GATE CLOUD Electromagnetics by RK Kanodia

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R. K. Kanodia

Ashish Murolia

JHUNJHUNUWALA

ELECTROMAGNETICS

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GATE CLOUD ELECTROMAGNETICS, 1e R. K. Kanodia, Ashish Murolia

CC1015

Information contained in this book has been obtained by author, from sources believes to be reliable. However, neither Jhunjhunuwala nor its author guarantee the accuracy or completeness of any information herein, and Jhunjhunuwala nor its author shall be responsible for any error, omissions, or damages arising out of use of this information. This book is published with the understanding that Jhunjhunuwala and its author are supplying information but are not attempting to render engineering or other professional services.

Copyright by Jhunjhunuwala ISBN 978-8192-34838-4

JHUNJHUNUWALA

B-8, Dhanshree Tower Ist, Central Spine, Vidyadhar Nagar, Jaipur – 302023 Ph : +91-141-2101150.

www.nodia.co.in

email : [email protected]

55, Suryalok Complex, Gunfoundry, Abids, Hyderabad - 500001. Phone : 040 - 64582577

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GATE CLOUD caters a versatile collection of Multiple Choice Questions to the students who are preparing

for GATE (Gratitude Aptitude Test in Engineering) examination. This book contains over 1200 multiple choice solved problems for the subject of Electromagnetics, which has a significant weightage in the GATE examination of Electronics and Communication Engineering. The GATE examination is based on multiple choice problems which are tricky, conceptual and tests the basic understanding of the subject. So, the problems included in the book are designed to be as exam-like as possible. The solutions are presented using step by step methodology which enhance your problem solving skills.

The book is categorized into ten chapters covering all the topics of syllabus of the examination. Each chapter contains :

Ÿ Exercise 1 : Level 1 Ÿ Exercise 2 : Level 2

Ÿ Exercise 3 : Mixed Questions Taken form Previous Examinations of GATE & IES. Ÿ Detailed Solutions to Exercise 1, 2 and 3.

Although we have put a vigorous effort in preparing this book, some errors may have crept in. We shall appreciate and greatly acknowledge the comments, criticism and suggestion from the users of this book which leads to some improvement.

You may write to us at [email protected] and [email protected]. Wish you all the success in conquering GATE.

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GATE ELECTRONICS & COMMUNICATION ENGINEERING

IES ELECTRONICS & TELECOMMUNICATION ENGINEERING

Networks:

Electromagnetic Theory:

Elements of vector calculus: divergence and curl; Gauss’ and Stokes’ theorems, Maxwell’s equations: differential and integral forms. Wave equation, Poynting vector. Plane waves: propagation through various media; reflection and refraction; phase and group velocity; skin depth. Transmission lines: characteristic impedance; impedance transformation; Smith chart; impedance matching; S parameters, pulse excitation. Waveguides: modes in rectangular waveguides; boundary conditions; cut-off frequencies; dispersion relations. Basics of propagation in dielectric waveguide and optical fibers. Basics of Antennas: Dipole antennas; radiation pattern; antenna gain.

Analysis of electrostatic and magnetostatic fields; Laplace's and Poisson's equations; Boundary value problems and their solutions; Maxwell's equations; application to wave propagation in bounded and unbounded media; Transmission lines : basic theory, standing waves, matching applications, microstrip lines; Basics of wave guides and resonators; Elements of antenna theory.

IES ELECTRICAL ENGINEERING

Electromagnetic Theory:

Electric and magnetic fields. Gauss's Law and Amperes Law. Fields in dielectrics, conductors and magnetic materials. Maxwell's equations. Time varying fields. Plane-Wave propagating in dielectric and conducting media. Transmission lines.

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CHAPTER 1

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EXERCISE 1.1

MCQ 1.1.1 The field lines of vector B=raq is

MCQ 1.1.2 Given the two vectors M=5ax-2ay+4az and N=-8ax-7ay+2az. The unit

vector in the direction of (M -N) will be

(A) 0.82ax+0.36ay-0.14az (B) 0.92ax-0.36ay+0.41az

(C) 0.92ax+0.36ay+0.14az (D) -0.92ax-0.36ay-0.14az

MCQ 1.1.3 A vector field is specified as G 4xya 2 2 x2 a 3z a

x y 2 z

= + ^ + h + . The unit vector in

the direction of G at the point ( 2, 1, 3)- will be

(A) 0.26ax+0.39ay+0.88az (B) -0.26ax+0.39ay+0.88az

(C) 0.36ax-0.29ay+0.24az (D) 0.88ax-0.29ay+0.36az

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relation between them?

(A) A#A=0 (B) A#^B+Ch=^A#Bh+^A#Ch

(C) (A#B)#C=A#^B#Ch (D) (B#A)=-^A#Bh

MCQ 1.1.5 The tips of three vectors A, B and C drawn from a point define a plane.

A:^B#Ch equals to

(A) +1 (B) -1

(C) zero

(D) can’t be determined as A, B and C are not given

MCQ 1.1.6 The component of vector A along vector B is

(A) A:_AB_(B) B A:B (C) A:B (D) AB A:B

MCQ 1.1.7 The vector fields are defined as A=ar+2af+3az and B=aar+baf-6az. If

the fields A and B are parallel then the value of a and b are respectively.

(A) -2, -2 (B) -2, -4

(C) -4, -2 (D) -2, -1

MCQ 1.1.8 Consider the vectors A =4ax+2kay+kaz and B =ax+4ay-4az. For what

value of k the two vectors A and B will be orthogonal ?

(A) 0 (B) +1

(C) -2 (D) -1

Common Data for Question 9 - 10 :

In a cubical region (x <2, y <2, z <2) a vector field is defined as

E =9zy2cos(2 )x ax+8zysin(2 )x ay+2y2sin(2 )x az.

