1401799646_electromagnetic field 2-2 set-1 a

S.1Electromagnetic Fields (March/April-2013, Set-1) JNTU-Anantapur

( JNTU-Anantapur ) B.Tech. II-Year II-Sem.

Code No.: 9A02404/R09

B.Tech. II Year II Semester Regular and Supplementary Examinations

April/May - 2013

ELECTROMAGNETIC FIELDS

( Electrical and Electronics Engineering )

Time: 3 Hours Max. Marks: 70

Answer any FIVE Questions

All Questions carry equal marks

- - -

1. An infinitely large cylinder has a radius and a uniform charge of one micro coulomb per meter. Calculate the potentialat a point 10 m away from the cylinder if zero potential point is taken to be at a radial distance of 1 m.(Unit-I, Topic No. 1.5)

2. (a) Explain,

(i) Dipole and

(ii) Dipole moment. (Unit-II, Topic No. 2.3)

(b) Find electric potential due to electric dipole. (Unit-II, Topic No. 2.4)

3. A parallel plate capacitor consists of two square metal plates with 500 mm side and separated by 10 mm. A slab ofsulphur (ε

r = 4) 6 mm thick is placed on the lower plate and air gap of 4 mm. Find capacitance of capacitor.

(Unit-III, Topic No. 3.4)

4. (a) State and explain Biot-Savart’s law. (Unit-IV, Topic No. 4.2)

(b) Develop an expression for the magnetic field at any point on the line through the centre at a distance ‘h’ fromthe centre and perpendicular to the plane of a plane circular loop of radius ‘a’ and carrying current ‘I’ amperes.(Unit-V, Topic No. 5.6)

5. A steady current of 1000 A is established in a long straight hollow aluminum conductor of inner radius 1 cm and outerradius 2 cm assume uniform resistivity and calculate B as a function of radius r from the axis of the conductor and alsoderive the formula used. (Unit-V, Topic No. 5.1)

6. (a) Derive the expression for the magnetic moment of a planar coil. (Unit-VI, Topic No. 6.7)

(b) A magnetic field B = 3.5 x 10– 2 az Wb/m2 exerts a force on a 0.3 m conductor along the x-axis. If the conductor

current is 5 A in the – az direction what force must be applied to hold the conductor in position? Also derive the

formula used. (Unit-VI, Topic No. 6.2)

7. (a) Determine the self inductance of a coaxial cable of inner radius a and outer radius b. (Unit-VII, Topic No. 7.7)

(b) A coaxial cable consists of an inner conductor of radius 1.2 cm and an outer conductor of radius 1.8 cm the twoconductors are separated by an insulating medium(µ

r = 4µ

0). If the cable is 3 m long and carries 25 mA current,

calculate the energy stored in the medium. (Unit-VII, Topic No. 7.9)

8. Write Maxwell’s equations in good conductors for time varying fields and static fields both in differential andintegral form. (Unit-VIII, Topic No. 8.5)

Set-1Solutions

S.2 Spectrum ALL-IN-ONE Journal for Engineering Students, 2014

B.Tech. II-Year II-Sem. ( JNTU-Anantapur )

SOLUTIONS TO APRIL/MAY-2013, SET-1, QPQ1. An infinitely large cylinder has a radius and

a uniform charge of one micro coulomb permeter. Calculate the potential at a point 10 maway from the cylinder if zero potential pointis taken to be at a radial distance of 1 m.

Answer : April/May-13, Set-1, Q1

For answer refer Unit-I, Q35.

Q2. (a) Explain,

(i) Dipole and

(ii) Dipole moment.

Answer : April/May-13, Set-1, Q2(a)

(i) Dipole

For answer refer Unit-II, Q7.

(ii) Dipole Moment


(b) Find electric potential due to electricdipole.

Answer : April/May-13, Set-1, Q2(b)


Q3. A parallel plate capacitor consists of twosquare metal plates with 500 mm side andseparated by 10 mm. A slab of sulphur(εεεεεr = 4) 6 mm thick is placed on the lowerplate and air gap of 4 mm. Find capacitanceof capacitor.


For answer refer Unit-III, Q17.

Q4. (a) State and explain Biot-Savart’s law.


For answer refer Unit-IV, Q3.

(b) Develop an expression for the magneticfield at any point on the line throughthe centre at a distance ‘h’ from thecentre and perpendicular to the planeof a plane circular loop of radius ‘a’ andcarrying current ‘I’ amperes.


