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DRILL PROBLEMS : CHAPTER 2 D2.1 (a) RAB = (5+6) ax + (8-4) ay + (-2-7) az = 11ax + 4ay - 9az (b) RAB = 112 + 42 + 92 = 14.76 m −20 × 10 −6 50 × 10 −6 (−11𝑎 𝑥 − 4𝑎 𝑦 + 9𝑎 𝑧 ) (c) FBA = 10 −9 𝑎𝑏𝑎 = −0.0413 = 30.78𝑎𝑥 + 11.195𝑎𝑦 − 25.18𝑎𝑧 mN 4𝜋 (14.76 2 ) 14.76 36 𝜋 −20 × 10 −6 50 × 10 −6 (−11𝑎 𝑥 − 4𝑎 𝑦...

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DRILL PROBLEMS : CHAPTER 2

D2.1 (a)

RAB = (5+6) ax + (8-4) ay + (-2-7) az = 11ax + 4ay - 9az

(b)

RAB = 112 + 42 + 92 = 14.76 m

(c)

FBA =

(d)

−20 × 10 −6 50 × 10 −6 4𝜋

10 −9 36 𝜋

𝑎𝑏𝑎 = −0.0413

(14.76 2 )

(−11𝑎 𝑥 − 4𝑎 𝑦 + 9𝑎 𝑧 )

−20 × 10 −6 50 × 10 −6

FBA = 4𝜋 ×8.854×10 −12 (14.76 2 ) 𝑎𝑏𝑎 = −0.04125

14.76

= 30.78𝑎𝑥 + 11.195𝑎𝑦 − 25.18𝑎𝑧 mN

(−11𝑎 𝑥 − 4𝑎 𝑦 + 9𝑎 𝑧 ) 14.76

= 30.74𝑎𝑥 + 11.18𝑎𝑦 − 25.15𝑎𝑧 mN

D2.2 𝒓 − 𝒓𝑨 = −25𝑎𝑥 + 30𝑎𝑦 − 15𝑎𝑧 , |𝒓 − 𝒓𝑨 | = 41.43 𝒓 − 𝒓𝑩 = 10𝑎𝑥 − 8𝑎𝑦 − 12𝑎𝑧 , |𝒓 − 𝒓𝑩 | = 17.54 𝑬𝑨 = −1.57 𝑬𝑩 = 14.61

(−25𝑎 𝑥 +30𝑎 𝑦 −15𝑎 𝑧 ) 41.43 (10𝑎 𝑥 −8𝑎 𝑦 −12𝑎 𝑧 ) 17.54

= 9480𝑎𝑥 − 11300𝑎𝑦 + 5600𝑎𝑧

= 83300𝑎𝑥 − 66600𝑎𝑦 − 99900𝑎𝑧

𝑬𝑻 = 𝑬𝑨 + 𝑬𝑩 = 92.48𝑎𝑥 − 77.9𝑎𝑦 − 94.3𝑎𝑧 (b)

𝑘𝑉 𝑚

𝒓 − 𝒓𝑨 = −10𝑎𝑥 + 50𝑎𝑦 + 35𝑎𝑧 , |𝒓 − 𝒓𝑨 | = 61.84 𝒓 − 𝒓𝑩 = 25𝑎𝑥 + 12𝑎𝑦 + 38𝑎𝑧 , |𝒓 − 𝒓𝑩 | = 47.04 𝑬𝑨 = −7050 𝑬𝑩 = 20300

(−10𝑎 𝑥 +50𝑎 𝑦 +35𝑎 𝑧 ) 61.84 (25𝑎 𝑥 +12𝑎 𝑦 +38𝑎 𝑧 ) 47.04

= 1140𝑎𝑥 − 5700𝑎𝑦 − 3990𝑎𝑧

= 10700𝑎𝑥 + 5180𝑎𝑦 + 16400𝑎𝑧

𝑬𝑻 = 𝑬𝑨 + 𝑬𝑩 = 11.84𝑎𝑥 − 0.52𝑎𝑦 + 12.41𝑎𝑧

D2.3 2

2

(a)

Sum = 2 + 0 + 5 + 0 + 17 + 0 = 2.517

(b)

Sum = 11.18 + 22.62 + 46.87 +

1.1

1.01

Made by Zaeem Ahmad Varaich Please inform if you find any error in any solution!

