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GATE-2012 ECE Q16 (electromagnetics)

Posted By Krishna Sankar On November 25, 2012 @ 7:59 pm In GATE | No Comments

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

## 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 [1]buy from Flipkart.com [2]).

## Finding the capacitance per unit length:

Using Gauss law [3] :

For a closed Gaussian surface $S$, the electrical flux [4] 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 [5]($\epsilon_0$ 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

$\Large{\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 [6] :

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 [7].

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,

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

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 [8] aided me in understanding the  integrals in electric field and magnetic field. Do checkout the play-lists : Gauss’s Law [9] , Sources of Magnetic Fields [10]

## The characteristic impedance :

From wiki entry on characteristic impedance [11], 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,

$\Huge{\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 [12]

[2] Fields and Waves in Communication Electronics, Simon Ramo, John R. Whinnery, Theodore Van Duzer (buy from Amazon.com [1]buy from Flipkart.com [2]).

[3] The youtube videos uploaded by user lasseviren1 [8] aided me in understanding the  integrals in electric field and magnetic field. Do checkout the play-lists : Gauss’s Law [9]

Sources of Magnetic Fields [10]

[4] Characteristic impedance [11]

[5] Ampere’s Law [6]

[6] Gauss law [3]

[7]  Permeability [7]

[8] Permittivity [5]

[9] Electrical Flux [4]

URL to article: http://www.dsplog.com/2012/11/25/gate-2012-ece-q16-electromagnetics/

URLs in this post:

[3] Gauss law: http://en.wikipedia.org/wiki/Gauss%27s_law

[4] electrical flux: http://en.wikipedia.org/wiki/Electric_flux

[5] permittivity : http://en.wikipedia.org/wiki/Permittivity

[6] Ampere’s Law: http://en.wikipedia.org/wiki/Amp%C3%A8re%27s_circuital_law

[7] permeability of vacuum: http://en.wikipedia.org/wiki/Vacuum_permeability