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VELAMMAL COLLEGE OF ENGINEERING AND TECHNOLOGY, Madurai – 625 009

Department of Electrical and Electronics EngineeringAcademic Year 2016-2017

Name of the Staff : K.SnehaName of the Subject : Electromagnetic Theory/EE 8391Class : II Year EEE A & B SectionDate :

UNIT IV-TUTORIAL – 3 PERIODS

1. A conductor of length 100cm moves at right angles to a uniform field of strength 10,000 lines per cm2 with a velocity of 50 m/sec. Calculate the emf induced init. Find also the value of induced emf when the conductor moves at an angle of 300 to the field direction.Solution:

Length = 100 x 10-2 m ; velocity = v = 50 m/sec.

B = 10,000 lines /m2 = 1 Wb /m2 ; = 300

Case I :

= 900

e = Blv sin

= 1 x 100 x 10-2 x 50

= 50 volts.

Case II:

= 300

e = 1 x 100 x 10-2 x sin 300

= 25 volts.

2. A conducting cylinders of radius 5cm and height 20cm rotates at 600rev / sec in a radial field B = 0.5Tesla. Sliding contacts at the top and bottom are connected to voltmeter. Find the induced voltage?Solution:

Induced voltage, V = ∫ E⋅d l

E=v×B

Velocity, u = x B ; B = 0.5a Tesla

= 2 N

u = 2 x 600 x 0.05 a x 0.5 a

E = 94.25 (-az)

As length is varying along z-axis

dl = dz az

V=∫0

0. 2

E (−az )⋅dz az

=−94 .25×0 . 2= −18 .85 volts .

3. The conduction current flowing through a wire with conductivity = 3 x 107 S/m and relative permittivity r = 1 is given by Ic = 3 sint (mA). If = 108 rad /sec. Find the Displacement current?Solution:

The displacement current density, JD = (dE / dt)

E is found from conduction current density as,

Jc = E

And Ic = E x A

E=I cσA

=3×10−3sinωt3×107×A

=1×10−10

Asinωt

∂E∂ t

=1×10−10

Aωcosωt

Jd=εω(1×10−10

A )cosωt = 8 .854×10−4

Acosωt

Displacement current, Id = ( Jd /A) = 8.854 x 10-14 cost Ampere.

4. Find JD:a) Next to your radio where local AM station provide E = 0.02 sin[0.01927(3 x 108t-z)]

ax

b) In a good conductor with = 107 V/m and J = 107 sin 120t ax

Solution:

a) JD =

∂D∂ t

=ε0∂E∂ t

= 8.854 x10-12 [ 0.02 cos[0.01927(3 x108t-z)] x 0.01927 x 3 x 108]

= 1.02 cos[0.01927(3 x 108t-z)]

b) J = EE = sin 120 t

JD =

∂E∂ t =3.33 cos 120t.

5. Let µ=10-5 H/m, = 4 x 10-9 F/m, = 0 and V = 0. Find K, so that each of the following pairs of fields satisfies Maxwell’s equations,

(i) D = 6 ax – 2y ay + 2z az nC / m2

H = Kx ax + 10y ay – 25z az A/m

(ii) E = (20y – kt) ax V/m.

H = (y + 2 x 106t) az A/m.

Solution:

µ=10-5 H/m ; = 4 x 10-9 F/m ; = 0 ; V = 0

Case (i):

D = 6 ax – 2y ay + 2z az nC / m2

H = Kx ax + 10y ay – 25z az A/m

Maxwell’s equation is

B = µH = 10-5(y + 2 x106t) az

-

∂B∂ t = -20az -----(1)

∇×E=−∂B∂ t

∇×E=¿|ax a y az ¿||∂∂ x

∂∂ x

∂∂ x ¿|¿

¿¿¿

-----(2)

From (1) and (2), equation is verified.

Similarly verify other equations.

Case (ii):

E = (20y – kt) ax V/m.

H = (y + 2 x 106t) az A/m.

Consider the Maxwell’s equation,

Here,

∂D∂ t = 0 --- (3)

∇×H=¿|ax ay az ¿||∂∂ x

∂∂ x

∂∂ x ¿|¿

¿¿¿

Hence verified.

