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GATE-2012 ECE Q52 (electromagnetics)

Posted By Krishna Sankar On December 1, 2012 @ 7:21 am In GATE | No Comments

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 [1].

## Ampere’s Law[1] :

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 [2].

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 [3]

[2] Fields and Waves in Communication Electronics, Simon Ramo, John R. Whinnery, Theodore Van Duzer (buy from Amazon.com [4]buy from Flipkart.com [5]).

[3] The youtube videos uploaded by user lasseviren1 [6] aided me in understanding the  integrals in electric field and magnetic field.

[4] Ampere’s Law [1]

[5] Permeability [2]

URL to article: http://www.dsplog.com/2012/12/01/gate-2012-ece-q52-electromagnetics/

URLs in this post:

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

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

[3] http://gate.iitm.ac.in/gateqps/2012/ec.pdf: http://gate.iitm.ac.in/gateqps/2012/ec.pdf