1.001

1.0001 89.44

= 0.1755

𝑘𝑉 𝑚

EE08.SOLUTIONS

(a)

D2.4 (a)

𝑄=

𝜌𝑣 𝑑𝑣 =

𝑣𝑜𝑙

−0.1 −0.1 −0.1 1 𝑑𝑥𝑑𝑦𝑑𝑧 −0.2 −0.2 −0.2 𝑥 3 𝑦 3 𝑧 3

1

1

+

0.2 0.2 0.2 1 𝑑𝑥𝑑𝑦𝑑𝑧 0.1 0.1 0.1 𝑥 3 𝑦 3 𝑧 3

1

1

= − 8 𝑥 2 −0.1 𝑦 2 −0.1 𝑧 2 −0.1 − − 8 𝑥 2 0.2 𝑦 2 0.2 𝑧 2 0.2 =8×(0.03) − 8×(0.03) = 0 −02

−02

(b)

𝜋 0.1 4 3 2 𝜌 𝑧 0 0 2

(c)

∞ 2𝜋 2𝜋 0 0 0

−02

0.1

0.1

sin 0.6𝜑 𝑑𝑧𝑑𝜌𝑑𝜑 = (−

𝑒 −2𝑟 sin 𝜃 𝑑𝜑 𝑑𝜃𝑑𝑟 = (−

0.1

cos 0.6𝜑 0.6

𝑒 −2𝑟 2

)

𝜋 𝜌4

0.1

0

0

(4)

∞

)

0

(cos 𝜃)

𝑧3

(3)

4 2

2𝜋 2𝜋 0 (𝜑) 0

= 1.018 𝑚𝐶

= −6.28 𝐶

D2.5 (a)

E=2 ×

(b)

Ex = 2𝜋𝜀

5×10 −9

𝒂 = 2𝜋𝜀 𝑜 (4) 𝒛 5×10 −9 (3𝒂𝒚 +4𝒂𝒛 ) 𝑜 (5)

5

44.95

𝑉 𝑚

= 10.788𝒂𝒚 + 14.384𝒂𝒛

E = Ex + Ey = 10.788𝒂𝒚 + 36.86𝒂𝒛

𝑉

5×10 −9

, Ey = 2𝜋𝜀 𝑚

𝑜 (4)

𝒂𝒛 = 22.4775 𝒂𝒛

𝑉 𝑚

𝑉 𝑚

D2.6 i)

Electric field due to 3 nC/m2: E1 =

3×10 −9

ii)

Electric field due to 6 nC/m2: E2 =

6×10 −9

iii)

Electric field due to -8 nC/m2: E3 =

2𝜀 𝑜 2𝜀 𝑜

𝒂𝑵 = 169.5 𝒂𝒛 𝒂𝑵 = 338.8 𝒂𝒛

−8×10 −9 2𝜀 𝑜

𝒂𝑵 = −451.76 𝒂𝒛

According to the direction of point relative to normal: (a) E = - E1 - E2 - E3 = −56.6 𝒂𝒛

(a) E = E1 + E2 - E3 = 961 𝒂𝒛

𝑉 𝑚 𝑉 𝑚

(a) E = E1 + E2 + E3 = 56.6 𝒂𝒛

Made by Zaeem Ahmad Varaich Please inform if you find any error in any solution!

𝑉 𝑚

EE08.SOLUTIONS

(a) E = E1 - E2 - E3 = 283 𝒂𝒛

𝑉 𝑚

D2.7 (a)

𝐸𝑦 𝐸𝑥

𝑑𝑦

= 𝑑𝑥

→

4𝑥 2 𝑦2

𝑦

𝑑𝑦

× 8𝑥 = 𝑑𝑥

→

𝑥𝑑𝑥 = − 2𝑦𝑑𝑦

Put x = 1 and y = 2 to get 𝑐 = 33 giving: (b)

𝐸𝑦 𝐸𝑥

𝑑𝑦

= 𝑑𝑥

→

𝑦 (5𝑥+1) 𝑥

𝑑𝑦

= 𝑑𝑥

→

1 5

×

→

𝑥 2 + 2𝑦 2 = 𝑐

𝒙𝟐 + 𝟐𝒚𝟐 = 𝟑𝟑 5𝑥+1−1 5𝑥+1

→

0.4𝑥 − 0.08 ln 5𝑥 + 1 + 𝑐 = 𝑦 2

𝟎. 𝟒𝒙 − 𝟎. 𝟎𝟖 𝐥𝐧 𝟓𝒙 + 𝟏 + 𝟏𝟓. 𝟕𝟒 = 𝒚𝟐

EE08.SOLUTIONS

Put x = 1 and y = 2 to get 𝑐 = 15.74 giving:

𝑑𝑥 = 𝑦𝑑𝑦

Made by Zaeem Ahmad Varaich Please inform if you find any error in any solution!

