DRILL PROBLEMS : CHAPTER 2

(1)

Made by Zaeem Ahmad Varaich

Please inform if you find any error in any solution!

E E 0 8 . S O L U T I O N S

DRILL PROBLEMS : CHAPTER 2

D2.1

(a) R

_AB

= (5+6) a

_x

+ (8-4) a

_y

+ (-2-7) a

_z

= 11a

_x

+ 4a

_y

- 9a

_z

(b) R

_AB

= 11

²

+ 4

²

+ 9

²

= 14.76 m

(c) F

BA

=

^{−20 × 10}⁻⁶^{50 × 10}⁻⁶

4𝜋^{10 −9}_{36 𝜋} (14.76²)

𝑎

_𝑏𝑎

= −0.0413

^(−11𝑎^𝑥^{− 4𝑎}^𝑦^{+ 9𝑎}^𝑧⁾

14.76

= 30.78𝑎

_𝑥

+ 11.195𝑎

_𝑦

− 25.18𝑎

_𝑧

mN (d) F

BA

=

4𝜋 ×8.854×10^{−20 × 10}⁻⁶^{50 × 10}⁻¹²(14.76⁻⁶²)

𝑎

_𝑏𝑎

= −0.04125

^(−11𝑎^𝑥^{− 4𝑎}^𝑦^{+ 9𝑎}^𝑧⁾

14.76

= 30.74𝑎

_𝑥

+ 11.18𝑎

_𝑦

− 25.15𝑎

_𝑧

mN

D2.2

(a) 𝒓 − 𝒓

_𝑨

= −25𝑎

_𝑥

+ 30𝑎

_𝑦

− 15𝑎

_𝑧

, |𝒓 − 𝒓

_𝑨

| = 41.43 𝒓 − 𝒓

_𝑩

= 10𝑎

_𝑥

− 8𝑎

_𝑦

− 12𝑎

_𝑧

, |𝒓 − 𝒓

_𝑩

| = 17.54

𝑬

_𝑨

= −1.57

^(−25𝑎^𝑥^+30𝑎_41.43^𝑦^−15𝑎^𝑧⁾

= 9480𝑎

_𝑥

− 11300𝑎

_𝑦

+ 5600𝑎

_𝑧

𝑬

_𝑩

= 14.61

^(10𝑎^𝑥^−8𝑎^𝑦^−12𝑎^𝑧⁾

17.54

= 83300𝑎

_𝑥

− 66600𝑎

_𝑦

− 99900𝑎

_𝑧

𝑬

_𝑻

= 𝑬

_𝑨

+ 𝑬

_𝑩

= 92.48𝑎

_𝑥

− 77.9𝑎

_𝑦

− 94.3𝑎

_𝑧 ^𝑘𝑉

𝑚

(b) 𝒓 − 𝒓

_𝑨

= −10𝑎

_𝑥

+ 50𝑎

_𝑦

+ 35𝑎

_𝑧

, |𝒓 − 𝒓

_𝑨

| = 61.84

𝒓 − 𝒓

_𝑩

= 25𝑎

_𝑥

+ 12𝑎

_𝑦

+ 38𝑎

_𝑧

, |𝒓 − 𝒓

_𝑩

| = 47.04 𝑬

_𝑨

= −7050

^(−10𝑎^𝑥^+50𝑎^𝑦^+35𝑎^𝑧⁾

61.84

= 1140𝑎

_𝑥

− 5700𝑎

_𝑦

− 3990𝑎

_𝑧

𝑬

_𝑩

= 20300

^(25𝑎^𝑥^+12𝑎_47.04^𝑦^+38𝑎^𝑧⁾

= 10700𝑎

_𝑥

+ 5180𝑎

_𝑦

+ 16400𝑎

_𝑧

𝑬

_𝑻

= 𝑬

_𝑨

+ 𝑬

_𝑩

= 11.84𝑎

_𝑥

− 0.52𝑎

_𝑦

+ 12.41𝑎

_𝑧 ^𝑘𝑉

𝑚

D2.3

(a) Sum = 2 + 0 +

²

5

+ 0 +

²

17

+ 0 = 2.517 (b) Sum =

^1.1

11.18

+

^1.01

22.62

+

^1.001

46.87

+

^1.0001

89.44

= 0.1755

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E E 0 8 . S O L U T I O N S

D2.4

(a) 𝑄 =

_𝑣𝑜𝑙

𝜌

_𝑣

𝑑𝑣 =

_−0.2^−0.1 _−0.2^−0.1 _−0.2^−0.1_𝑥₃_𝑦¹₃_𝑧₃

𝑑𝑥𝑑𝑦𝑑𝑧 +

¹

𝑥³𝑦³𝑧³

𝑑𝑥𝑑𝑦𝑑𝑧

0.2 0.1 0.2 0.1 0.2 0.1

= −

_{8 𝑥}₂ ¹

−02−0.1 𝑦²₋₀₂^−0.1 𝑧²₋₀₂^−0.1

− −

¹

8 𝑥²_0.1^0.2 𝑦²_0.1^0.2 𝑧²_0.1^0.2

=

_8×(0.03)¹

−

¹

8×(0.03)