Similarly for other Maxwell’s equations.

6. (a) An area of 0.65 m2 in the z = 0 plane is enclosed by a filamentary conductor. Find the induced voltage, given that, B = 0.05 cos 103t (ay+az)/2.Solution:

v = −∫

S

∂B∂ t

⋅ds az

=∫S

50 sin 103 t (ay+az

√2 )⋅ds az

¿23 . 0 sin103 t volts .

∇×H=∂D∂ t

(b) Find the induced voltage in the conductor along x-axis of length 0.2m from the origin where B = 0.04 ay Tesla and U = 2.5 sin 103t az.

Solution:

Em = U ¿ B

= 0.10 sin 103t (-ax) V/m

v = −∫

0

0 .2

0 . 10sin 103 t(−ax )⋅dx ax

= -0.02 sin103t volts.




UNIT III-TUTORIAL – 3 PERIODS

1. Determine H at P1(0.4, 0.3, 0) in the field of an 8A filamentary current directed inward from infinity to the origin on the positive x-axis and then outward to infinity along y-axis.Solution:

Magnetic Field Intensity for a finite length current element is given as,

H x=I4 πρx

(sinα 2 x−sinα1 x ) aφ

H y=I4 πρ y

(sin α 2 y−sinα 1 y ) aφ

H x=84 π (0 .3 )

(sin 53 .1−sin(−90)) az

= 12π

az A /m .

H x=I4 πρx

( sin(36 . 9)−sin(90 ))(− az )

=−8π

az A/m .

H=H x+H y=−6 .37 az A /m .

2. Consider a Square loop of sides (1, 0, 0), (1, 2, 0), (3, 0, 0) and (3, 2, 0) in the z = 0 plane carrying 2mA in the field of an infinite filament on the y-axis. And a current of 5A flows towards the origin along the y-axis. Calculate the total force on the loop.Solution:

The Magnetic Field Intensity is given as,

H=I2πx

az=152πx

az A /m

and B=μ0 H =3×10−6

xazTesla

Force equation is, F=−I∮B×dl

=−2×103×3×10−6 [∫1

3 az

x dx ax+∫0

2 az

3 dy ay+∫3

1 az

x dx ax+∫2

0 az

1 dy ay ]on solving,

F = -8 axNewton.

3. An Air-cored toriod of cross-sectional area 6cm2, mean radius of 15 cm with an sir gap of 2mm. The magnetic field intensity of steel is 200A-t and B =1T and find:

a. Vm,air

b. Vm,steel

c. The current required in a 1300 turn coil linking the left leg.Solution:

(a) Rair =

dair

μS= 2×10−3

4 π×10−7×6×10−4 =2. 65×106 A−t /Wb .

= BS = 1(6 x 10-4) = 6 x 10-4 Wb.

Vm, air = (6 x 10-4) (2.65 x 106) = 1590 A-t.

(b) Hsteel = 200 A-t.Vm, steel = Hsteel dsteel = 200 x 0.3 =188 A-t.

(c) Total mmf = 1778 A-tCoil Current = (total mmf / number of turns) = 3.56 A.

4. State Ampere’s circuital law and explain how it can be used for magnetic field calculations:Ans:

It states that the line integral of Magnetic Field Intensity about any closed path is exactly equal to the direct current enclosed by that path.

∮ H⋅dl=I enclosed

1

2

(0, 4)(8, 4)

(8, 0)(0, 0) 10A

2m

4m

In Ampere’s law , to determine H there must be a degree of symmetry and the following conditions is to be met:

1. At each point of the closed path H is either tangential or normal to the path.2. H has the same value at all points of the path where H is tangential.

Application of Ampere’s Law:

Infinite long filament: Infinitely long coaxial transmission line:

5. A rectangular loop carrying 10A current is placed on Z = 0 plane. Find the Field Intensity at (4, 2, 0).Solution:

(i) H1 at P due to Horizontal side of 8m is,

H1=1

4 πh (cosα 1+cos α2)

h = 2m ; cos1 = cos2 = 0.894.