DRILL PROBLEMS 3

D3.1 (a) Evaluate the triple volume integral to find the total volume enclosed by the portion of sphere / surface and then just multiply it with the given charge to find the total change within it: 𝜋 𝜋 0.26 2 2

1 × 𝑞 = 7.5𝜇𝐶 8

𝑟 2 𝑠𝑖𝑛𝜃 𝑑𝜃𝑑𝜙𝑑𝑟 × 𝑞 = 0

0 0

(b) This surface encloses the whole charge q, so answer is 60 µC (c) Only the upper half of the flux lines pass through the plane at z = 26 cm, so D = 0.5 x 60 = 30 µC

D3.2 (a) 𝐸 =

𝑘𝑄 4𝑎 𝑥 −6𝑎 𝑦 +12𝑎 𝑧 𝑟2

16+36+144

= 0.72𝑎𝑥 − 1.08𝑎𝑦 + 2.16𝑎𝑧

𝑀𝑉 𝑚

𝜇𝐶 𝑠𝑜, 𝐷 = 𝜀𝑜 𝐸 = 6.38𝑎𝑥 − 9.56𝑎𝑦 + 19.125𝑎𝑧 2 𝑚 −3𝑎 𝑦 +6𝑎 𝑧 20 𝑀𝑉 (b) 𝐸 = = −23.97𝑎𝑦 + 47.94𝑎𝑧 45

𝑚

𝑠𝑜, 𝐷 = 𝜀𝑜 𝐸 = −212𝑎𝑦 + 424𝑎𝑧 (c) 𝐸 =

120 2𝜀 𝑜

𝑎𝑧 =

60 𝜀𝑜

𝑎𝑧

𝜇𝑉 𝑚

𝜇𝐶 𝑚2

, 𝑠𝑜 𝐷 = 𝜀𝑜 𝐸 = 60𝑎𝑧

𝜇𝐶 𝑚2

D3.3 (a) 𝐸 =

𝐷 𝜀𝑜

= 33.88𝑟 2 𝑎𝑟 , so at P: 𝐸 = 33.88(2)2 𝑎𝑟 =135.5𝑎𝑟

Solved by Saad & Zaeem. Please report if you find any mistake!

EE08.SOLUTIONS

2𝜋𝜀 𝑜 .45

(b) 𝑄 = 2𝜋 𝜋 0 0

𝐷. 𝑑𝑠 = 2𝜋 0

48.6

𝑎2 𝑠𝑖𝑛𝜃 𝑑𝜃𝑑𝜙𝑎𝑟 × 0.3𝑟 2 𝑎𝑟 = 24.3

2𝜋 0

−𝑐𝑜𝑠𝜃|𝜋0 𝑑𝜙 =

𝑑𝜙 = 305 . 208 𝑛𝐶 2𝜋

(c) On same steps: 𝑄 = 76.8 0

−𝑐𝑜𝑠𝜃|𝜋0 𝑑𝜙 = 964.608 μC

D3.4

10 3

10

(a) 𝑄 = 𝑄1 + 𝑄2 = 0.243 𝜇𝐶 (b) 𝑄 = 𝑙𝑒𝑛𝑔𝑡𝑕 × 𝜌𝐿 = 31.4 𝜇𝐶

𝑦 = 3𝑥 10

(c) Area = 𝑙𝑒𝑛𝑔𝑡𝑕 𝑜𝑓 𝑙𝑖𝑛𝑒 (𝑕𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒) × 𝑤𝑖𝑑𝑡𝑕 𝑎𝑐𝑟𝑜𝑠𝑠 𝑧 = 10.53 × 10

So, 𝑄 = 𝑎𝑟𝑒𝑎 × 𝜌𝐴 = 10.53 𝜇𝐶

𝜌𝑠3

D3.5 (a) 𝐷 =

0.25 4𝜋(0.005)2

= 795.77 𝜇𝐶

𝜌𝑠2

(b) 𝑄2 = 4𝜋 0.01 2 × 𝜌𝑠2 = 2.51 𝜇𝐶, 0.25+2.51 4𝜋(0.015)2 2

(c) 𝑄3 = 4𝜋 0.018

𝑠𝑜, 𝐷 = (d) 𝐷 =

= 977 𝜇𝐶 × 𝜌𝑠3 = −2.44 𝜇𝐶,

0.25+2.51−2.44 4𝜋 (0.025)2

= 40.74 𝜇𝐶

0.25+2.51−2.44+4𝜋 0.03 2 ×𝜌 𝑠 4𝜋(0.035)2

= 0 , so 𝜌𝑠4 = -28.29 𝜇𝐶

D3.6 (a) 𝜓 =

𝐷. 𝑑𝑠 =

3 2 16𝑥 2 𝑦𝑧 3 𝑑𝑥𝑑𝑦 1 0

Solved by Saad & Zaeem. Please report if you find any mistake!