MCQ 1.1.9 The vector field component Ex will be zero in the plane

(A) z=0 (B) y=0

(C) x=p/4 (D) All of the above

MCQ 1.1.10 In the plane y-4z=0, the vector field components Ey and Ez are related as

(A) 2Ey=Ez (B) Ey+Ez =0

(C) Ey=2Ez (D) Ey=Ez

MCQ 1.1.11 The plane surface on which the vector field E will be zero is

(A) x=0 (B) y=0

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MCQ 1.1.12 Distance between the points P x( =2,y=3,z=-1) and Q(r=4,f=-50,z=2) is

(A) 3.74 (B) 4.47

(C) 6.78 (D) 8.76

MCQ 1.1.13 The uniform vector field A=ay can also be expressed as

(A) sinfar+sinfaf

(B) cosfar+sinfaf

(C) sin cosq far+cos cosq faq+sinfaf

(D) sin cosq far+cos cosq faq-sinfaf

MCQ 1.1.14 The vector filed F=12ax can be expressed in spherical coordinates at the point

(x=3, y=2, z=-1) as

(A) 8ar-2aq+5af (B) 8ar-2.2aq-5.5af

(C) -8ar+2.2aq+5.5af (D) 8ar+2.2aq+5.5af

MCQ 1.1.15 The angle formed between A=-5ar+10af+3az and surface z=5 is

(A) 10c (B) 15c

(C) 45c (D) 75c

MCQ 1.1.16 The component of vector A=-4ar-20af+4az parallel to the line x=6, z=-2

at the point P 3 90 2( , c, )is

(A) -4ar+2af (B) ax-ay

(C) -2ay (D) -2az

MCQ 1.1.17 In the cartesian co-ordinate system the co-ordinates of a point P is ( , , )a b c .Now consider the whole cartesian system is being rotated by 145c about an axis from the origin through the point (1, 1, 1) such that the rotation is clockwise when looking down the axis towards the origin. What will be the co-ordinates of the point P in the transformed cartesian system ?

(A) (a 2/ , b 2/ , c 2/ ) (B) (-a, -b, -c)

(C) (c, b, a) (D) (c, a, b)

MCQ 1.1.18 Consider R be the position vector of a point P x y z( , , ) in cartesian co-ordinate system then gradR equals to

(A) 1 (B) 4_RR

(C) R_R (D) ₂R_R

MCQ 1.1.19 Given the vector filed A=y2ax+(2xy+x2+z2)ay+(4x+2 )yz az. The divergence

of the vector field is

(A) 2(x+y) (B) x2₊y2_{+ +}z2 6x₊2y

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For View Only Shop Online at www.nodia.co.in MCQ 1.1.20 A vector field F is defined as F =rsinfar+2r2zaf+zcosfaz. The 4#F at

point P 1_^ , / ,p 2 2_h equals to

(A) ar+6az (B) -ar+2az

(C) 3ar+6az (D) -5ar+6az

MCQ 1.1.21 Which one of the following vector function has divergence and curl both zero ? (A) A=2xzax-yzay-z2az (B) B=xyax-yzaz

(C) C=xyax-xzay-yzaz (D) D=yzax+xzay+xyaz

MCQ 1.1.22 The curl of the unit vectors ar, af and az in cylindrical co-ordinate system is listed

below. Which of them is correct ?

ar af az ar af az

(A) 0 r1az 0 (B) 0 1rar af

(C) raf ar 0 (D) raf 1rar 0

MCQ 1.1.23 In a certain region consider f and g are the two scalar fields where as A is a vector field. Which of the following is not a correct relation ?

(A) d:(fA)=f^d:Ah-A:^dfh (B) d#^fAh=f^d#Ah-A#^dfh (C) _gf g g f f g 2 d d dc m= -(D) _g ( ) ( ) g g g A A A 2 # d c m= d# - # d

MCQ 1.1.24 Laplacian of the scalar field f₌2_rzsin_f₊4z2cos2_f₊4_r2_{at the point}_{P 3}_{( , / , )}_p _{2 6}

is

(A) 16 (B) 0

(C) 46 (D) 40

MCQ 1.1.25 A conservative field M is given by M =(zcos( )xz +y)ax+5kxay+xcos( )xz az.

The value of k will be

(A) 1 (B) 0

(C) -1 (D) 1 2/

MCQ 1.1.26 A scalar field g₌_^1₊5k x y_h 2 ₊xyz_{will be harmonic at all the points for the value}

of k equals to

(A) 1 2/ (B) 0

(C) -1 2/ (D) can’t be determined

MCQ 1.1.27 Curl of the gradient of any scalar field is

(A) zero everywhere (B) zero at origin only

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MCQ 1.1.28 A vector field is given as A=(x+4 )z ax+(3x-7z)ay+(4x+3y-cz)az. The

value of c for which A will be solenoidal is

(A) 1 (B) -1

(C) 0 (D) 3

MCQ 1.1.29 For a vector function A=(4x+k z1 )ax+(k x2 -5 )z ay+(4x-k y3 +2 )z az to be

irrotational value of k1, k2 and k3 will be respectively.

(A) -5, 0, 4 (B) 0, 4, 5

(C) 4, 0, 5 (D) 1, 4, 3

MCQ 1.1.30 The unit vector normal to the plane 2x+3y+6z=7 is

(A) a 2a 3a 24 1 x+ y+ z ^ h (B) a a a 14 1 ₄ ₄ ₆ x+ y+ z ^ h (C) a a a 14 1 ₂ ₃ x+ y+ z ^ h (D) 2a 4a 6a 58 1 x+ y+ z ^ h

MCQ 1.1.31 Consider C is a certain closed path and dl is the differential displacement along the path then the contour integral dl

C

#

is (A) zero

(B) 1 (C) -1

(D) can’t be determined as C is not defined.