H1 = (10 / (4(2)) x 2 x 0.984) az = 0.72 az AT / m.

(ii) H2 at P due to vertical side of 4m.h = 4m ; cos1 = cos2 = 0.447.

H2=104 π×4

(2×0 .447 )

=0. 178 az AT /m .

H = H1 + H2

21

P

4m

= 2 x 0.172 + 2 x 0.178

= 1.779 az.

6. Find H at the center of an Equilateral triangular loop of side 4m carrying current of 5A?

Solution:

Magnetic Field Intensity at appoint due to finite length of conductor,

H1=1

4 πh (cosα 1+cos α2) az

tan 300 = (h / 2) ; h = 1.154 m.

cos 1 = cos2 = cos 300

H1=54 π (1 .154 )

(cos30+cos30 )az

= 0 . 597 az AT /m .

Magnetic Field due to all the three sides is, 3 x 0.597 = 1.79 AT/m.




UNIT V-TUTORIAL – 3 PERIODS

1. A free Space- sliver interface has Eio = 100 V/m on the free space side. The frequency is 15MHz and the = 61.7 Ms/m. Determine Ero and Eto at the interface.Solution:

For a Free space conductor interface, τ = -1

Ei 0

Er 0=−1 ⇒ E r 0=−100 V /m .

τ=Et 0

Ei 0=

2η2

η1+η2

η2 is for sliver medium,

η2=√ωμσ

∠ 450

=√2 π×150×106×4 π×10−7×161 .7×10−6 = 1.38×10−3Ω

η1 at the free space is 120.

./451035.7

1038.13771038.12

040

3

3

0

0

mVE

EE

t

i

t

2. A Steel pipe is constructed of a material for which µr = 180 and = 4 x 106 S/m. The two radii are 5 and 7mm and length is 75m. If the total current I(t) carried by the pipe is 8 cos t , where = 1200 rad/s.Find (i) Skin Depth. (ii) Effective Resistance.

Solution:

µr = 180 and = 4 x 106 S/m.

a1 = 5mm ; a2 = 7mm ; L = 75m

I(t) = 8 cost Ampere ; = 1200 rad/sec.

(a) Skin Depth:

=

1√πf μσ

= 1√πfμr μ0 σ

=1¿ √π×1200π

2 π×180×4 π×10−7×4×106 ¿

¿¿

= 0.766 mm.

(b) Effective Resistance:

R =

L2πa2σδ

=752 π×7×10−3×4×106×0 .766×10−3

= 0.557 ohms.

3. In a non – Magnetic medium, E = 4 sin (2 x 107t – 0.8x) az V/m. Find (a) r and . (b) The time average power carried by the wave.Solution:

E = 4 sin (2 x 107t – 0.8x) az V/m

= 0 ; = 0

The wave is traveling along x- axis, H along (-ay) and E along az direction.

= 2 x 107t rad/sec; E0 =4 V/m.

= 0.8 rad / m.

β=ωμ

⇒ μ=2π×107

0 . 8=7 . 85×107m /sec

u=1√μ0 μr ε0ε r

=3×108

√ μr εrFor non−magentic material⇒μr=1

εr=3×108

7 . 85×107 =14 . 6

Intrinsic Im pedance⇒η=√μεη=√ μ0 μr

ε 0εr=120π √1

14 .6=98 . 7 Ω

Average Power , Pav=E0

2

2|η|e−2αx cosθη ax

As α=0 and θη=0

Pav=E0

2

2ηax=

42

2×98 . 7=81 ax mW /m2

4. The Electric Field Intensity associated with a plane wave traveling in a perfect dielectric medium is givn by,

Ex(z, t) = 10 cos(2 x 10-7t – 0.1z) V/m.

(i) What is velocity of propagation.(ii) Write down an expression for magnetic field intensity associated with the

wave µ = µ0.Solution:

Ex(z, t) = 10 cos(2 x 10-7t – 0.1z) V/m.

Here, = 2 x 10-7 ; = 0.1.

