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B.E DEGREE EXAMINATION, NOV/DEC 2006

EC 1253 ELECRTOMAGNETIC FIELDS-KEY(4

thsemester)

Answer ALL questions

Part-A (10*2=20 marks)

1. State divergence theorem.

Ans: The integral of the normal component of any vector field over a closed surface is

equal to the integral of the divergence of this vector field throughout the volume

enclosed by that closed surface.

s F.dS = v (.F)dV

2. Convert the point P(3, 4, 5) from Cartesian to spherical co-ordinates.

Ans: i. r = x2 + y2 + z2 = 9 + 16 + 25 = 50ii. = Cos -1(z/ x2 + y2 + z2) = Cos -1(1 / 2) = 450

iii. = tan -1(y / x) = tan -1(5 / 4) = 51.340

3. State the Biot-Savart s Law for a current element.

Ans: The Biot-Savart s Law states that,

The magnetic field intensity dH produced at a point P due to a differential current

element IdL is,

i. Proportional to the product of the current I and differential length dL.

ii. The sine of the angle between the element and the line joining point P to the element

iii. Inversely proportional to the square of the distance R between the point P and the

element.

Mathematically the Biot-Savart s Law can be stated as,

dH = KIdLSin/R2

4. Write an expression for torque in vector form.Ans: T = R F in Nm Where, T = Torque, F = Force, R = Moment arm

5. A current carrying of 2A flowing is an inductor of inductance 100mH. What is

the energy stored in the inductor?

Solution: Given data: I =2a and L 100mH ;

Wm = LI2

= 200mW.

6. Define Magnetization.

Ans: The magnetization is defined as the magnetic dipole moment per unit volume.Its

unit is A/m. M = Limv

0 (1/v ).a= 1 to nvma

7. State the Faradays Law for a moving charge in constant magnetic field.Ans: Consider that a charge Q is moved in a magnetic field B at a velocity v. then the

force on a charge is given by,

F = Qv B

The induced e.m.f is given by,

Em.dL = (v B).dLIt represents total e.m.f induced when a conductor is moved in a uniform constant

magnetic field.

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8. Write down the Maxwells equations derived from Amperes Law.

Ans: H.dL = s[J + D/t].dSThis is the Maxwells equations derived from Amperes Law in integral form.

H = J + D/tThis is the Maxwells equations derived from Amperes Law in point form.

9. What are the conditions to be satisfied for a linearly polarized uniform planewave?

Ans: When Ex and Ey components are in phase with either equal or unequal amplitudes,

for a uniform plane wave traveling in z-direction, the polarization is linear.

10. What is Brewster angle?

Ans: B =tan-1(2/1) = tan

-1(n2/n1).The significance of Brewster angle is that, whenan unpolarized wave is incident obliquely at the Brewster angle B, only the componentwith perpendicular polarization will be reflected, while component with parallel

polarization will not be reflected. Hence it is also referred as polarizing angle.

Part-B (5*16=80 marks)

11. (a)(i). Derive an expression for the electric field due to a straight uniformly

charged wire of length L meters and with a charge density of +c/m at the pointP which lies along the perpendicular bisector of wire. (10)

Ans:

Consider the line charge distribution having a charge density L as shown inFig.dQ = LdE = Ld40R

2.aR

The aR and dis to be obtained depending upon the co-ordinate system used.

(ii). Given A = (yCosx)ax + (y + ex)az. FindA at the origin. (6)

Solution: A = (yCosx)ax + (y + ex)az

Ax = yCosx, Ay = 0, Az = y + ex

A = det of [ax, ay, az, /x, /y, /z, yCosx, 0, y + ex

](33)

A = ax - exay - Cosxaz

At origin, x = 0, y = 0, z = 0A = ax - ay - az

(OR)

(b)(i). State and prove Gausss Law. (8)

Ans:

Statement of Gausss Law: The electric flux passing through any closed surface is

equal to the total charge enclosed by that surface.

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Proof: d = DndSDn = component of D in the direction of normal to the surface dS

d = DCosdSA.B = ABCosAB = s D.dS Q = Charge enclosed.