MCQ 1.1.32 Consider C is any closed path and U is a scalar field. So, the contour integral

U dl C : d ^ h

#

is (A) 1 (B) -1 (C) zero

(D) Can’t be determined as C and U is not given

MCQ 1.1.33 A vector field is defined as A=3yzax+z x2 ay+2xyaz.The surface integral of the

field over a closed surface S is (A) 1

(B) 5 (C) zero

(D) can’t be determined as surface S is not given ***********

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EXERCISE 1.2

MCQ 1.2.1 If the edge of a cube is 3 units then the angle formed between it’s body diagonals will be

(A) 70.53c (B) 53 70c.

(C) 66 21c. (D) 61c

Consider a triangle ABC, whose vertex A, B and C are located at the points

(-4 2 5, , ), (16, 20,-3) and ( 14, 10,- 15) respectively.

MCQ 1.2.2 The unit vector perpendicular to the plane of the triangle is (A) 0.61ax+0.42ay-0.37az

(B) 0.66ax-0.38ay+0.65az

(C) -0.54ax+0.11ay-0.19az

(D) 0.43ax-0.21ay+0.11az

MCQ 1.2.3 The unit vector in the plane of the triangle which is perpendicular to AC is (A) -0.55ax-0.832ay+0.077az

(B) 0.23ax-0.11ay-0.43az

(C) -0.51ax+0.41ay+0.76az

(D) 0.49ax-0.23ay-0.44az

MCQ 1.2.4 The unit vector in the plane of the triangle which bisects the interior angle at A is (A) 0.11ax-0.81ay+0.44az (B) 0.21ay-0.41ax+0.52az (C) 0.23az+0.12ax+0.11ay (D) 0.37az+0.92ay+0.17ax MCQ 1.2.5 A vector field x y x y F ax 2 ay 2 2 = + + ^ h at the point P(r =2, f=p/4, z=0 1. ) is (A) 2ax+3ay (B) -2ax-3ay (C) 0.5ar (D) -0.5ar

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MCQ 1.2.6 The component of vector A rar r3sin cos a r₄a

2

q f

= - q+ f tangential to the spherical

surface r=20 at point P 20 150 330( , c, c) is

(A) 0.043aq+100af (B) =0.43aq+10af

(C) 4.3aq+100af (D) -0.043aq-10af

A vector field has the value A=-12ar-4aq+9af at the point P 9 150 45( , c, c).

MCQ 1.2.7 The vector component of A, that is tangent to the cone q =150c is

(A) -12af+9ar (B) -12ar-8aq

(C) -8aq+9af (D) -12ar+9af

MCQ 1.2.8 The unit vector that is perpendicular to A and tangent to the cone q =135c is

(A) a a 10 1 3 + q f a k (B) 1 4₅^ ar+3afh (C) ₅1 3^ ax+4afh (D) (9a 2 )a 85 1 r+ f

MCQ 1.2.9 Consider R is a separation vector from a fixed point (a, b, c) to a varying point

( , , )x y z in the Cartesian coordinate system. The value of grad ^ hR1 equals to

(A) R R 3 - (B) R R 6 -(C) R12 - (D) RR3

MCQ 1.2.10 A certain hill located in Udaipur is of height h that varies as :

( , )

h x y ₌6x2₊8y2_-3xy₊36x_-56y₊100

Where x is the distance (in miles) in north and y is the distance (in miles) in east of Udaipur railway station. The top of the hill will be located at

(A) 3 miles north, 2 miles west of Railway station. (B) 2 miles south, 3 miles east of Railway station. (C) 2 miles north, 3 miles east of Railway station. (D) 6 miles south, 2 miles east of Railway station.

MCQ 1.2.11 At any point P x y z( , , ) a vector field is given by F= _R152aR, where R is the position

vector of the point P. The divergence of the vector field F will be

(A) zero, everywhere (B) zero, at all points excluding origin

(C)

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Statement for Linked Question 12 - 14 :

Given a vector field A =3x2ax+3yzay+3y2az.

MCQ 1.2.12 The line integral of A from origin to the point (2,2,2) by the route

( , , )0 0 0 "( , , )2 0 0 "( , , )2 2 0 "( , , )2 2 2 will be

(A) 32 units (B) 8 units

(C) 24 units (D) 6 units

MCQ 1.2.13 The line integral of the vector field A from the origin to the point (2, 2, 2) along the direct straight line is

MCQ 1.2.14 The line integral of the field A around the closed loop that goes out along the route defined in Question 12 and back along the route defined in Question 13 is

MCQ 1.2.15 A wedge defined by 0#r#5_,₄₅#f#₁₈₀_c_,_z=₂_{is shown in figure}

Circulation of A =rsinfar+z2cosfaz along the edge L of the wedge is

(A) 10 units (B) 5/2 units

(C) -25 4/ units (D) 25 4/ units

MCQ 1.2.16 Volume integral of the function f₌30z2_{over the tetrahedron with corners at}

( , ,0 0 -1); ( ,0 -1 0, ), (-1 0 0, , ), ( , , )0 0 0 is

(A) 1/2 (B) -1 2/

(C) 31 2/ (D) 1 60/

MCQ 1.2.17 Total outward flux of a vector field A₌ 1₄_r2cos2_fa ₊2z2a

r f through the closed

surface of a cylinder 0#z#2, r =2 is

(A) 4p (B) 16p

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MCQ 1.2.18 A quarter cylinder of radius 2 and height 5 exists in the first octant of a cartesian coordinate system as shown in the figure below. Surface integral of a vector field

(4 sin ) sin2 6z

A= r+3r 2f ar+r faf+ az over the surface of the cylinder will be

(A) 12p (B) 30p

(C) 40p (D) 80p

MCQ 1.2.19 A vector function is given by G=6x y2 ax-5y2ay+2zaz. If L is a closed path

defined by the sides of a triangle as shown in the figure then, G dl L

:

#

equals to

(C) 2 units (D) 10 6/ units

Consider S1 and S2 are respectively the top and slanting surfaces of an ice cream

cone of slant height 2 m and angle 60c as shown in figure, where a vector field F

is defined as F x y x y z x_a x_a y_a y_a 4 5 4 2 x 32 y 2 x 2 y 2 2 2 2 2 = + + + _: ₊ _- ₊ _D