= 0 (b)

₀^𝜋 ₀^0.1

𝜌

₂⁴ ³

𝑧

²

sin 0.6𝜑 𝑑𝑧𝑑𝜌𝑑𝜑 = (−

^{cos 0.6𝜑}_0.6

)

0 𝜋

(

^𝜌⁴

4

)

0 0.1

(

^𝑧³

3

)

2

4

= 1.018 𝑚𝐶

(c)

₀^∞ ₀^2𝜋 ₀^2𝜋

𝑒

^−2𝑟

sin 𝜃 𝑑𝜑 𝑑𝜃𝑑𝑟 = (−

^𝑒^−2𝑟₂

)

0

∞

(cos 𝜃)

₀^2𝜋

(𝜑)

₀^2𝜋

= −6.28 𝐶

D2.5

(a) E = 2 ×

_2𝜋𝜀^5×10⁻⁹

𝑜(4)

𝒂

_𝒛

= 44.95

^𝑉

𝑚

(b) E

_x

=

^5×10⁻⁹

2𝜋𝜀𝑜(5)

(3𝒂𝒚+4𝒂𝒛)

5

= 10.788𝒂

_𝒚

+ 14.384𝒂

_𝒛 ^𝑉

𝑚

, E

_y

=

^5×10⁻⁹

2𝜋𝜀𝑜(4)

𝒂

_𝒛

= 22.4775 𝒂

_𝒛 ^𝑉

𝑚

E = E

_x

+ E

_y

= 10.788𝒂

_𝒚

+ 36.86𝒂

_𝒛 ^𝑉

𝑚

D2.6

i) Electric field due to 3 nC/m

²

: E

1

=

^3×10_2𝜀 ⁻⁹

𝑜

𝒂

_𝑵

= 169.5 𝒂

_𝒛

ii) Electric field due to 6 nC/m

²

: E

₂

=

^6×10_2𝜀 ⁻⁹

𝑜

𝒂

_𝑵

= 338.8 𝒂

_𝒛

iii) Electric field due to -8 nC/m

²

: E

₃

=

^−8×10⁻⁹

2𝜀𝑜

𝒂

_𝑵

= −451.76 𝒂

_𝒛

According to the direction of point relative to normal:

(a) E = - E

₁

- E

₂

- E

₃

= −56.6 𝒂

_𝒛 ^𝑉

𝑚

(a) E = E

₁

- E

₂

- E

₃

= 283 𝒂

_𝒛 ^𝑉

𝑚

(a) E = E

₁

+ E

₂

- E

₃

= 961 𝒂

_𝒛 ^𝑉

𝑚

(a) E = E

₁

+ E

₂

+ E

₃

= 56.6 𝒂

_𝒛 ^𝑉

𝑚

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E E 0 8 . S O L U T I O N S

D2.7

(a)

^𝐸^𝑦

𝐸𝑥

=

^𝑑𝑦

𝑑𝑥

→

^4𝑥²

𝑦²

×

^𝑦

8𝑥

=

^𝑑𝑦

𝑑𝑥

→ 𝑥𝑑𝑥 = − 2𝑦𝑑𝑦 → 𝑥

²

+ 2𝑦

²

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

^𝟐

+ 𝟐𝒚

^𝟐

= 𝟑𝟑

(b)

^𝐸_𝐸^𝑦

𝑥

=

^𝑑𝑦

𝑑𝑥

→

^{𝑦 (5𝑥+1)}

𝑥

=

^𝑑𝑦

𝑑𝑥

→

¹

5

×

^5𝑥+1−1

5𝑥+1

𝑑𝑥 = 𝑦𝑑𝑦 → 0.4𝑥 − 0.08 ln 5𝑥 + 1 + 𝑐 = 𝑦

²

Put x = 1 and y = 2 to get 𝑐 = 15.74 giving: 𝟎. 𝟒𝒙 − 𝟎. 𝟎𝟖 𝐥𝐧 𝟓𝒙 + 𝟏 + 𝟏𝟓. 𝟕𝟒 = 𝒚

^𝟐

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

E E 0 8 . S O L U T I O N S

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:

𝑟 ² 𝑠𝑖𝑛𝜃 𝑑𝜃𝑑𝜙𝑑𝑟 ×

𝜋 2

0 𝜋 2

0 0.26

0 𝑞 = 1

8 × 𝑞 = 7.5𝜇𝐶

(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

𝑟

²

4𝑎

_𝑥

−6𝑎

_𝑦

+12𝑎

_𝑧

16+36+144 = 0.72𝑎 _𝑥 − 1.08𝑎 _𝑦 + 2.16𝑎 _𝑧 ^𝑀𝑉

𝑚

𝑠𝑜, 𝐷 = 𝜀 _𝑜 𝐸 = 6.38𝑎 _𝑥 − 9.56𝑎 _𝑦 + 19.125𝑎 _𝑧 𝜇𝐶 𝑚 ²

(b) 𝐸 = ²⁰

2𝜋𝜀

_𝑜

.45

−3𝑎

_𝑦

+6𝑎

_𝑧

45 = −23.97𝑎 _𝑦 + 47.94𝑎 _𝑧 ^𝑀𝑉

𝑚

𝑠𝑜, 𝐷 = 𝜀 _𝑜 𝐸 = −212𝑎 _𝑦 + 424𝑎 _𝑧 𝜇𝐶 𝑚 ²

(c) 𝐸 = ¹²⁰

2𝜀

_𝑜

𝑎 _𝑧 = ⁶⁰

𝜀

_𝑜

𝑎 _𝑧 ^{𝜇 𝑉}

𝑚 , 𝑠𝑜 𝐷 = 𝜀 _𝑜 𝐸 = 60𝑎 _𝑧 ^𝜇𝐶

𝑚

²

D3.3

(a) 𝐸 = ^𝐷

𝜀

_𝑜

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

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E E 0 8 . S O L U T I O N S

(b) 𝑄 =

𝐷. 𝑑𝑠 = ₀ ^2𝜋 𝑎 ₀ ^𝜋 ² 𝑠𝑖𝑛𝜃 𝑑𝜃𝑑𝜙𝑎 _𝑟 × 0.3𝑟 ² 𝑎 _𝑟 = 24.3 −𝑐𝑜𝑠𝜃| ₀ ^2𝜋 ₀ ^𝜋 𝑑𝜙 = 48.6 ₀ ^2𝜋 𝑑𝜙 = 305 . 208 𝑛𝐶

(c) On same steps: 𝑄 = 76.8 ₀ ^2𝜋 −𝑐𝑜𝑠𝜃| ₀ ^𝜋 𝑑𝜙 = 964.608 μC

D3.4

(a) 𝑄 = 𝑄 ₁ + 𝑄 ₂ = 0.243 𝜇𝐶

(b) 𝑄 = 𝑙𝑒𝑛𝑔𝑡𝑕 × 𝜌 _𝐿 = 31.4 𝜇𝐶

(c) Area = 𝑙𝑒𝑛𝑔𝑡𝑕 𝑜𝑓 𝑙𝑖𝑛𝑒 (𝑕𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒) × 𝑤𝑖𝑑𝑡𝑕 𝑎𝑐𝑟𝑜𝑠𝑠 𝑧 = 10.53 × 10 So, 𝑄 = 𝑎𝑟𝑒𝑎 × 𝜌 _𝐴 = 10.53 𝜇𝐶

D3.5

(a) 𝐷 = ^0.25

4𝜋(0.005)

²

= 795.77 𝜇𝐶

(b) 𝑄 ₂ = 4𝜋 0.01 ² × 𝜌 _𝑠2 = 2.51 𝜇𝐶, 𝑠𝑜, 𝐷 = ^0.25+2.51

4𝜋(0.015)

²

= 977 𝜇𝐶

(c) 𝑄 ₃ = 4𝜋 0.018 ² × 𝜌 _𝑠3 = −2.44 𝜇𝐶, 𝑠𝑜, 𝐷 = 0.25+2.51−2.44

4𝜋 (0.025)

²

= 40.74 𝜇𝐶

(d) 𝐷 = 0.25+2.51−2.44+4𝜋 0.03

²

×𝜌

_𝑠

4𝜋(0.035)