=

3 2 16𝑥 2 𝑦. 8𝑑𝑥𝑑𝑦 1 0

= 1365 𝑝𝐶

EE08.SOLUTIONS

𝑠𝑜, 𝐷 =

(b) 𝐸 =

𝐷 𝜀𝑜

𝑎𝑛𝑑 𝑒𝑣𝑎𝑙𝑢𝑎𝑡𝑒 𝑖𝑡 𝑎𝑡 𝑃

(c) 𝐸𝑣𝑎𝑙𝑢𝑎𝑡𝑒 𝑡𝑕𝑒 𝑓𝑜𝑟𝑚𝑢𝑙𝑎 8 𝑎𝑡 𝑃 𝑎𝑛𝑑 ∆𝑉 = 10−12 𝑚3

D3.7 𝐸𝑣𝑎𝑙𝑢𝑎𝑡𝑒 𝑡𝑕𝑒 𝑐𝑜𝑟𝑟𝑒𝑠𝑝𝑜𝑛𝑑𝑖𝑛𝑔 𝑓𝑜𝑟𝑚𝑢𝑙𝑎𝑒 𝑓𝑜𝑟 div 𝐃 𝑖. 𝑒. 15, 16 & 17 𝑎𝑡 𝑡𝑕𝑒 𝑔𝑖𝑣𝑒𝑛 𝑝𝑜𝑖𝑛𝑡𝑠 𝑃

D3.7 𝑇𝑎𝑘𝑒 div 𝐃 𝑎𝑠 𝑠𝑡𝑎𝑡𝑒𝑑 𝑏𝑦 𝑀𝑎𝑥𝑤𝑒𝑙𝑙 ′ 𝑠1𝑠𝑡 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 𝑡𝑜 𝑔𝑒𝑡 𝑡𝑕𝑒 𝑒𝑥𝑝𝑟𝑒𝑠𝑠𝑖𝑜𝑛𝑠 𝑓𝑜𝑟 𝜌𝑣 , 𝑤𝑖𝑡𝑕 𝑝𝑟𝑜𝑝𝑒𝑟 𝑡𝑟𝑖𝑔𝑛𝑜𝑚𝑒𝑡𝑟𝑖𝑐 𝑚𝑎𝑛𝑖𝑝𝑢𝑙𝑎𝑡𝑖𝑜𝑛

D3.8 R.H.S: 𝜑 𝜑 𝜑 ∇. D = 12𝑠𝑖𝑛 − 0.75𝑠𝑖𝑛 = 11.25 𝑠𝑖𝑛 2 2 2 𝜋

2

5

11.25 0

0

𝜑 𝑠𝑖𝑛 2

𝜌𝑑𝑧𝑑𝜙𝑑𝜌 =

11.25 × 20 = 225

0

L.H.S: 𝜋 5

(𝑫)𝜌=2 𝜌𝑑𝑧𝑑𝜑𝑎𝜌 + 0 0

(𝑫)𝜑=0 𝑑𝑧𝑑𝜌𝑎𝜑 0 0

2 5

+

(𝑫)𝜑=𝜋 𝑑𝑧𝑑𝜌(−𝑎𝜑 ) 0 0

Now, (𝑫)𝜌=2 = 12𝑠𝑖𝑛 Solved by Saad & Zaeem. Please report if you find any mistake!

𝜑 𝑎 2 𝜌

EE08.SOLUTIONS

𝐷. 𝑑𝑠 =

2 5

2 5

𝑫 𝜑=𝜋 𝑑𝑧𝑑𝜌𝑎𝜑 = 0 0 0

𝜋 5

𝑠𝑜,

𝐷. 𝑑𝑠 = 24 0

= 24

2 5

𝜑 𝑠𝑖𝑛 𝑑𝑧𝑑𝜑 + (𝑫)𝜑=0 𝑑𝑧𝑑𝜌𝑎𝜑 2 0 0 0 𝜑 𝜋 𝜌2 2 5 −2𝑐𝑜𝑠 | × 𝑧 |0 − 1.5 | × 𝑧 2 0 2 0

|50

= −48 0 − 1 5 − 15

EE08.SOLUTIONS

= 225

Solved by Saad & Zaeem. Please report if you find any mistake!