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MCQ 1.2.20 The surface integral of the vector field F over the surface S1 will be

(A) 2.3 units (B) 24 units

(C) 0.18 units (D) 1.9 units

MCQ 1.2.21 The surface integral of the vector field over the surface S2 will be

(A) 4 3p/ (B) 3p/3

(C) 4p 3 3/ (D) 0

Negative gradient of a scalar field f is A=-4f =(x+z)ax-4zay+(x-3y-z)az

MCQ 1.2.22 The vector A is

(A) irrotational but not solenoidal (B) both irrotational and solenoidal (C) solenoidal but not irrotational (D) neither solenoidal nor irrotational

MCQ 1.2.23 The scalar field, f equals to

(A) x₂2+xz+z₂2 (B) -x₂2-2xz+6yz+z₂2

(C) -xz+3yz+z₂2 (D) -x₂2-xz+3yz+z₂2

MCQ 1.2.24 Line integral of a vector field A=5(yax+xay) from a point P(2, 1, 3) to the point

Q(8, 2, 3) along the curve y= x 2/ will be

MCQ 1.2.25 A vector field F=2^_r13+_r13cos2fhar exists in the region between the two spherical

shells of radius 1 m and 2 m centred at the origin. The total outward flux of F

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(A) p (B) -p (C) -2p (D) 2p

MCQ 1.2.26 Two vectors A and B make an angle 30c between them as shown in figure. Magnitude of vector A and B are 4 units and 3 units respectively. If a third vector

R is defined such that R=6A-4B then it’s graphical construction will be

MCQ 1.2.27 A certain, vector R is defined as R=A#^B#Ch. Directions of A, B and C

are mentioned in the list below. Which of the following gives the correct direction of R for the given direction of the three vectors.

Direction of A Direction of B Direction of C Direction of R

(A) North South East West

(B) South North West East

(C) East West North West

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For View Only Shop Online at www.nodia.co.in MCQ 1.2.28 Which of the following vectors are equal :

A =-ax+ay+3az at ^1 2 3, , h in cartesian co-ordinates B =ar+af+3az at ^2, / ,p 2 3h in cylindrical co-ordinates C = 2ar+3az at ^3 3 4 9, p/ , h in cylindrical co-ordinates

(A) A and B only (B) B and C only

(C) A and C only (D) all the three vectors A, B and C

MCQ 1.2.29 Two vectors are defined as A =ax+5ay+3az and B =3ax+2ay+az. Which of

the following vector is perpendicular to ^A+Bh

(A) -4ax+4ay (B) 4ay+4az

(C) ax+az (D) none of these

MCQ 1.2.30 The gradient of a scaler function is d ^V x y z, , h =1.5x yz2 2ax+0.5x z3 2ay+x yz3 az

The scalar function is

(A) x yz3 2_(B)x yz

2

3 2

(C) x yz2₂ 3 (D) xy z3 2

MCQ 1.2.31 The equation of the plane tangential to the surface xyz=1 at the point _b_{2 4 4}, ,1_l is

(A) 16x+32y+ =z 24 (B) 2x+ +y 32z=12

(C) x+2y+32z=12 (D) x+16y+32z=24

MCQ 1.2.32 Consider a volume v is defined as the part of a spherical volume of radius unity lying in the first octant. The volume integral 2xdv

v

#

is equal to

(A) 2p (B) p/16

(C) p/4 (D) p/8

MCQ 1.2.33 Consider a vector field A=rcosfar+₃raf. If C is the contour shown in the figure

then the contour integral A dl C

:

#

is equal to

(A) p +4 (B) p +₂ 1

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MCQ 1.2.34 A vector field is defined as A₌2 cos_r 2_fa ₊_ra

r f. Consider the two contours C1 and

C2 as shown in the figure.

The ratio of the contour integrals

d d A l A l C C 2 1 : :

#

is (A) -₉1 (B) ₉1 (C) 9 (D) -9

MCQ 1.2.35 Consider the contour C as shown in the figure

If a vector field is defined as A= e_rr aq

-b l then the contour integral A dl C :

#

is (A) p₂e-1_(B)_e 2 1 p - -(C) p_{2 1}_^ ₊e-1_h_(D)_e 2 1 1 p_^ _- -_h

MCQ 1.2.36 The divergence of the unit vectors ar,aq and af in spherical co-ordinate system is

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ar aq af

(A) r 2/ r/ tanq 0

(B) 2/r cotq/r 0

(C) r 2/ r/ tanq 1

(D) 2/r cotq/r 1

MCQ 1.2.37 Which of the following vector can be expressed as the gradient of a scalar ?

(A) 2yzax+2xzay+2xyaz (B) e a_r

r f

-(C) 2₂ cos a sin a

r ^ f r+ f fh (D) Both (A) and (C)

MCQ 1.2.38 Which of the following vector can be expressed as curl of another vector ? (A) 1₂^x2-y2hax-xyay+2az (B) e a_r r f -(C) cos sin r a r a 2 r

3 q + 3q q (D) All of the above

MCQ 1.2.39 The vector field pattern of A=3yax is

MCQ 1.2.40 A two dimensional vector field in Cartesian coordinate system is defined as

,

x y

A^ h=Axax+2Ayay

The curl and divergence of the vector field are both zero. Which of the following differential equation satisfies Ax and Ay

(A) d2Ax=0 (B) d2Ay=0

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MCQ 1.2.41 The circulation of F=x2ax-3xzay-y2az around the path shown below is

(A) 3- (B) 1 ₆1

(C) 6- (D) 1 1₃

MCQ 1.2.42 If R=xax+yay+zaz is the position vector of point ( , , )P x y z and R= R then