²

= 0 , so 𝜌 _𝑠4 = -28.29 𝜇𝐶

D3.6

(a) 𝜓 = 𝐷. 𝑑𝑠 = 16𝑥 ₁ ³ ₀ ² ² 𝑦𝑧 ³ 𝑑𝑥𝑑𝑦 = 16𝑥 ₁ ³ ₀ ² ² 𝑦. 8𝑑𝑥𝑑𝑦 = 1365 𝑝𝐶

10 10

10 3

𝑦 = 3𝑥

𝜌 _𝑠2

𝜌 _𝑠3

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E E 0 8 . S O L U T I O N S

(b) 𝐸 = ^𝐷

𝜀

_𝑜

𝑎𝑛𝑑 𝑒𝑣𝑎𝑙𝑢𝑎𝑡𝑒 𝑖𝑡 𝑎𝑡 𝑃

(c) 𝐸𝑣𝑎𝑙𝑢𝑎𝑡𝑒 𝑡𝑕𝑒 𝑓𝑜𝑟𝑚𝑢𝑙𝑎 8 𝑎𝑡 𝑃 𝑎𝑛𝑑 ∆𝑉 = 10 ⁻¹² 𝑚 ³

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 𝜌 _𝑣 , 𝑤𝑖𝑡𝑕 𝑝𝑟𝑜𝑝𝑒𝑟 𝑡𝑟𝑖𝑔𝑛𝑜𝑚𝑒𝑡𝑟𝑖𝑐 𝑚𝑎𝑛𝑖𝑝𝑢𝑙𝑎𝑡𝑖𝑜𝑛

D3.8 R.H.S:

∇. D = 12𝑠𝑖𝑛 𝜑

2 − 0.75𝑠𝑖𝑛 𝜑

2 = 11.25 𝑠𝑖𝑛 𝜑 2

11.25 𝑠𝑖𝑛 𝜑

2 ^{𝜌𝑑𝑧𝑑𝜙𝑑𝜌 =} 11.25 × 20 = 225

5 0 𝜋

0 2

0 L.H.S:

𝐷. 𝑑𝑠 = (𝑫) _𝜌=2 𝜌𝑑𝑧𝑑𝜑𝑎 _𝜌

5 0 𝜋

0 + (𝑫) _𝜑 ₌₀ 𝑑𝑧𝑑𝜌𝑎 _𝜑

5 0 2

0 + (𝑫) _𝜑 _=𝜋 𝑑𝑧𝑑𝜌(−𝑎 _𝜑 )

5 0 2

0 Now,

(𝑫) _𝜌=2 = 12𝑠𝑖𝑛 𝜑

2 ^𝑎 ^𝜌

(7)

E E 0 8 . S O L U T I O N S

𝑫 _𝜑 _=𝜋 𝑑𝑧𝑑𝜌𝑎 _𝜑

5 0 2

0 = 0

𝑠𝑜, 𝐷. 𝑑𝑠 = 24 𝑠𝑖𝑛 𝜑

2 𝑑𝑧𝑑𝜑 + (𝑫) _𝜑 ₌₀ 𝑑𝑧𝑑𝜌𝑎 _𝜑

5 0 2

0 5

0 𝜋

0 = 24 −2𝑐𝑜𝑠 𝜑

2 ^| ⁰ ^𝜋 ^{× 𝑧 |} ⁰ ⁵ ^{− 1.5}

𝜌 ²

2 | ₀ ² × 𝑧 | ₀ ⁵

= −48 0 − 1 5 − 15

= 225

(8)

Solved by Aqeel Anwar (aqeelanwar.co.cc) Digitized by Zaeem (ee08.net.tc)

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

E E 0 8 . S O L U T I O N S

DRILL PROBLEMS 5

D5.1

(a) At P:

ܬ= 10ሺ 3ሻ ² ሺ 2ሻ ܽ _ߩ െ4ሺ 3ሻ ܿ ݋ݏ ² ሺ 30ሻ ܽ _߮ = 180ܽ _ߩ െ9ܽ _߮

(b) Using formula (2):

ܫ= න න 10ߩ ³ ݖ ܽ _ߩ .݀ݖ ݀߶

2.8 2 2ߨ

0 ܽ _ߩ = 27 න න 10ݖ ݀ݖ ݀߶

2.8 2 2ߨ

0 = 27 ቆ ݖ ²

2 ቇ| ₂ ^2.8 ሺ ߶ሻ | ₀ ^2ߨ

= 325.72 ݉ܣ ࢕࢘ 3.25 ܣ

D5.2

(a) Using formula (2):

ܫ= െන න 10 ⁶ ݖ ^1.5 ܽ _ݖ .ߩ݀ߩ݀߶

20ߤ

0 2ߨ

0 ܽ _ݖ = െන න 10 ⁶ ሺ 0.1ሻ ^1.5 ߩ݀ߩ݀߶

20ߤ

0 2ߨ

0 = െ10 ⁶ ሺ 0.1ሻ ^1.5 ቆ ߩ ²

2 ቇ| ₀ ^20ߤ ሺ ߶ሻ | ₀ ^2ߨ = െ39.7ߤܣ

(b) Using formula (3):