Solved by Aqeel Anwar (aqeelanwar.co.cc)

Digitized by Zaeem (ee08.net.tc)

Solved by Aqeel Anwar (aqeelanwar.co.cc)

Digitized by Zaeem (ee08.net.tc)

Solved by Aqeel Anwar (aqeelanwar.co.cc)

Digitized by Zaeem (ee08.net.tc)

Solved by Aqeel Anwar (aqeelanwar.co.cc)

Digitized by Zaeem (ee08.net.tc)

Solved by Aqeel Anwar (aqeelanwar.co.cc)

Digitized by Zaeem (ee08.net.tc)

Solved by Aqeel Anwar (aqeelanwar.co.cc)

Digitized by Zaeem (ee08.net.tc)

Solved by Aqeel Anwar (aqeelanwar.co.cc)

Digitized by Zaeem (ee08.net.tc)

Solved by Aqeel Anwar (aqeelanwar.co.cc)

Digitized by Zaeem (ee08.net.tc)

Solved by Aqeel Anwar (aqeelanwar.co.cc)

Digitized by Zaeem (ee08.net.tc)

Solved by Aqeel Anwar (aqeelanwar.co.cc)

Digitized by Zaeem (ee08.net.tc)

Solved by Aqeel Anwar (aqeelanwar.co.cc)

Digitized by Zaeem (ee08.net.tc)

Solved by Aqeel Anwar (aqeelanwar.co.cc)

Digitized by Zaeem (ee08.net.tc)

DRILL PROBLEMS5

D5.1 (a) At P: 2ሺ ‫ =ܬ‬10ሺ 3ሻ 2ሻ ܽߩെ4ሺ 3ሻ ܿ ‫ݏ݋‬2 ሺ 30ሻ ܽ߮ = 180ܽߩെ9ܽ߮

(b) Using formula (2): 2ߨ 2.8

2ߨ 2.8

‫ݖ‬2 2.8 ‫ =ܫ‬න න 10ߩ ‫ݖ‬ ܽߩ.݀‫ݖ‬ ݀߶ܽߩ = 27 න න 10‫ݖ݀ݖ‬ ݀߶= 27 ቆ ቇ| 2 ሺ ߶ሻ | 2ߨ 0 2 3

0

2

0

2

= 325.72 ݉‫࢘࢕ܣ‬3.25 ‫ܣ‬

D5.2 (a) Using formula (2): 2ߨ20ߤ

2ߨ20ߤ

1.5 ‫ =ܫ‬െන න 106 ‫ݖ‬1.5 ܽ‫ݖ‬.ߩ݀ߩ݀߶ܽ‫ =ݖ‬െන න 106 ሺ 0.1ሻ ߩ݀ߩ݀߶ 0

0

0

0

2

(b) Using formula (3): 6 1.5 ‫ܬ‬ 0.1ሻ ݉‫ܥ‬ ‫ ݖ‬െ10 ሺ ߩ‫= ݒ‬ = = െ 15.81 ‫ݒ‬ 2 × 106 ݉3 ‫ݖ‬

(c) Same formula: 1.5 ‫ܬ‬ െ106 ሺ 0.15ሻ ݉ ‫ݖ‬ ‫ݒ‬ = = = 29.04 ‫ݖ‬ ߩ‫ݒ‬ െ2000 ‫ݏ‬

Solved by Zaeem. Please report if you find any mistake!

EE08.SOLUTIONS

ߩ 20ߤ 1.5 = െ106 ሺ 0.1ሻ ߶ሻ | 2ߨ ‫ܣ‬ ቆ ቇ| 0 ሺ 0 = െ39.7ߤ 2

D5.5 (a) Putting point P in given V, while evaluating trigonome tric functions using radians:

ܸ= 48.848 ܸ (b) Using formula: ࡱ= െ݃‫ݎ‬ ܸܽ݀= െ100 coshሺ 5‫ݔ‬ሻ.5 sinሺ 5‫ݕ‬ሻܽ‫ݔ‬െ100 sinhሺ 5‫ݔ‬ሻ5 cosሺ 5‫ݕ‬ሻܽ‫ݕ‬

ܸ ݉

At P: E = െ474.43ܽ‫ݔ‬െ140.77ܽ‫ݕ‬ ܸ

(c) ȁ Eȁ= ඥ474.432 + 140.772 = 494.87 ݉

݊‫ܥ‬

݊‫ܥ‬

(d) ɏs= ‫ =ܰܦ‬ȁ ࡰࡼȁ , so asࡰࡼ= ߝ ࡰࡼȁ = 4.38 ݉2 ‫݋‬E= െ4.2ܽ ‫ݔ‬െ1.246ܽ ‫݉ݕ‬2 , so ɏ s= ȁ