(I) Velocity of Propagation = ( / ) = (2 x 10-7 / 0.1) = 2 x 10-6 m / sec.(II) Hy = (Ex / )

= √ με=√ μ0

ε =√ 4 π×10−7

8 . 854×10−12=376 . 73

Hy = 0.027 cos(2 x 10-7t – 0.1z) A/m.

5. Find the values of phase shift constant, attenuation constant and intrinsic impedance of the medium with r = 2.25 and = 10-4 V/m for a wave of frequency 2.5 MHz.Solution:

Propagation constant = √ j ωμ (σ+ jωε )

= 0.025 + j 0.0786

= 0.025 rad / m and = 0.0786 rad / m.

= √ jωμσ+ j ωε

= 240 .34∠17 .72 Ω

6. Compare the conduction and displacement current densities in copper ( = 0 , µ = µ0 and = 5.8 x 107 S/m) at a frequency of 1MHz. Repeat for Teflon which has = 2.10, µ = µ0 and = 3 x 10-8 S/m at 1MHz.Solution:

Assuming sinusoidal variation of the electric field in the material,

E = E0 sint V/m.

Where, = 2f and f = 106

Conduction current density, J = E

= E0 sint A/m2

Displacement current density, J =

∂D∂ t

=ε ∂E∂ t

=ωε E0cos ωt A /m2

| JJ d|= σ

ωε

For copper,

σωε

= 5.8×107

2π×106 136 π

×10−9=1012

(conduction current dominates the displacement current)

For Teflon,

σωε

= 3×10−8

2π×106×2 .1× 136π

×10−9=2. 57×10−4




UNIT II-TUTORIAL – 3 PERIODS

1. Find the total electric Field intensity at point P(0, 6, 5)m due to a charge of 20 C located at (2, 0, 6)m and charge of 60C located at (0, -1, 2)m and charge of 100C located at (2, 3, 4)m.Solution:

Q1 = 20 C at (2, 0, 6)m;

Q2 = 60C at (0, -1, 2)m

Q3 = 100C at (2, 3, 4)m.

Electric Field Intensity at (0, 6, 5)

Eo=E1+ E2+ E3

Eo=Q1

4πε 0r1ar 1 +

Q2

4 πε0r2ar2 +

Q3

4 πε0r3ar 3

R1=−2 ax+6 a y−az

|R1|=√4+36+1=√101=6 . 4m

R2=7 a y+3 az

|R2|=√49+9=7. 615m

R3=−2 ax+3 a y+ az

|R1|=√4+9+1=3 .74 m

E=9×109¿[20×10−6

(6 . 4 )3(−2 ax+6 a y− az ) ¿]¿

¿

¿

E=−3781.15 { ax+64291.82 { a¿ y+20185 . 9 az V /m .¿

2. A charge q1is located at each of the four corners of a square of side ‘a’ meter. Find the charge which is to be located at center of the square to stabilize the system.Solution:

FA = FBA + FDA + FCA + FEA

FBA=1

4 πεo

q12

R2 aBA = ka2 q1

2 aBA

FDA=1

4 πε o

q12

R2 aDA = ka2 q1

2 aDA

FCA=1

4 πεo

q12

(√2a2 )2aCA = k

2a2 q12 aCA

FEA=1

4 πε o

q12

(√2a2

2 )2 aEA = k

a2 /2q2

2 aEA

The system is stable (i.e.) FA= 0

0 = FBA + FDA + FCA + FEA

0=kq1

a2 [q1 aBA+q1 aDA+q1

2aCA+2q2 aEA ]

Here , aBA = cos 45o⋅aCA ; aDA=cos 45o⋅aCA; aEA= aCA

0=kq1

a2 [q1 cos 45o aCA+q1 cos45o aCA +q1

2aCA+2q2 aCA]

0=kq1

a2 [q1 (1√2 ) aCA+q1 (1√2 ) aCA +q1

2aCA+2q2 aCA ]

q2=−[√2q1+q1

22 ] = −

(√2+1/2)2

= −0 .957q1

3. An uniform line charge is lying along y = 3, z = 5 if L = 30nC/m. Find E at (a) the origin (b) PB (0, 6, 1) (c) PC (5, 6, 1).Solution:

(a) E=

ρL

2 πεoRaR

=ρL

2πε oRR|R|

R=−3 a y−5 az |R|=√9+25=√34

E=30×10−9

2π×8 .854×10−12×34[−3a y−5az ]

E=−47 . 6 a y−79 . 31 { a z

V /m . ¿

(b) PB (0, 6, 1) :

R=3 ay−4 az |R|=√9+16=5

E=30×10−9

2π×8 .854×10−12×25[3ay−4az ]

E=64 . 8 ay−86 . 4 az V /m .

(c) PC (5, 6, 1):

R=3 ay−4 az |R|=√9+16=5

E=30×10−9

2π×8 .854×10−12×25[3ay−4az ]

E=64 . 8 ay−86 . 4 az V /m .

4. A positive point charge of 100C is located in air at x = 0, y = 0.1m and another such charge at x = 0, y = -0.1m. What is the magnitude and direction of E and what is absolute potential V at x = 0.2m and y = 0.Solution:

Q = 100C at (0, 0.1) and (0, -0.1)

E1=Q4πε or1

2ar 1

ar1 =0.2 ax−0 .1 a y

0.2236

Er 1=100 ×10−12

4 πε0 (0 . 2236 )2 [0 .2 ax−0 .1 a y

0 .2236 ] = 16 .1 ax−8 .05 { ay

¿

Similarly,

E2=Q4πε or2

2ar 2

ar 2 =0 .2 ax+0 . 1 ay

0 .2236

Er 2=100 ×10−12

4 πε 0(0 . 2236 )2 [0 . 2ax+0 .1 a y

0 . 2236 ] = 16 .1 ax+8 . 05 { ay

¿

The Electric Field Intensity at (0.2, 0) is,

E=E1+E2=32 . 2 ax V/m.

Absolute Potential is,

V 1=Q4 πεo r1

Er 2=100 ×10−12

4 πε 0(0 .2236 )= 4 .025 V

V 2=Q4 πε or2

Er 2=100 ×10−12

4 πε 0(0 .2236 )= 4 .025 V

Absolute Potential at (0.2, 0) is,

V = V1+ V2

= 8.05 V.

5. Conducting spherical shells with radii a = 10cm and b = 30cm are maintained at a potential difference of 100V such that V(r = b) = 0 and V(r = a) = 100V. Determine V and E in the region between the shells:Solution:

a =10 cm ; b = 30cm ; v =100 V ;

Boundary conditions are,

r = b at v = 0V.

r = a at v = 100V

Laplace equation in spherical co-ordinates is,

∇2V= 1r 2

∂∂r (r2 ∂V

∂r )+ 1r 2 sinθ

∂∂r (sin θ ∂V

∂θ )+ 1r2 sin2θ (∂

2V∂ ϕ2 )

V varies only with respect to r,

1r2

∂∂ r (r2 ∂V

∂r ) = 0

V = -(A/r) + B ----- (1)

A and B are constants.

Applying the boundary conditions,

r = b at v = 0V.

0 = - (A / b) + B

B = A / b

(1) ----- V = -(A / r) + (A / b)

r = a at v = 100V

V0 = - (A / a) + (A / b)

A =

V 0

[ 1b−1

a ]

V=−V 0

[1b−1a ]

1r+

V 0

[1b− 1a ]

1b

= V 0 [ 1r−1

b1a−1

b ] = 15[1r−103 ] volts

E=−∇V =− dVdr

ar =− Ar2 ar =

V 0

r2 [1a−1b ]

= 15r2 ar V /m .

6. As shown in figure below, charge distributed along the z-axis between Z = 5m with a uniform density, L = 20nC/m. Determine E at (2, 0, 0)m in Cartesian co-ordinates. Also explain the answer in cylindrical co-ordinates.

R

(2, 0, 0)Y

+5

-5

1cm

Glass

Air

Solution:

L = 20nC /m.

