(B) for and for

(C) for and for

(D) for and for

Solution

Ampere’s Law :

Magnetic field in the region :

Magnetic field in the region :

References

Q16. A coaxial cable with an inner diameter of 1mm and outer diameter of 2.4mm is filled with a dielectric of relative permittivity 10.89. Given , the characteristic impedance of the cable is

(A)

(B)

(C)

(D)

Solution

Finding the capacitance per unit length:

Finding the inductance per unit length:

The characteristic impedance :

References

Q28. A transmission line with a characteristic impedance of 100 ${\Omega}$ is used to match a 50 ${\Omega}$ section to a 200 ${\Omega}$ section. If the matching is to be done both at 429MHz and 1GHz, the length of the transmission line can be approximately

An infinitely long uniform solid wire of radius ${a}$ carries a uniform dc current of density ${\vec{j}}$ .

Q52. The magnetic field at a distance ${r}$ from the center of the wire is proportional to

(A) ${r}$ for ${r<a}$ and ${1/r^2}$ for ${r>a}$

(B) ${0}$ for ${r<a}$ and ${1/r}$ for ${r>a}$

(C) ${r}$ for ${r<a}$ and ${1/r}$ for ${r>a}$

(D) ${0}$ for ${r<a}$ and ${1/r^2}$ for ${r>a}$

Magnetic field in the region ${r<a}$ :

Magnetic field in the region ${r>a}$ :

Q16. A coaxial cable with an inner diameter of 1mm and outer diameter of 2.4mm is filled with a dielectric of relative permittivity 10.89. Given ${\mu_0=4\pi\mbox{x}10^{-7}\mbox{ H/m, }\epsilon_0={10^{-9}}/{36\pi}\mbox{ F/m}}$ , the characteristic impedance of the cable is

(A) ${330\mbox{\Omega}}$

(B) ${100\mbox{\Omega}}$

(C) ${143.3\mbox{\Omega}}$

(D) ${43.4\mbox{\Omega}}$