R Rn : d is equal to (A) nrn_(B)₍_n₊₃₎_rn (C) (n₊2)rn_(D)₀ MCQ 1.2.43 If A₌_rsin_fa ₊_r2a

r r, and L is the contour of figure given below, then circulation

d A l L :

#

is (A) 7p + (B) 2 7p -2 (C) 7p (D) 0 ***********

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EXERCISE 1.3

MCQ 1.3.1 The direction of vector A is radially outward from the origin, with A ₌krn_.

where r2₌x2₊y2₊z2_and_k_{is a constant. The value of}_n_{for which} _A ₀

: d = is (A) -2 (B) 2 (C) 1 (D) 0 MCQ 1.3.2 If A=xyax+x2ay, then A dl C $

#

o over the path shown in the figure is

(A) 0 (B)

3 2

(C) 1 (D) 2 3

MCQ 1.3.3 If a vector field V is related to another vector field A through V=d #A, which

of the following is true? (Note : C and SC refer to any closed contour and any surface whose boundary is C . )

(A) V dl A dS C SC $ =

##

$

#

(B) A dl V dS C SC $ =

##

$

#

(C) V dl A dS C SC # : : d = d# ^ h

##

^ h

#

(D) V dl V dS C SC : : d # = ^ h

##

#

MCQ 1.3.4 Consider points A, B, C and D on a circle of radius 2 units as in the shown figure below. The items in List II are the values of af at different points on the circle.

Match List I with List II and select the correct answer using the code given below the lists : GATE 2012 GATE 2012 GATE 2010 GATE 2009 IES EC 2010

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List I List II a. A 1. ax b. B 2. ay c. C 3. -ax d. D 4. (ax+ay)/ 2 5. -(ax+ay)/ 2 6. (ax-ay)/ 2 Codes : a b c d (A) 3 4 5 2 (B) 1 6 5 2 (C) 1 6 2 4 (D) 3 5 4 2

MCQ 1.3.5 If V₌2sinh cosx kyepz_{is a solution of Laplace’s equation, what will be the value}

of k ? (A) p 1 1 2 + (B) 1 p 2 + (C) p 1 1 2 - (D) 1 p 2

-MCQ 1.3.6 The electric field intensity E at a point P is given by 10ax+10ay+10az where a ax, y

and az are unit vectors in x y, and z directions respectively. If a b g, , respectively

the angles the E vector makes with x y, and z axes respectively, they are given by which of the following ?

(A) a= = =b g 30c (B) a= = =b g 60c (C) cos 3 1 1 a= =b g - _(D) cos 311 a= = =b g

-MCQ 1.3.7 Which one of the following potentials does NOT satisfy Laplace’s Equation ?

(A) V=10xy (B) V=rcosf

(C) V=10/r (D) V=rcosf+10

MCQ 1.3.8 Laplacian of a scalar function V is (A) Gradient of V

(B) Divergence of V

(C) Gradient of the gradient of V

(D) Divergence of the gradient of V

IES EC 2009

IES EC 2007

IES EC 2003

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For View Only Shop Online at www.nodia.co.in MCQ 1.3.9 A field A=5x yz2 ax+x z3 ay+(x y3 -2 )z az can be termed as

(A) Harmonic (B) Divergence less

(C) Solenoidal (D) Rotational

MCQ 1.3.10 Laplace equation in cylindrical coordinates is given by

(A) V V V 0 z V 1 1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 r r r r r f d = e o+ e o+ = (B) V xV yV zV 2 2 2 2 2 2 2 2 2 2 2 2 2 d = + + (C) 2V v e r d =-(D) 0 sin V _{r r} r V_r r 1 1 2 2 2 2 2 d 2 2 2 2 2 q f = c m+ =

MCQ 1.3.11 What is the value of the integral dl

c

#

along the curve c ?( c is the curve ABCD in the direction of the arrow ) ?

(A) 2 (R ax+ay)/ 2 (B) -2 (R ax+ay)/ 2

(C) 2Rax (D) -2Ray

MCQ 1.3.12 Given a vector field A=4rcosfar in cylindrical coordinates. For the contour as

shown below,

#

A:dl is (A) 1 (B) 1-( /2)p (C) 1+( /2)p (D) -1 IES EC 2002 IES EC 2001 IES EE 2005 IES EE 2003

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MCQ 1.3.13 If a vector field B is solenoidal, which of these is true ?

(A) B dl 0 L : =

#

(B) B ds 0 s : =

#

(C) d#B= (D) 0 d:B=Y 0

MCQ 1.3.14 Which of the following equations is correct ? (A) ax#ax= ax 2

(B) (ax#ay)+(ay#ax)=0 (C) ax#(ay#az)=ax#(az#ay) (D) ar:aq+aq:ar =0

MCQ 1.3.15 Match List I with List II and select the correct answer :

List I (Term) List II (Type)

a curl ^ hF =0 1. Laplace equation

b div ^ hF =0 2. Irrotational

c div Grad ^ hf =0 3. Solenoidal

d div div ^ hf =0 4. Not defined

Codes : a b c d (A) 2 3 1 4 (B) 4 1 3 2 (C) 2 1 3 4 (D) 4 3 1 2

MCQ 1.3.16 If A=2ar+af+az, the value of

#

A:dl around the closed circular quadrant shown in the given figure is

(A) p (B) p +₂ 4 (C) p + (D) 4 p +₂ 2 *********** IES EE 2002 IES EE 2002 IES EE 2002 IES EE 2001

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SOLUTIONS 1.1

SOL 1.1.1 Option (D) is correct.

Given the vector field has the only component in aq direction and its magnitude is

r so as r increase from origin to the infinity field lines will be larger and directed along aq as shown in option (A).

SOL 1.1.2 Option (B) is correct.