ߩ _ݒ = ܬ _ݖ

ݒ _ݖ = െ10 ⁶ ሺ 0.1ሻ ^1.5

2 × 10 ⁶ = െ15.81 ݉ܥ

݉ ³

(c) Same formula:

ݒ _ݖ = ܬ _ݖ

ߩ _ݒ = െ10 ⁶ ሺ 0.15ሻ ^1.5

െ2000 = 29.04 ݉

ݏ

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(22)

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

E E 0 8 . S O L U T I O N S

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 5ݕሻܽ

_ݔ

െ100 sinhሺ 5ݔሻ5 cosሺ 5ݕሻܽ

_ݕ

At P: E = െ474.43ܽ

_ݔ

െ140.77ܽ

_ݕ

ܸ

݉

(c) ȁ E ȁ= ඥ 474.43

²

+ 140.77

²

= 494.87 _݉ ^ܸ

(d) ɏ

_s

= ܦ

_ܰ

= ȁ ࡰ

_ࡼ

ȁ , so as ࡰ

_ࡼ

= ߝ

_݋

E= െ4.2ܽ

_ݔ

െ1.246ܽ

_ݕ

_݉ ^݊ܥ

₂

, so ɏ

_s

= ȁ ࡰ

_ࡼ

ȁ = 4.38 _݉ ^݊ܥ

₂

D5.6

(a) For original line charge, with ߩ

_ܮ

= 40

^݊ܥ

݉

ܸ= െනܧ.݈݀ ,

ܽݏ ܧ= ߩ

_ܮ

2ߨߝ

_݋

ߩ ܽ

_ߩ

, ݏ ݋ ܸ= െ ߩ

_ܮ

2ߨߝ

_݋

න ܽ

_ߩ

ߩ ݈݀

(7,െ1,5)

4

ܽ

_ߩ

ߩ = ሺ ݔെ6ሻ ܽ

_ݔ

+ (ݕെ3)ܽ

_ݕ

(ݔെ6)

²

+ (ݕെ3)

²

ܽݏ , ݈݀= ݀ݔ ܽ݊݀ ݕ= െ1,ݏ ݋:

ܸ= െ720 න ሺ ݔെ6ሻ

(ݔെ6)

²

+ 16 ݀ݔ= െ360 ݈ ݊ȁ (ݔെ6)

²

+ 16ȁ

₄⁷

7

4

= 58.50 ܸ

For mirror line charge, with ߩ

_ܮ

= െ40

^݊ܥ

݉

ܸ= െනܧ.݈݀ ,

ܽݏ ܧ= ߩ

_ܮ

2ߨߝ

_݋

ߩ ܽ

_ߩ

, ݏ ݋ ܸ= െ ߩ

_ܮ

2ߨߝ

_݋

න ܽ

_ߩ

ߩ ݈݀

(7,െ1,5)

4

(23)

(24)

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,

ρ_v

=

−∇²V,

so we first calculate

∇²V

:

∇V

=

−8_(x^yzx₂₊₁₎2

+

_x₂^4z₊₁

+

_x₂^4y₊₁

⇒ ∇²V

= 32

_(x^yzx₂₊₁₎²3 −

8

_(x₂^yz₊₁₎2

⇒ ∇²V|P(1,2,3)

= 12;

so,

ρv

=

−∇²V

=

−o

(12) =

−106.25_m^pC3

(b)V|_P(3,^π

3,2)

=

−22.5V

As,

ρv

=

−∇²V,

so we first calculate

∇²V

:

∇²V

= 20 cos (2

φ)−

20 cos (2

φ)

⇒ ∇²V|P(3,^π₃,2)

= 0;

so,

ρv

=

−∇²V

=

−o

(0) = 0

^pC_m3

(c) V|_P_(0.5,45o,60^o)

= 4

V

As,

ρv

=

−∇²V,

so we first calculate

∇²V

:

∇²V

= 4

^cos(φ)_r₄ −

2

_r₄_(sin(θ))^cos(φ) 2

⇒ ∇²V|P(0.5,45^o,60^o)

= 0;

so,

ρv

=

−∇²V

=

−o

(0) = 0

^pC_m3

D7.2

Apply the formulae & concepts to find the answers!