D5.6 ݊‫ܥ‬

(a) For original line charge, withߩ‫ =ܮ‬40 ݉ ܸ= െන‫ܧ‬.݈݀ , ܽ‫=ܧݏ‬

ߩ‫ܮ‬ ܽߩ, 2ߨߝ ‫ߩ݋‬ (7,െ1,5)

ߩ‫ܮ‬ ‫ݏ‬ ‫ =ܸ݋‬െ න 2ߨߝ ‫݋‬ 4

ܽߩ ݈݀ ߩ

ܽߩ ሺ ‫ݔ‬െ6ሻ ܽ‫ݔ‬+ (‫ݕ‬െ3)ܽ‫ݕ‬ = ߩ (‫ݔ‬െ6) 2 + (‫ݕ‬െ3) 2 ܽ‫ݏ‬ , ݈݀= ݀‫ =ݕ݀݊ܽݔ‬െ1,‫ݏ‬ ‫݋‬: ܸ= െ720 න 4

ሺ ‫ݔ‬െ6ሻ 7 ݀‫ =ݔ‬െ360 ݈ ݊ȁ (‫ݔ‬െ6) 2 + 16ȁ 4 = 58.50 ܸ (‫ݔ‬െ6) 2 + 16 ݊‫ܥ‬

For mirror line charge, with ߩ‫ =ܮ‬െ40 ݉ ܸ= െන‫ܧ‬.݈݀ , ܽ‫=ܧݏ‬

ߩ‫ܮ‬ ܽߩ, 2ߨߝ ‫ߩ݋‬ (7,െ1,5)

ߩ‫ܮ‬ ‫ݏ‬ ‫ =ܸ݋‬െ න 2ߨߝ ‫݋‬ 4

ܽߩ ݈݀ ߩ

Solved by Zaeem. Please report if you find any mistake!

EE08.SOLUTIONS

7

1

CHAPTER 7 DRILLS Solved by Zaeem A. Varaich www.ee08.net.tc

D7.1 (a) V |P (1,2,3) =

4(2)(3) (1)2 +1

= 12 V

As, ρv = −∇2 V, so we first calculate ∇2 V : ∇V = −8 (x2yzx + +1)2

4z x2 +1

+

4y x2 +1

2

yz ⇒ ∇2 V = 32 (xyzx 2 +1)3 − 8 (x2 +1)2

⇒ ∇2 V |P (1,2,3) = 12; pC so, ρv = −∇2 V = −o (12) = −106.25 m 3

(b) V |P (3, π3 ,2) = −22.5 V As, ρv = −∇2 V, so we first calculate ∇2 V : ∇2 V = 20 cos (2 φ) − 20 cos (2 φ) ⇒ ∇2 V |P (3, π3 ,2) = 0; pC so, ρv = −∇2 V = −o (0) = 0 m 3

(c) V |P (0.5,45o ,60o ) = 4 V As, ρv = −∇2 V, so we first calculate ∇2 V : cos(φ) ∇2 V = 4 cos(φ) − 2 r4 (sin(θ)) 2 r4

⇒ ∇2 V |P (0.5,45o ,60o ) = 0; pC so, ρv = −∇2 V = −o (0) = 0 m 3

D7.2 Apply the formulae & concepts to find the answers!

2

D7.3 (a) The solution to Laplace’s equation

1 ∂ ρ ∂ρ



 ∂ ρ ∂ρ = 0 is:

V = A ln ρ + B Putting the given values of V & ρ and solving the simultaneous equations, we get: A = −73.9 & B = 101.28 so, V = −73.9 ln ρ + 101.28 Now, E = −∇V = ρ1 (73.9) aρ = √ (∵ ρ = 32 + 12 )

√1 (73.9) aρ 10

= 23.36 aφ

V so, |E| = 23.36 m

(b) The solution to Laplace’s equation

1 ∂2V ρ2 ∂φ2

= 0 is:

V = Aφ + B Putting the given values of V & φ and solving the simultaneous equations, we get: A = −85.9 & B = 64.9 so, V = −85.9φ + 64.9 Now, E = −∇V = ρ1 (85.9) aφ = √ (∵ ρ = 32 + 12 ) V so, |E| = 27.16 m

D7.4 & D7.5 Not included in the course!