M-N =5ax-2ay+4az- -( 8ax-7ay+2 )az

5ax 2ay 4az 8ax 7ay 2az

= - + + +

-13ax 5ay 2az

= + +

So, the unit vector in the direction of (M-N) is

a = M_M_--_NN ₁₃13a_a ₅5a_a ₂2_aa x y z x y z = ₊+ ₊+ 0.92ax 0.36ay 0.14az = + +

SOL 1.1.3 Option (C) is correct. Vector G at (-2 1 3, , ) :

G =4( 2)(1)- ax+2(2+ -( 2) )2 ay+3(3)2az

8ax 12ay 27az

=- + +

So, unit vector in the direction of G at Q :

aG = _GG 8₈a_a ₁₂12a_a ₂₇27_aa x y z x y z = -- ++ ++ 0.26ax 0. 94 ay 0.88az =- + +

Option (A), (B), (D) are the properties of vector product.

Now we check the relation defined in option (C). Since the triple cross product is not associative in general so, the given relation is incorrect. This inequality can be explained by considering vector A=B and C ,perpendicular to A as shown in the figure.

According to right hand rule we determine that ^B#Ch points out of the page

and so A#^B#Ch points down that has magnitude ABC.

But in L.H.S. of the relation, since A=B

So we have (A#B)=0

and hence (A#B)#C =0

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As the direction of cross vector is normal to the plane. So, direction of B#C will

be normal to the plane defined by the three vectors.

Now the dot product of two mutually perpendicular vectors is always zero and since the direction of B#C will be perpendicular to the plane of vector A. So A:^B#Ch will be zero.

SOL 1.1.6 Option (C) is correct.

Consider the two vectors A and B are as shown below.

As the angle between the two vectors is a. So component of vector A along B is

A1 =^cosahA

cosine of the angle between the two vectors is defined as

cos a ₌A_AB:B

So, A1 =bA_AB:BlA

B

A:B

=

Cross product of two parallel vector fields is always zero since the angle between them is q=0c.

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For View Only Shop Online at www.nodia.co.in a a a 1 2 3 6 z a b -r f 0 = a a a 12 3b 3a 6 b 2a z - - r+ + f+ -^ h ^ h ^ h =0

Solving it we have, b =-4 and a =-2

SOL 1.1.8 Option (A) is correct.

Dot product of the two orthogonal vectors is always zero.

i.e. A:B =0

(4)(1)+(2 )(4)k +( )( 4)k - =0 4+8k-4k =0

4k =-4

k =-1

From the given field vector we have the component

Ex =3zy2cos2x.

So for the given condition Ex=0

We have, 9zy2cos(2 )x ₌₀

This condition met when, z =0

or, y =0

or, cos x2 =0 &2x=p/2 &x=p/4

Therefore the planes on which field component Ex will be zero are

z=0 y=0 and x=p/4

From the given field vector we have the field components

Ey =2zysin2x

and Ez =2y2sin2x

Now, in the plane y-4z=0 &y=4z

So, Ey =8 4z z^ hsin2x =32z2sin2x

Ez =2 4^ hz 2sin2x =32z2sin2x

Thus Ey =Ez

For the given condition E =0, we must have

Ex =Ey=Ez=0

i.e. 9zy2cos2x ₌₈_zy_sin₂_x₌₂_y2_sin₂_x₌₀

This condition met in the plane y=0.

Since the two points are defined in different coordinate system so we represent the point Q in Cartesian system as

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y =rsinf=4sin(-50)=-3 064. and z =2

So, the distance between the two points P 2 3( , ,-1) and Q 2 57( . ,-3 064 2. , ) is given as

PQ ₌ _{( .}_{2 57}_-₂₎2_{+ -}₍ _{3 064}_. _-₃₎2₊₍₂₊₁₎2

6.78 units =

Since, the options include spherical as well as cylindrical representation of A, so, we will transform the vector in both the forms to check the result.

The components of vector field A are

Ax=1, Ay=0 and Az =0

Now, we transform the vector components in cylindrical system as

A A Az r f R T S S SS V X W W WW cos sin sin cos A A A 0 0 0 0 1 x y z f f f f = -R T S S SS R T S S SS V X W W WW V X W W WW Ar =(cosf)(1)=cosf Af = -( sinf)( )1 =-sinf Az =0

So, the vector field in cylindrical system is

( , , )z

A r f =cosfar-sinfaf

Hence, both the options (A) and (B) are incorrect.

Again, we transform the vector components in spherical system as

A A A r q f R T S S SS V X W W WW sin cos cos cos sin sin sin cos sin cos cos sin A A A 0 x y z q f q f f q f q f f q q = R T S S S S R T S S SS V X W W W W V X W W WW

Ar =(sin cosq f)( )1 =sin cosq f

Aq =(cos cosq f)( )1 =cos cosq f

Af = -( sinf)(1)=-sinf

So, the vector field in spherical system is

( , , )r

A q f =sin cosq far+cos cosq qaq-sinfaf

We transform the given vector field in spherical system. Since the given vector field is F=10ax

The Cartesian components of the field are Fx=10, Fy=0, Fz=0.

So, the spherical components of vector field can be determined as

F F F r q f R T S S SS V X W W WW sin cos cos cos sin sin sin cos sin cos cos sin F F F 0 x y z q f q f q q f q f f q q = -R T S S S S R T S S SS V X W W W W V X W W WW

So, we get Fr =10sin cosq f

Fq =10cos cosq f

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Now, for the given point (x=3, y=2, z=-1) we have

r ₌ ( )3 2₊( )2 2_{+ -}( 1)2 ₌ 14 q ₌tan-1 x2_z+y2 ( ) ( ) ( ) 105.5 tan 1 3 2 ₁ 2 2 c = - _f _-+ _p= f ₌tan-1_{a k}_xy _tan _33.7 3 2 1 c = - _{b l}=

Putting all the values in the matrix transformation, we have

Fr =5sin(105.5 )c cos(33.7 )c =8

Fq =10cos(105.5 )c cos(33.7 )c =-2 2.

Ff =-10sin(33.7 )c =-5 5.