This is the mathematical representation of Gausss Law.

(ii). Describe any two applications of Gausss Law. (8)

Ans: i. Used to find line charge, infinite sheet of charge and a spherical distribution of

charge

ii. Used to find the charge enclosed

12.(a)(i). State and prove Amperes circuital Law. (8)

Ans:

Statement of Amperes circuital Law:

The Amperes circuital law states that, the line integral of the magnetic field

intensity vector H around a closed path is exactly equal to the current enclosed by that

path. Mathematically it is given by,

H.dL = IProof:

H = I / 2r.aH.dL = I / 2r.a.rd.a H.dL = I = current carried by conductor

This proves that the integral H.dL along the closed path gives the direct current enclosed

by that closed path.

(ii). A single-phase circuit comprises two parallel conductors A and B, 1cm

diameter and spaced 1 metre apart. The conductors carry currents of +100 and

100 amps respectively. Determine the field intensity at the surface of each

conductor and also in the space exactly midway between A and B. (8)

Solution:

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H at P = H1 + H2 = I1/ 2d1.a+ I2/ 2d2.ad1 =d2 = 0.5m, I1 = I2 = 100A

H at P = 63.6619 aA/m

H1 = I/ 2r.aH2 = I/ 2*(1 0.5*10

-2).a is due to other conductor

HA = [100 / 2*0.5*10-2

+ 100 / 2*0.995]a= 3199.09aA/m

Same is the value of H on the surface of conductor B but in opposite directionHB = - 3199.09aA/m

(OR)

(b)(i). Find an expression for torque acting on a square loop carrying a current I.

(8)

Ans:

For side 2-3, the force exerted is given by,

F1 = I(ay Bax) = - BIazFor side 4-1, the force exerted is given by,

F2 = I(- ay Bax) = - BIazMoment arm for side 2-3 is,R1 = - w/2.ax

Moment arm for side 4-1 is,

R2 = + w/2.ax

Torque T = BIS(-ay)

(ii). What is a scalar magnetic potential? And also derive an expression for vector

magnetic potential. (8)

Ans:

Scalar magnetic potential:

Vm = 0

H = - VmH = J ; J = 0Vm a, b = - a to b

H.dL specified path.Vector magnetic potential:

. (A)= 0A =Vector magnetic potential. B = 0B = AB = 0HJ = 1/0[A] = 1/0[(.A) -

2A]

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13.(i) Explain Poissons and Laplaces equations. (5)

Ans: From Gausss Law,

.D = vD = Flux density and v = Volume charge densityD = EE = - .V

Poissons equation is 2

.V = - v/.2.V = 0 is called Laplaces equation.

(ii). Derive an expression for the capacitance for the capacitance of a spherical

capacitor consisting of two concentric spheres of radii a and b. (6)

Ans:

E = Q/4r2.ar V/m

V = r = b to a Q/4r2.ar .dL

dL = dr.ar

V = Q/V = - Q/[Q/4(1/a 1/b)]C = 4 / (1/a 1/b) in F

(iii). The radii of two spheres differ by 4cm and the capacity of the spherical

condenser is 53.33 PF.If the outer sphere is earthed, calculate the radii assuming

air as dielectric. (5)Solution:

Given data: C = 54pF and b a = 4cm, r = 1C = 4 / (1/a 1/b) in F b > aab = 1.9413*10

-2

b = 4* 10-2

+ a

a2+ 0.04a 0.019413 = 0

a = 0.1207 m

and b = 0.1607 m

(OR)

(b)(i). Derive the expressions for energy stored and energy density in the magnetic

field. (8)

Ans:

Energy stored by an inductor is Wm = * LI2

L = /I = = sB.dSWm = * H

2(x.y.z)

v = (x.y.z)

wm = Limv

0wm/v = * H2

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wm = B2

/.dv

(ii). Derive the magnetic boundary condition at the interface between two magnetic

medium. (8)

Ans: Case: 1. Boundary conditions for normal component

According to Gausss law for the magnetic field,sB.dS = 0

Let the area of the top and bottom is same, equal toS.

topB.dS + bottomB.dS + lateralB.dS = 0For top surface topB.dS = BN1. topdS = BN1.SFor bottom surface bottomB.dS = BN2. bottomdS = BN2.SFor lateral surface lateralB.dS = 0

BN1.S - BN2.S = 0BN1 =BN2; B = H;HN1 / HN2= 2/1= r2/r1

Case: 2 Boundary conditioned for tangential component.