For answer refer Unit-V, Q25 [Note: Replace ‘r’ by‘a’].

Q5. A steady current of 1000 A is established in along straight hollow aluminum conductor ofinner radius 1 cm and outer radius 2 cmassume uniform resistivity and calculate Bas a function of radius r from the axis of theconductor and also derive the formula used.


Derivation

Consider a hollow conductor, of inner radius ‘R1’ and

outer radius ‘R2’ placed along the z-axis as shown in the

figure (1). Let ‘I’ denotes the current carrying by theconductor.

Z

R1

R2

��

��

��

��

Z

R1

R2

��

��

��

��

Figure (1): Hollow Conductor

Case (1)

Construct an Ampere’s loop of radius ‘r’ inside theconductor such that r < R

1 as shown in the figure (2).

Z

R1r

��

��

��

��

��

Z

R1r

��

��

��

��

��

Figure (2): Ampere’s Loop for r < R1

According to Ampere’s circuital law, the line integral

of the magnetic field intensity ( H ) about any closed path

is equal to the current enclosed by that path.

Mathematically, ∫ ldH . = I ... (1)



Where, H = A/m)( zzaHaHaH ++ φφφφ .

ld is the differential length along the Ampere’s loopas shown in the figure (3). Since, the differential length isvarying only along the circumference, in cylindricalco-ordinates, it is expressed as,

dlr

�� adld�

�� addldlr

�� adld�

�� addl

Figure (3): Ampere’s Loop

ld = φφ ad

∴ From equation (1), we have

⇒ φφφ ∫ φ++ adaHaHaH zzrr )( = I

⇒ φφφ π++ araHaHaH zzrr )2)((r

= I

⇒ Hφ (2πr) = 0 [Q I = 0 inside the hollowconductor]

⇒ Hφ = 0 A/m

H = 0 A/m, for r < R1

⇒ B = 0 Wb/m2 for r < R1 [Q B = µ. H ].

Case (2)

Construct an Ampere’s loop of radius ‘r’ such that R1

< r < R2 as shown in the figure (4).

Applying Ampere’s circuital law, we get

Z

R1

r

��

��

��

��

��

R2

Z

R1

r

��

��

��

��

��

R2

Figure (4): Ampere’s Loop for R1 < r < R

2

∫ ldH . = I

Where,

H = aHaHaH zzrr )( ++ φφ A/m

ld = φφ ad [From figure (3)]

⇒ ∫ ldH = Current density (J) × Area. ].[ AJI =Q

⇒ ∫ φφφ φ++ adaHaHaH zzrr .)(

= )( 2

122 RR

I

π−π × (πr2 – πR

12)

⇒ φφφ π++ araHaHaH zzrr )2)(( = )(

)(21

22

21

2

RR

RrI

−π−π

⇒ Hφ (2πr) = I × )(

)(21

22

21

2

RR

Rr

−−

⇒ Hφ = )(

)(

2 21

22

21

2

RR

Rr

r

I

−−

π

⇒ H = φ−−

πa

RR

Rr

r

I

)(

)(

2 21

22

21

2

A/m. For R1 < r < R

2.

⇒ B = φ−−

πµ

aRR

Rr

r

I

)(

)(

2 21

22

21

20 Wb/m2 for R

1 < r < R

2

[ BQ = µ0 H ]

Case (3)

Construct an Ampere’s loop of radius ‘r’ such thatr > R

2 as shown in the figure (5).

Z

R1

r

��

��

��

��

��

��

R2

Z

R1

r

��

��

��

��

��

��

R2

Figure (5): Ampere’s Loop for r > R2

According to Ampere’s law, we have



∫ ldH . = I

Where,

H = A/m)( zzrr aHaHaH ++ φφ

ld = φφ ad [From figure (3)]

⇒ ∫ ldH = I

⇒ )( zzrr aHaHaH ++ φφ ∫ φφ ad = I

⇒ Hφ (2πr) = I

⇒ Hφ = r

I

π2

⇒ H = r

I

π2 φa A/m for r > p2.