(25)

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 lnρ

+ 101.28

Now,

E

=

−∇V

=

¹_ρ

(73.9)

aρ

=

^√¹

10

(73.9)

aρ

= 23.36

aφ

(

∵ρ

=

√

3

²

+ 1

²

)

so,

|E|

= 23.36

_m^V

(b)

The solution to Laplace’s equation

_ρ¹2

∂²V

∂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

=

¹_ρ

(85.9)

aφ

=

^√¹

10

(85.9)

aφ

= 27.16

aφ

(

∵ρ

=

√

3

²

+ 1

²

) so,

|E|

= 27.16

_m^V

D7.4 & D7.5

Not included in the course!

(26)

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

(27)

4

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

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(28)

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.

∆H

2

=

^I¹^∆L_4πR¹^×a2^R12 12

Here,

aR12

=

^(4−0)a^x^√^+(2−0)a₄₂₊₂₂₊₂^y^+(0−2)a₂ ^z

= 0.816a

x

+ 0.408a

y−

0.408a

z

R²₁₂

= 4

²

+ 2

²

+ 2

²

= 24 so, ∆H

2

=

^I¹^∆L_4πR¹^×a2^R12

12

=

^2πa^z^×(0.816a^x_301.59^+0.408a^y^−0.408a^z⁾µ

=

^5.12a_301.59^y^−2.56a^xµ

=

−8.5ax

+ 17.0a

ynA m

(b)

As, ∆H

2

=

^I¹^∆L_4πR¹^×a2 ^R12 12

Here,

aR12

=

^(4−0)a^x^√^+(2−2)a^y^+(3−0)a^z

4²+0²+3²

= 0.8a

_x

+ 0.6a

_z R²₁₂

= 4

²

+ 0

²

+ 3

²

= 25

so, ∆H

₂

=

^I¹^∆L_4πR¹^×a₂^R12

12

=

^2πa^z^×(0.8a_100π^x^+0.6a^z⁾µ

=

^5.02a_100π^yµ

= 16a

_y^nA_m

(c)

As, ∆H

2

=

^I¹^∆L_4πR¹^×a2 ^R12 12

Here,

aR12

=

^(−3−1)a^x^√^+(−1−2)a^y^+(2−3)a^z

4²+3²+1²

=

−0.78ax−

0.58a

z−

0.19a

z

R²₁₂

= 4

²

+ 3

²

+ 1

²

= 26 so, ∆H

₂

=

^I¹^∆L_4πR¹^×a2^R12

12

=

^2π(−a^x^+a^y^+2a^z^)×(−0.78a_104π ^x^−0.58a^z^−0.19a^z⁾µ

=

^(1.96â^x^−3.53_104πâ^y^+2.74â^z^)πµ

⇒

∆H

2

= 18.85

ax−

33.94

ay

+ 26.40

aznA m

(29)

2 D8.2

Using,

H₂

=

_2πρ^I a_φ (a)

For

P_A

(

√

20, 0, 4), we have

ρ

=

√

20 + 0 = 4.47, so:

H₂

=

₂_π(4¹⁵_.47)a_φ

=

0.533a_φ

= (0.533

× −

sin

φ)a_x

+ (0

.533×

cos

φ)a_y

, where

φ

= tan

⁻¹ ^y_x

= tan

⁻¹

√0 20

= 0

^o

, so

H2

= 0.533a

y A

m

(b)

For

PB

(2,

−4,

4), we have

ρ

=

√

2

²

+ 4

²

= 4.47, so:

H2

=

₂_π(4¹⁵_.47)aφ

=

0.533aφ

= (0.533

× −

sin

φ)ax

+ (0

.533×

cos

φ)ay

, where

φ

= tan

⁻¹ ^y_x

= tan

⁻¹ ⁻⁴₂

=

−63.43^o

, so

H2

= (0

.533×0.89)ax

+ (0

.533×0.44)ay

= 0.474a

x

+ 0.238a

y

D8.3

(a)

For infinitely long filament;

H

=

_2πρ^I aφ

Here,

ρ

=

p

(0.1)

²

+ (0.1)

²

=

√2 10

H

=

_2πρ^I aφ

=

^(2.5)

2π(

√ 2

10)aφ

= 2.8134a

_φ

= (2

.8134× −

sin

φ)a_x

+ (2

.8134×

cos

φ)a_y

Now,

φ

= 270

^o−θ

= 270

^o−

tan

⁻¹ ^0.1_0.1

= 270

^o−

45

^o

= 225

^o

,

so,

H

= (2

.8134×0.707)ax

+ (2

.8134× −

0.707)a

y

= 1.989a

x−

1.989a

y A m

(30)

3

(b)

For

ρ < a,

H

=

_2πa^Iρ₂a_φ

=

^(2.5)(0.2)_2π(0.3)₂a_φ

= 0.884a

_φ

= (0.884× − sin

φ)ax

+ (0

.884×

cos

φ)ay

Now,

φ

= tan

⁻¹ _x^y

= tan

⁻¹ ^0.2₀

= 90

^o

,

so,

H

=

−0.884ax A m

(c)