Field Point (2, 0, 0)m

Electric Field Intensity at a point due to a line charge (for finite line charge)

E=ρl

2πε0 lρ√ρ2+ l2

= 2×10−9

2×π×8 . 854×10−12×25√52+22

= 166 . 898 ax V /m .

7. A parallel plate capacitor with a separation d = 0.1cm has 29000V applied when free space is the only dielectric. Assume that air has a dielectric strength of 30,000 V/cm. Show why 290,000V/cm and thickness d2 = 0.2cm is inserted as sown in figure:

Solution:

d = 1 cm ; v =29000 V ; E1 = 30 KV / cm

E2 = 290 KV /cm ; d2 = 0.2 cm;

V = potential across the capacitor = 29KV.

When glass dielectric is inserted,

V = V1 + V2

= E1 d1 + E2 d2

= 0.8 E1 + 0.2 E2

V 1

V 2=d1

ε0

ε0ε r

d2⇒ V 1=26 V 2

V 1+V 2=29000026V 1+V 2=29000V 1=27925 .925 volts . V 2= 1074 .074 volts .

Then , E1=V 1

d1= 34 .907 KV /cm

8. Find the work done in moving a point charge Q = 5C from the origin to (2m, /4, /2) spherical co-ordinates, in the field,

E = 5e-r/4 ar + (10 / r sin)a V / m.

Solution:

Q = 5C

Work done = ∮SE⋅dl .

dl = dr ar+ r d a + r sin d a

E = 5e-r/4 ar + (10 / r sin) a V/m.

Work done = ∫0

2

5e−r /4dr+∫0

π /210r sin θ

r sin θ dφ

= [ 5e−r/4

−(1/4 ) ]02

+10 [φ ]0π /2

= - 20 [ 0.6065 -1] +5

= 7.869 + 5.

Workdone = 23.5769.

9. A hallow sphere is charged to 12C of electricity. Find the potential (a) at its surface (b) inside the sphere (c) at a distance 0.3m from the surface. The radius of sphere is 0.1m.Solution:

(a) Potential at the surface:

V =

Q4 πε0 r

If r = a and so V =

12×10−6

4 π×8 .854×10−12×0. 1=1 .08×10Volts .

(b) Potential Inside:E = -V

E = 0 and so V = constant

(c) Outside:r = 0.3m (r > a)

V =

Q4 πε0 r

r = 0.4m

V =

12×10−6

4 π×8 .854×10−12×0. 4=2 .7×106Volts .




UNIT I-TUTORIAL 3 PERIODS

1. Given A = 2ax + 4ay and B = 6ay – 4az. Find the smaller angle between them using, (a) the cross product (b) the dot product.Solution:

(a) the cross product:

A x B =

|ax a y az ¿||2 4 0 ¿|¿¿

¿¿

|A|=√(2 )2+( 4 )2+(0)2=4 . 47|B|=√(0 )2+(6 )2+(−4 )2=7 .21|A×B|=√(−16 )2+(8)2+(12)2=21 .54

Then, |A×B|=|A||B|sinθ

Sin = 21.54 / ((4.47)(7.21)) = 0.688.

= 41.9o

(b) the dot product:

A. B =(2)(0) +(4)(6)+(0)(-4) = 24.

|A⋅B|=|A||B|cos θ

cos = 24/(4.47)(7.21) = 0.745

= 41.9o

2. Use the spherical co-ordinate system to find the area of the strip on the spherical shell of radius ‘a’. What results when =0 and = ?Solution:

The differential surface element (in spherical co-ordinate system) is,

dS = r2sin d d

Then,

A = ∫0

2 π

∫α

β

a2 sinθ dθ dφ

= 2a2 (cos - cos)

here, =0 and = then,

A = 2a2 (cos 0o – cos )

= 4a2.

This is the surface area of the entire sphere.