Therefore, the vector field in spherical coordinate is

F =Frar+Fqaq+Ffaf

8ar 2.2a 5.5a

= - q- f

Since z-axis is normal to the surface z=5, so first of all we will find the angle between z-axis and A which can be easily obtained from the figure shown below :

i.e. cos f ₌ A_Az ( 5) ( )10 ( )3 3 134 3 2 2 2 = - + + = f cos . 134 3 _{74 98} ₇₅ 1 c. c = - = c m

Therefore, the angle between surface z=5 and vector A is (90c-f)=15c.

The given line x=6, z=-2 is parallel to y-axis. So, the component of A parallel to the given line is

Ay =(A:a ay) y ( 2a 20a 4 )az :ay ay = -6 r+ f+ @ ( 2sinf 20cosf)ay = - + At point P, f =90c, so, Ay =-2ay

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After 120c rotation looking down the axis the new co-ordinate axes (xl, yl, zl) will be as shown below :

So, the rotation carries z axis into y ; y-axis into x and x into z. therefore the new co-ordinates of point P are :

xl= =z c yl= =x a zl= =y b

i.e. (c, a, b) is the co-ordinates of point P in the transformed system.

The position vector can be defined as :

R =xax+yay+zaz R ₌ x2₊y2₊z2 So, gradR R_x ax R_y ay R_z az 2 2 2 2 2 2 = + + x y z x x y z y x y z z a a a 2 1 2 2 1 2 2 1 2 x y z 2 2 2 2 2 2 2 2 2 = + + + + + + + + x y z xax yay zaz 2 2 2 = + + + + R R =

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For View Only Shop Online at www.nodia.co.in SOL 1.1.19 Option (D) is correct.

A : d _x( )y2 _y(2xy x2 z2) _z(4x 2yz) 2 2 2 2 2 2 = + + + + + x y 0 2 2 = + + =2(x+y)

We have the vector field components as

sin Fr=r f , Ff=r2z and Fz=zcosf Now, d#F F F F a a a 1 z z z r r r = ₂2 2 2 2 2 r r r f f f

cos cos sin

z z z a z z a 1 2 1 2 2 2 2 2 2 2 2 r f f rr r r r f r f = ; + E r- ; - E f 1 2z sin az 2 2 2 2 r r rr fr f + _; - _E sin cos z a a z a 1 ₀ ₀ _{1 3} z 3 2 r f r r r r f = 6- - @ r-6 - @ f+ 6 - @ (zsin )a (3 z cos )a 1 z 3 r f r r f =- + r+ -At point P 1( , / , )p 2 2 F # d =-1(2#1+1 )3 ar+(3#1#2-0)az 3a 6az =- r+

D =yzax+xzay+xyaz D : d _x( )yz _y( )xz _z( )xy 2 2 2 2 2 2 = + + =0 D # d yz xz xy a a ax x y y z z = ₂2 2 2 2 2 (x x)ax (y y)ay (z z)az = - - - + - =0

SOL 1.1.22 Option (D) is correct. Curl of the unit vector ar is

a # d r a a a 1 1 0 0 0 z z r r = ₂2 = 2 2 2 2 r r f f

The curl of unit vector af is a # d f a a a a a 1 0 1 0 1 z z ₂ z z 2 r r r r r r r = ₂2 = = 2 2 2 2 r r f f ^ h

and curl of unit vector az is az # d a a a 1 0 0 1 0 z z r r = ₂2 = 2 2 2 2 r r f f

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Options (A), (B) and (C) are properties of d operator where as :

g A # d c m ( ) ( ) g g A A g 2 # # d d = +

In a cylindrical coordinate system Laplacian of a scalar field is defined as

f 2 4 f f z f 1 1 2 2 2 2 2 2 2 2 2 2 2 2 2 r r r r r f = e o+ +

( zsin ) zsin z sin cos

1 ₂ ₈ 1 ₂ ₆ 2 2 2 2 2 2 r r r f r r r f f f f = + + _c- - _m _z(2 sin 6zcos2 ) 2 2 _r _f _f + +

(2zsin 16 ) (2 zsin 6z cos2 ) cos

1 1 ₆ 2 2 2 r f r _r r f f f = + - + + cos z cos 16 6 2 6 2 2 2 f r f = + -At point P 3^ , / ,p 2 6h f 2 d =16+ -0 6#₉36#( 1)- =40

A vector field is called conservative (irrotational) if its curl is zero.

i.e. d #M =0 cos cos z xz y kx x xz a a a 2 x x y y z z + 2 2 2 2 2 2 =₀

(0 0) (cosxz xzsinxz cosxz xzsinxz) (2k ) 0

ax - -ay - - + + -1 az=

k

2 -1 =0

or, k =1₂

For a scalar field to be harmonic,

g 2 d =0 x g y g z g 2 2 2 2 2 2 2 2 2 2 2 2 + + =0 2^1+2k yh =0 which results in k=-₂1

Consider V is a scalar field. So the gradient of the field is

V 4 x V y V z V 2 2 2 2 2 2 = + +

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( V) 4# 4 a a ax x x V y y y V z z z V = ₂2 2 2 2 2 2 2 2 2 2 2 R T S S S S V X W W W W z yV y zV ax x zV z xV ay x yV y xV az 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 =c - m +c - m +c - m =0

So the curl of the gradient of any scalar field is zero everywhere.

For vector A to be solenoidal its divergence must be zero.

i.e. d:A=0 ( 4 ) (2 3 ) (4 3 ) x x z y x z z x y cz 2 2 2 2 2 2 + + - + + - =0 c 1+ -0 =0 c =1

For a vector function to be irrotational its curl must be zero.

i.e. d#A =0 A A A a a ax x x y y y z z z 2 2 2 2 2 2 ₌₀ ( x k z) (k x z) ( x k y z) a a a 4 5 4 2 x x y y z z 1 2 3 +2 - - + 2 2 2 2 2 =₀ (-k3- -( 5))ax-(4-k1)ay+( )k2 az =0 ( 5) k3 - - - =0 &k₃=5 4-k1=0 &k1=4 and k2-0=0 &k2=0

So k1, k2 and k3 are 4, 0 and 5 respectively. SOL 1.1.30 Option (C) is correct.