Now,

H1

=

¹₂K1×aN

=

¹₂

(2.7)a

x×ay

= 1.35a

z

H2

=

¹₂K2×aN

=

¹₂

(−1.4)a

x×ay

=

−0.7az

H3

=

¹₂K3×aN

=

¹₂

(−1.3)a

x× −ay

= 0.65a

z

so,

H

=

H₁

+

H₂

+

H₃

= 1.300a

_z_m^A

(31)

4

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(32)

CHAPTER 8 DRILLS

CHAPTER 8 DRILLS

(D8.4

onwards)

Solved by

Zaeem A. Varaich www.ee08.net.tc

D8.4

(a) For path 1:

´

(3za

x−

2x

³

a

z

).(dxa

x

+ dya

y

+dza

z

) =

´4

2

(3zdx) = 3(4)(4

−

2) = 24 A For path 2:

´

(3za

x−

2x

³

a

z

).(dxa

x

+ dya

y

+dza

z

) =

´1

4

(−2x

³

dz) =

−2(4³

)(1

−

4) = 384 A For path 3:

´

(3za

x−

2x

³

a

z

).(dxa

x

+ dya

y

+dza

z

) =

´2

4

(3zdx) = 3(1)(2

−

4) =

−6

A For path 4:

´

(3za

x−

2x

³

a

z

).(dxa

x

+ dya

y

+dza

z

) =

´4

1

(−2x

³

dz) =

−2(2³

)(4

−

1) =

−48

A So,

¸

H.dL = 24 + 384

−

6

−

48 = 354 A

(b)

4SN

= 3

×

2 = 6 m

²

, so (∇ × H)

_y

=

¸H.dL

4SN

=

³⁵⁴₆

= 59

_m^A₂

(c) (∇ × H) =

a

_x

a

_y

a

_z

∂

∂x

∂

∂y

∂

∂z

H

x

H

y

H

z

=

a

_x

a

_y

a

_z

∂

∂x

∂

∂y

∂

∂z

3z 0

−2x³

= (0

−

0)a

x−(−6x²−

3)a

y

+(0

−

0)a

z

= (6x

²

+ 3)a

y

At the center, x = 3, z = 2.5, so (∇ × H)

y

= [6(3)

²

+ 3]=57

_m^A2

1

ee08.net.tc

(33)

CHAPTER 8 DRILLS

D8.5

(a) J =

∇ ×

H =

a

x

a

y

a

z

∂

∂x

∂

∂y

∂

∂z

H

_x

H

_y

H

_z

=

a

x

a

y

a

z

∂

∂x

∂

∂y

∂

∂z

0 x

²

z

−y²

x

= (−2yx

−

x

²

)a

_x−(−y²−

0)a

_y

+(2xz

−

0)a

_z

At P: J =

−16ax

+9a

y

+16a

z A m

(b) J =

∇ ×

H =

1 ρ

∂H_z

∂φ −^∂H_∂z^φ

a

_ρ

+

_∂H

ρ

∂z −^∂H_∂ρ^z

a

_φ

+

1 ρ

∂(ρHφ)

∂ρ −¹_ρ^∂H_∂φ^ρ

a

_z

=

1 ρ

.0

−

0 a

_ρ

+ (0

−

0) a

_φ

+

1

ρ

.(2 cos 0.2φ)

−¹_ρ²_ρ

(−0.2) sin 0.2φ a

_z

At P: J =0.055a

_z_m^A₂

(c) Using equation (26):

=

_r_sin¹ _θ

(0

−

0) a

r

+

¹_r

(0

−

0) a

θ

+

¹_r _∂r^∂ _sinθ^r −

0 a

φ

=

_r_sinθ¹

a

φ

At P: J =a

_φ _m^A₂

D8.6

(a)

¸

H.dL : For path 1:

´

(6xya

x−

3y

²

a

y

).(dxa

x

+ dya

y

+dza

z

) =

´5

2

(6xydx) = 6(−1)

⁽⁵²⁻²₂ ²⁾

=

−63

A For path 2:

´

(6xya

x−

3y

²

a

y

).(dxa

x

+ dya

y

+dza

z

) =

´1

−1

(−3y

²

dy) =

−3⁽¹³⁺¹₃ ³⁾

=

−2

A For path 3:

´

(6xya

x−

3y

²

a

y

).(dxa

x

+ dya

y

+dza

z

) =

´2

5

(6xydx) = 6(1)