3. Transform,

A = yax + xay +

x2

√x2+ y2az from Cartesian to cylindrical co-ordinates:

Solution:

We know that,

x = r cos ; y = r sin ; z = z ; r = √ x2+ y2

Hence, A = r sin ax + r cos ay + r cos2 az

Now, the projections of the Cartesian unit vectors on ar, a, and az are,

Dot (or) a a az

ax cos -sin 0

ay sin cos 0

az 0 0 1

Therefore,

ax=cosφ ar−sinφ aφa y=sinφ ar+cosφ aφaz= az

and A = 2r sin cos ar + (r cos2 - r sin2)a + rcos2az.

4. Consider A = 5r ar + 2 sin a + 2 cos ain spherical co-ordinates. Transform the above vector to Cartesian co-ordinates.Solution:

The variables r, , , can be written interms of Cartesian variables as,

r=√x2+ y2+z2 ; cosθ= z√ x2+ y2+z2

; tanφ= yx

The vector can be written as,

A=5√x2+ y2+z2 ar+2 y

√x2+ y2aθ+

2 z√ x2+ y2+z2

aφ

The unit vectors ar , a, a can also transformed into Cartesian equivalents,

ar=x√x2+ y2+z2

ax+y√x2+ y2+z2

a y+z√ x2+ y2+z2

az

aθ=xz√x2+ y2+z2√x2+ y2

ax+yz√ x2+ y2+z2√ x2+ y2

ay−√ x2+ y2

√ x2+ y2+z2az

ar=− y√x2+ y2+z2

ax+x√x2+ y2

a yOn

combining the transformed component vector is,

A=(5 x+2 xyz√ x2+ y2+z2 (x2+ y2)

−2 yz√ x2+ y2+z2√ x2+ y2 )ax

+(5 y+2 y2 z( x2+ y2 )√x2+ y2+z2

+2 xy√ x2+ y2+z2 √x2+ y2 )a y+(5 z−2 y

√ x2+ y2+z2 )az

5. Prove that div(curl F) = 0 where, F = Fx ax + Fy ay + Fz az . Solution:

div(curl F) = ∇⋅(∇×F )

curl F=∇×F=¿|ax ay az ¿|¿¿

¿¿

¿¿

div(curl F) = ∇⋅(∇×F )

= ( ∂∂ x

ax+∂∂ y

ay+∂∂ z

az)⋅(∇×F )

=( ∂∂ x

ax+∂∂ y

ay+∂∂ z

az)⋅¿ ¿

(∂F z

∂ y−∂F y

∂ z )ax−(∂ F z

∂ x−∂F x

∂ z )a y+(∂ F y

∂ x−∂Fx

∂ y )az

=0.

Hence proved.

6. Given that D = (10 3/4) a in cylindrical co-ordinates, evaluate both sides of divergence theorem for the volume enclosed by = 2, z = 0 and z =10.Solution:

Divergence theorem,

∮SD⋅ds=∫

V(∇⋅D )dv

L.H.S:

∮SD⋅ds=∫

z=0

10

∫φ=0

2 π10 ρ3

4aρ⋅ρ dφ dz aφ

we have, = 2,

∮SD⋅ds=∫

z=0

10

∫φ=0

2 π10×23

4aρ⋅(2)dφ dz aφ

= 40 x 2 x 10 = 800.

R.H.S:

∫v(∇⋅D)dv= ∫

z=0

10

∫φ=0

2 π

∫ρ=0

2

(∇⋅D) ρ dρ dφ dz

∇⋅D=1ρ

∂∂ ρ

( ρ Δρ )

=1ρ

∂∂ ρ

(10 ρ3

4) = 1

ρ(10 ρ3) = 10 ρ2

∫v(∇⋅D)dv= ∫

z=0

10

∫φ=0

2 π

∫ρ=0

2

(∇⋅D) ρ dρ dφ dz

=∫0

10

∫0

2 π

(10 ρ4 /4 )02 dφ dz

= 40 x 2π x 10 = 800π.

Hence verified.

7. Show that A = 4ax–2 ay–az and B = ax+ 4ay–4az are perpendicular. To Prove:

* A and B are Perpendicular. (i.e) AB = 0

Proof:

AB = (4ax–2ay–az) (ax+ 4ay–4az)

= (4)(1) + (-2)(4) + (-1)(-4)

= 0.

Hence proved.

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