√1 (85.9) aφ 10

= 27.16 aφ

3

D7.6

The solution to this problem depends on how you proceed in each iteration and on your initial estimate. The dotted lines show how the initial estimate was found.

(a) 20.8 V

(b) 44.47 V

(c) 89.8 V

4 Please report to the following e-mail, if you find any mistake:

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1

CHAPTER 8 DRILLS (Upto D8.3)

Solved by Zaeem A. Varaich www.ee08.net.tc

D8.1 (a) Using ∆- form of equation (2), i.e. ∆H2 = Here, aR12 =

(4−0)ax +(2−0)ay +(0−2)az √ 42 +22 +22

I1 ∆L1 ×aR12 2 4πR12

= 0.816ax + 0.408ay − 0.408az

2 R12 = 42 + 22 + 22 = 24

so, ∆H2 =

I1 ∆L1 ×aR12 2 4πR12

(b) As, ∆H2 = Here, aR12 =

=

2πaz ×(0.816ax +0.408ay −0.408az ) µ 301.59

=

5.12ay −2.56ax µ 301.59

= −8.5ax + 17.0ay

I1 ∆L1 ×aR12 2 4πR12

(4−0)ax +(2−2)ay +(3−0)az √ 42 +02 +32

= 0.8ax + 0.6az

2 R12 = 42 + 02 + 32 = 25

so, ∆H2 =

I1 ∆L1 ×aR12 2 4πR12

(c) As, ∆H2 = Here, aR12 =

=

2πaz ×(0.8ax +0.6az ) µ 100π

=

5.02ay 100π µ

= 16ay

nA m

I1 ∆L1 ×aR12 2 4πR12

(−3−1)ax +(−1−2)ay +(2−3)az √ 42 +32 +12

= −0.78ax − 0.58az − 0.19az

2 = 42 + 32 + 12 = 26 R12

so, ∆H2 =

I1 ∆L1 ×aR12 2 4πR12

=

2π(−ax +ay +2az )×(−0.78ax −0.58az −0.19az ) µ 104π

⇒ ∆H2 = 18.85 ax − 33.94 ay + 26.40 az

nA m

=

(1.96 ax −3.53 ay +2.74 az )π µ 104π

nA m

2

D8.2 Using, H2 =

I 2 πρ aφ

√ √ (a) For PA ( 20, 0, 4), we have ρ = 20 + 0 = 4.47, so: H2 =

15 2 π(4 .47 ) aφ

φ = tan−1

= 0 .533 aφ = (0 .533 × − sin φ)ax + (0 .533 × cos φ)ay , where  
y −1 √0 = 0o , so = tan x 20

H2 = 0.533ay

A m

(b) For PB (2, −4, 4), we have ρ = H2 =

√

22 + 42 = 4.47, so:

15 2 π(4 .47 ) aφ

φ = tan−1


y x

= 0 .533 aφ = (0 .533 × − sin φ)ax + (0 .533 × cos φ)ay , where
= tan−1 −4 = −63.43o , so 2

H2 = (0 .533 ×0.89)ax + (0 .533 ×0.44)ay = 0.474ax + 0.238ay

D8.3 (a)

For infinitely long filament; I H = 2πρ aφ Here,p ρ = (0.1)2 + (0.1)2 = H=

I 2πρ aφ

=

(2.5) √

2π(

2 10 )

√

2 10

aφ = 2.8134aφ = (2 .8134 × − sin φ)ax + (2 .8134 × cos φ)ay

Now, φ = 270o − θ = 270o − tan−1

0.1 0.1




= 270o − 45o = 225o ,

so, H = (2 .8134 ×0.707)ax + (2 .8134 × − 0.707)ay = 1.989ax − 1.989ay

A m

3 (b)

For ρ < a, H=

Iρ 2πa2 aφ

=

(2.5)(0.2) 2π(0.3)2 aφ

Now, φ = tan−1

y x

so, H = −0.884ax




= 0.884aφ = (0 .884 × − sin φ)ax + (0 .884 × cos φ)ay

= tan−1

0.2 0




= 90o ,

A m

(c)

Now, H1 = 12 K1 × aN = 12 (2.7)ax × ay = 1.35az H2 = 12 K2 × aN = 12 (−1.4)ax × ay = −0.7az H3 = 12 K3 × aN = 12 (−1.3)ax × −ay = 0.65az so, H = H1 + H2 + H3 = 1.300az