The unit vector normal to a given plane f=0 is

an f f d d =

The given equation is

x y z 2 +4 +6 =7 x y z 2 +4 +6 -7 =0 So, f =2x+4y+6z-7 and gradient of f is f d =-2ax+3ay+6az f d = 22+42+62 = 56 So, an 2a 4a 6a 56 1 x y z = ^ + + h a 2a 3a 14 1 x y z = ^ + + h

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Consider the differential displacement,

dl =dxax+dyay+dzaz So dl C

#

dx a dy a dz a C x C y C z =_c

#

_m +_c

#

_m +_c

#

_m

For a contour the initial and final points are same. So, all the individual integrals described above will be zero. Therefore,

dl

C

#

=0

According to stoke’s theorem.

d A l L :

#

=

#

^d#Ah:dS So U dl C : d ^ h

#

=

#

6d#^dUh@:dS

Since d#^dUh =0 (curl of the gradient of a scalar field is always zero)

So the contour integral is zero.

According to the divergence theorem surface integral of vector over a closed surface is equal to the volume integral of its divergence inside the region defined by closed surface. i.e. A dS S :

#

A dv v : d =

#

^ h Now, d:A x 2yz y z x2 z 2xy 2 2 2 2 2 2 = ^ h+ ^ h+ ^ h 0 0 0 = + + So

#

A:dS =0 ***********

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SOLUTIONS 1.2

Consider that the cube has its edges on the x y, and z- axes respectively as shown in the figure. As the angle between any of the two body diagonals of the cube will be same so we determine the angle q between the diagonals OB and AC of the cube.

From the figure we get the co-ordinates of points A, B and C as :

A"^0 0 3, , h B"^3 3 3, , h and C"^3 3 0, , h

So, the vector length, OB =3ax+3ay+3az

and AC =3ax+3ay-3az

For determining the angle q between them, we take their dot product as

OB : AC

^ h ^ h= OB AC cosq

9+ -9 9 =^3 3 3 3h^ hcosq

So, the angle formed between the diagonals is

q cos 1 ₄1 70.53 c = - _{b l}=

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AB =20ax+18ay-10az and AC =-10ax+8ay+15az

Since the cross product of two vectors is always perpendicular to the plane of vectors. So the unit vector perpendicular to the plane of triangle is given by

an _ABAB _ACAC # # = now, AB#AC a a a 20 10 18 8 10 15 x y z = -R T S S SS V X W W WW 10 8 a 20 15 ( 10) ( 10)a 18#15 # x # # y =6 - -^ h@ -6 - - - @ +620#8- -^ 10#18 ah@ z 350ax 200ay 340az = - + So, an ₃₅₀350_aa ₂₀₀200_aa ₃₄₀340_aa x y z x y z = _-- ₊+ ( ) ( ) ( ) 350a 200a 340a 350 200 340 x y z 2 2 2 = + - + - + 0.664ax 0.379ay 0.645az = - +

The unit vector in the direction of vector AC is given by

aAC = _ACAC ( ) ( ) ( ) 10a 8a 15a 10 8 15 x y z 2 2 2 = - + + - + + 0.507ax 0.406ay 0.761az =- + + 0. 16 ax 0.41ay 0.76az =- + +

Since the cross product of two vectors is always perpendicular to the plane of vectors. So, the unit vector in the plane of the triangle which is perpendicular to

AC is given by cross product of the unit vector perpendicular to the plane of the triangle and the unit vector aAC.

i.e. aP=an#aAC =-0.550ax-0.832ay+0.077az

Unit vector in the direction of AB is given by

aAB ( ) ( ) ( ) a a a 20 18 10 10 x 18 y 10 z 2 2 2 = + + -+ -0.697ax 0.627ay 0.348az = +

-A non unit vector in the direction of bisector of interior angle at A is defined as

(a a ) 2

1

AB+ AC =1 0 697₂6 . ax+0 627. ay-0 348. az-0 507. ax+0 406. ay+0 761. az@

=0.095ax+0.516ay+0.207az

So the unit vector in the direction of bisector of interior angle at A is given by

abis (0.095) (0.516) (0. ) 0.095a 0.516a 0.207a 207 x y y 2 2 2 = + + + + 0.168ax 0.915ay 0.367az = + +

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0.17ax 0.92ay 0.37az

= + +

The vector field F can be written in cartesian system as

( , , )x y z F x y xax yay 2 2 = + + x x y x y y ax ay 2 2 2 2 = + + + ( , , )x y z F cos_r2 ax sin2 ay r f r r f = + (x=rcosf,y=rsinf) (cos a sin a) 1 x y r f f = +

The components of vector field F are

1 cos

Fx= _r f, Fy= 1_rsinf and Fz=0

So the components of vector field F in cylindrical system can be expressed as

F F Fz r f R T S S SS V X W W WW cos sin sin cos F F F 0 0 0 0 1 x y z f f f f = -R T S S SS R T S S SS V X W W WW V X W W WW Fr = _r16cos2f+sin2f@ = _r1

Ff = _r16cosf(-sinf)+sin cosf f@=0

Fz =0

So the vector field F r f( , , )z = _r1ar

At the point P(r =2, f=p/4, z=0 1. )

F = ₂1ar=0.5ar

Any vector field can be represented as the sum of its normal and tangential component to any surface as

A =At+An

where At is tangential component and An is normal component to the surface

r=20 at point P 20 150 330( , c, c). So, An =rar=20ar and therefore, At =A-An sin cos r2 a r4a 4 q f =- q+ f 0.043a 100a = q+ f

Any vector field can be represented as the sum of its normal and tangential components to any surface as

A =At+An