(No Ratings Yet)

# GATE-2012 ECE Q52 (electromagnetics)

by on December 1, 2012

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

## 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 $\Large{r:

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 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 $\Large{r>a}$:

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 is,

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

Summarizing,

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

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

## 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.combuy 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

D id you like this article? Make sure that you do not miss a new article by subscribing to RSS feed OR subscribing to e-mail newsletter. Note: Subscribing via e-mail entitles you to download the free e-Book on BER of BPSK/QPSK/16QAM/16PSK in AWGN.

{ 1 comment… read it below or add one }

trx training May 7, 2014 at 8:50 pm

That was it. And there was nothing I could do about it. Now. But other ears than mine had heard us talking. The wall was thin, the inside doors flimsy. Bolivar must have heard every word spoken here. He would know that I could take care of myself, know at the same that he had to follow Chaise, run him to ground-and find out where Angelina was being kept. If her freedom were secured then Kaia’s plans were instantly worthless. Which meant that all I could do was mark time, do as I was told. And find a way to open some lines of communication with Bolivar.
trx training http://www.udzungwa.org/trx/