A m

4 Please report to the following e-mail, if you find any mistake:

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CHAPTER 8 DRILLS

CHAPTER 8 DRILLS (D8.4 onwards) Solved by Zaeem A. Varaich www.ee08.net.tc

D8.4

(a) For path 1: ´ ´4 (3zax − 2x3 az ).(dxax + dyay +dzaz ) = 2 (3zdx) = 3(4)(4 − 2) = 24 A For ´ path 2: 3 ´1 (3zax − 2x az ).(dxax + dyay +dzaz ) = 4 (−2x3 dz) = −2(43 )(1 − 4) = 384 A For ´ path 3: 3 ´2 (3zax − 2x az ).(dxax + dyay +dzaz ) = 4 (3zdx) = 3(1)(2 − 4) = −6 A For ´ path 4: 3 ´4 (3zax − 2x az ).(dxax + dyay +dzaz ) = 1 (−2x3 dz) = −2(23 )(4 − 1) = −48 A ¸ So, H.dL = 24 + 384 − 6 − 48 = 354 A (b) 4SN = 3 × 2 = 6 m2 , so ¸

(∇ × H)y =

H.dL 4SN

=

ax ∂ (c) (∇ × H) = ∂x Hx

354 6

ay ∂ ∂y

Hy

= 59 mA2 az ax ∂ ∂ ∂z = ∂x 3z Hz

ay ∂ ∂y

0

∂ = (0 − 0)ax −(−6x2 − 3)ay +(0 − 0)az = (6x2 + 3)ay ∂z 3 −2x az

At the center, x = 3, z = 2.5, so (∇ × H)y = [6(3)2 + 3]=57 mA2

1

ee08.net.tc

CHAPTER 8 DRILLS

D8.5 ax ∂ (a) J = ∇ × H = ∂x Hx

ay ∂ ∂y

Hy

At P: J = −16ax +9ay +16az

az ax ∂ ∂ ∂z = ∂x 0 Hz

ay ∂ ∂y 2

x z

∂ = (−2yx − x2 )ax −(−y 2 − 0)ay +(2xz − 0)az ∂z 2 −y x az

A m

     ∂Hφ ∂(ρH ) ∂H z z (b) J = ∇ × H = ρ1 ∂H aρ + ∂zρ − ∂H aφ + ρ1 ∂ρ φ − ∂φ − ∂z ∂ρ     = ρ1 .0 − 0 aρ + (0 − 0) aφ + ρ1 .(2 cos 0.2φ) − ρ1 ρ2 (−0.2) sin 0.2φ az At P: J =0.055az



az

A m2

(c) Using equation (26): 1 1 = r sin θ (0 − 0) ar + r (0 − 0) aθ + At P: J =aφ

1 ∂Hρ ρ ∂φ

1 r

∂ r ∂r sinθ


− 0 aφ =

1 r sin θ aφ

A m2

D8.6

(a)

¸

H.dL :

For path 1: ´ ´5 2 2 ) (6xyax − 3y 2 ay ).(dxax + dyay +dzaz ) = 2 (6xydx) = 6(−1) (5 −2 = −63 A 2 For path 2: ´ ´1 3 3 ) = −2 A (6xyax − 3y 2 ay ).(dxax + dyay +dzaz ) = −1 (−3y 2 dy) = −3 (1 +1 3 For path 3: ´ ´2 2 2 ) (6xyax − 3y 2 ay ).(dxax + dyay +dzaz ) = 5 (6xydx) = 6(1) (2 −5 = −63 A 2 For path 4: ´ ´ −1 3 3 (6xyax − 3y 2 ay ).(dxax + dyay +dzaz ) = 1 (−3y 2 dy) = −3 (−1 3−1 ) = 2 A ¸ So, H.dL = −63 − 2 − 63 + 2 = −126 A ´

(∇ × H).dS : ax ay az ∂ ∂ ∂ (∇ × H) = ∂x ∂y ∂z Hx Hy Hz

(b) Now,

S

ax ∂ = ∂x 6xy

ay ∂ ∂y

−3y 2

az ∂ ∂z = (0 − 0)ax − (0 − 0)ay + (0 − 6x)az = −6xaz 0

(∇ × H).dS =(−6xaz )(dydzax + dxdza z ) = −6x dxdy,  y +dxdya  5  ´ ´´ 1 x2 (∇ × H).dS = −6 x dxdy = −6 2 y|−1 = −6 (25−4) (1 + 1) = −126 A 2 2

D8.7 (a) Hφ =

Iρ 2πa2

=

(20)(0.5m) 2π(1m)2

A = 1592 m

(b) Bφ = µo Hφ = 4π × 10−7 (20)(0.8...