⇒ B = r

I

πµ2

0φa Wb/m2 for r > R

2

[Q B = µ0 H ]

The results of 3 cases can be summarized as below,

B =

<π

µ

<<−−

πµ

<

φ

φ

20

2121

22

21

20

1

;2

;)(

)(

2

;0

Rrforar

I

RrRforaRR

Rr

r

I

Rrfor

Problem

Given that,

Current in the conductor, J = 1000 A

Inner radius, R1 = 1 cm = 1 × 10–2 m

Outer radius, R2 = 2cm = 2 × 10–2 m

Radius of the conductor, r = r cm = r × 10–2 m

Case (1)

For r < R1 ⇒ r < 0.01 m, we know that

H = 0

⇒ B = 0 [Q B = µ0 H ]

Case (2)

For 0.01 m < r < 0.02, We know that

B = 2

21

22

21

20 Wb/m

)(

)(

2 φ−−

πµ

aRR

Rr

r

I

⇒ B = )101104(

)10110(

102

100010444

442

2

7

−−

−−

−

−

×−××−×

××π×π r

r

⇒ B = 4

425

10)14(

10)1(102000−

−−

×−×−× r

r

⇒ B = r

r )1(

3

102000 25 −×× −

⇒ B = 666.67 × 105 r

r )1( 2 −

⇒ B = 6.67 r

r )1( 2 − m Wb/m2

Case (3)

For r > 0.02 m, we have

B = r

I

πµ2

0 φa Wb/m2

⇒ B = φ−

−

×π××π

ar )10(2

10001042

7

⇒ B = φ

−×a

r

5102000

⇒ B = φar

2Wb/mm20

∴ The result can be summarized as below,

B =

>

<<−<

m0.02rfor;Wb/m2

m0.02rm0.01for;Wb/mm1)(

6.67

m0.01rfor;0

20

22

mr

r

r



Q6. (a) Derive the expression for the magneticmoment of a planar coil.


For answer refer April/May-11, Set-1, Q6(a).

(b) A magnetic field B = 3.5 x 10– 2 az Wb/m2

exerts a force on a 0.3 m conductoralong the x-axis. If the conductorcurrent is 5 A in the – az direction whatforce must be applied to hold theconductor in position. Also derive theformula used.


For answer refer Unit-VI, Q9.

Q7. (a) Determine the self inductance of a co-axial cable of inner radius a and outerradius b.


Consider a coaxial cable of length ‘d’ carrying current‘I’ amperes. Let the radius of the inner conductor be ‘a’ andthe radius of the outer conductor be ‘b’ as shown in thefigure below.

a

b

d

z

a

b

d

z

Figure: Co-axial Cable

Assume that the current of I amp is along z-axis.

The magnetic field intensity (H) between the innerand outer conductors at any point is given by,

H = r

I

π2a < r < b

The magnetic field intensity (b) is given by,

B = µ H

= r

I

πµ2

If the cable is along z-axis, then the flux densityextends from r = a to b and z = 0 to d

∴ B = r

I

πµ2 φa

But, the total magnetic flux produced (φ) is given by,

φ = sdBS

.∫Where,

ds is the differential current element and isgiven as,

ds = dr dz φa

∴ φ = ∫ ∫= =

φ

d

z

b

ar

adzdrB0

.

= ∫ ∫= =

φφπµ

d

z

b

ar

adzdrar

I

0

.2

= ∫ ∫= =

πµ

d

z

b

ar

dzdrr

I

0

1

2 [Q φφ aa . = 1]

= [ ]∫=

=πµ

d

z

bar dzr

I

0

ln2

= ∫=

−π

µd

z

dzabI

0

)ln(2

= ∫=

πµ

d

z

dza

bI

0

ln2

= ∫=

πµ

d

z

dza

bI

0

.1ln2

= ( )dzza

bI0ln

2 =

πµ

= )0(ln2

−

πµ

da

bI



φ =

πµ

a

bIdln

2

The inductance of a co-axial cable is given by,

L = )(current Total

)(flux Total

I

φ

= I

a

bId

πµ

ln2

A/mln2

πµ=∴

a

bdL

(b) A coaxial cable consists of an inner conductor of radius 1.2 cm and an outer conductor ofradius 1.8 cm the two conductors are separated by an insulating medium(µµµµµr = 4µµµµµ0). If thecable is 3 m long and carries 25 mA current, calculate the energy stored in the medium.


Unsolved.

Q8. Write Maxwell’s equations in good conductors for time varying fields and static fields both indifferential and integral form.


For answer refer Unit-VIII, Q17.

1401799646_electromagnetic field 2-2 set-1 a

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