H.dL = IH.dL = a to bH.dL + b to 1H.dL + 1 to cH.dL + c to d H.dL + d to 2H.dL + 2 to aH.dL = Itan 1/ tan 2 = Btan 1/ Btan 2 = 1/2 = r1/r2

14.(a)(i). Derive Maxwells equation derived from Faradays law in integral and

point forms. (8)

Ans:

i.Maxwells equation derived from Faradays law in integral form is given by,

E.dL =- sB/t.dS

ii.Maxwells equation derived from Faradays law in point form is given by,

E = -B/t

(ii). Explain about displacement current and displacement current density. Also

find displacement current density for the field E = 300sin109t V/m (8)

Solution: JD = D/t = /t(E) = 0r and r = 1JD = /t(0E) = 2.6562Cos10

9A/m

2

(OR)

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(b)(i). State and prove Poynting Theorem. (8)

Ans:

Statement of Poynting Theorem:

When electromagnetic waves propagate through space from their source to distant

receiving points, there is a transfer of energy from the source to the receivers. There

exists a simple and direct relation between the rate of this energy transfer and the

amplitudes of electric and magnetic field intensities of the electromagnetic wave.Proof:

P = E H

H = J + D/t.EH = H..EH - E.HUsing Divergence theorem,

v.EH.dV = s EH.dSBy solving we get,

The power flow per unit area is

P = E H

(ii). Briefly explain about the power flow in co-axial cable. (8)

Ans:Consider a co-axial cable in which the power is transferred to the load resistance

R along a cable. There are two conductors namely inner conductor and outer conductor

concentric to each other.

H.dS = IH = I / 2rV = q / 2r.log(b/a)E = V/r.log(b/a)

According to Poynting theorem. The Poynting vector is given by,

P = E H

Poynting vector is given by,

P = E.H

W = Total power flow = surfaceP.daW = V.I

15(a)(i). Derive the general wave equation. (8)

Ans: Consider the Maxwells equation expressed in E and H as

H = J + D/tLet us assume that a free space is perfect dielectric, then J =0,

H = D/tExpressing D in rectangular co-ordinate system,

H = /t( Dx.ax + Dy.ay + Dz.az)

- Hy/t = /t( Dx.ax + Dy.ay + Dz.az)D = EE = - B/t

(Ey/t - Ez/t )ax + (Ex/t - Ez/t )ay + (Ex/t -Ey/t)az =-/t( Bx.ax + By.ay + Bz.az)

2Ex/t2

= v2.2Ex/z

2

Above equation is the wave equation and it is differential equation of second order.

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(ii). Discuss about the plane waves in lossy dielectrics. (8)

Ans: The propagation constant in lossy dielectric is given by, = ( + j)j = j.1- j(/) = j/ + j =

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According to Poynting theorem average power density is given by,

Piavg = Er2/1 W/m

2

The average power transmitted in medium 1 is given by,

Pr avg = Ei2/W/m2

The average power reflected in medium 1 is given by,

Pavg = Er2/ W/m2

Pt avg + Pt avg = Pt avg

ii. Oblique incidence at a plane Dielectric Boundary.

r = iOB/ AO

= OO

Sint / OO

Sini = v2 / v1

Sint / Sini = v2 / v1 = 1/2Sint / Sini = n1 / n2

This is Snells law of refraction.v1 = 1/1.1 = 1/0.1 m/s andv2 = 1/2.2 = 1/0.2 m/s1 =0/1 and 2 =0/2

Sint / Sini = n1 / n2 = v2 / v1 = 1/2 = 1/2