⁽²²⁻⁵₂ ²⁾

=

−63

A For path 4:

´

(6xya

x−

3y

²

a

y

).(dxa

x

+ dya

y

+dza

z

) =

´−1

1

(−3y

²

dy) =

−3⁽⁻¹³₃⁻¹³⁾

= 2 A So,

¸

H.dL =

−63−

2

−

63 + 2 =

−126

A (b) Now,

´

S

(∇ × H).dS :

(∇ × H) =

a

x

a

y

a

z

∂

∂x

∂

∂y

∂

∂z

H

_x

H

_y

H

_z

=

a

x

a

y

a

z

∂

∂x

∂

∂y

∂

∂z

6xy

−3y²

0 = (0

−

0)a

x−

(0

−

0)a

y

+ (0

−

6x)a

z

=

−6xaz

(∇ × H).dS =(−6xa

_z

)(dydza

_x

+ dxdza

_y

+ dxdya

_z

) =

−6x dxdy,

´

(∇ × H).dS =

−6´ ´

x dxdy =

−6

x² 2

5 2

y|

¹₋₁

=

−6⁽²⁵⁻⁴⁾₂

(1 + 1) =

−126

A

D8.7

(a) H

φ

=

_2πa^Iρ₂

=

^(20)(0.5m)_2π(1m)₂

= 1592

_m^A

(b) B

_φ

= µ

_o

H

_φ

= 4π

×

10

^{−7 (20)(0}_2π(1m)^.8m)₂

= 3.2 mT

2

ee08.net.tc

(34)

CHAPTER 8 DRILLS

(c) φ =

´

S

B.dS =

´d 0

´a 0

Iρ

2πa²

dρdz =

_2πa^Iµ^O2

d

^ρ₂²

a 0

, so

φ

d

= 4π

×

10

^{−7 (20)}_4π

= 2 µ

^{W b}_m

(d) φ =

´

S

B.dS =

´d 0

´0.5m 0

Iρ

2πa²

dρdz =

_2π(1m)^Iµ^o 2

d

^ρ₂²

0.5m 0

, so φ = 4π

×

10

^{−7 (20)}_4π(1m)^d(0.5m)2 ²

= 0.5d µ

^{W b}_m

(e) φ =

´

S

B.dS =

´d 0

´∞ 0

Iρ

2πa²

dρdz =

_2π(1m)^Iµ^o^d2

ρ² 2

∞ 0

, so φ = 4π

×

10

^{−7 (20)}_4π(1m)^d(∞)2²

=

∞^{W b}_m

D8.8

(a) H =

_2πρ^I

a

φ

, so first we find I:

I =

´

KdN =

´2π

0

(2.4)(ρdφ) = 18.09 A H =

^18.09_2πρ

a

φ

=

^2.88_ρ

a

φ

(b) H =

−∇Vm

, so

∇V_m

=

−_(1.5)^2.88

a

_φ

=

−1.92a_φ

1 ρ

∂Vm

∂φ

=

−1.92⇒

V

m

=

−2.88´0.6π

0

dφ =

−5.43V

(c) Proceeding similar as above:

⇒

V

m

=

−2.88´0.6π

2π

dφ =

−2.88(0.6−

2)π = 12.66V (d) Similarly,

⇒

V

m

=

−2.88´0.6π

π

dφ =

−2.88(0.6−

1)π = 3.62V (e) Similarly,

⇒

V

_m

=

h

−2.88´π

−0.6π

dφ

i

+ 5 =

−14.47 + 5 =−9.47V

D8.9

We know that, B = µ

_o

H

For solid conductor along z-axis:

H =

_2πa^Iρ₂

a

φ

, so B = µ

o Iρ

2πa²

a

φ

Also, B =

∇ ×

A

Comparing both sides’ a

_φ

components:

µ

o Iρ

2πa²

=

−^∂A_∂ρ^z⇒

A

z

=

−´

µ

o Iρ

2πa²

dρ +

C

From given conditions, at ρ = a for A

z

:

3

ee08.net.tc

(35)

CHAPTER 8 DRILLS

C =

´

µ

o Iρ

2πa²

dρ + µ

oIln5

2πa²

= µ

o Iρ² 4πa²

a

0

+ µ

oIln5

2πa²

= 4.218

×

10

⁻⁷

Now,

A

z

=

−µo Iρ²

4πa²

+ (4.39

×

10

⁻⁷

) =

h

−10⁻⁷^ρ_a²2

+ (4.218

×

10

⁻⁷

)

i

I

(a) At ρ = 0 A = 0.4218I a

z

µ

^Wb_m

(b) At ρ = 0.25a A = 0.415I a

_z

µ

^Wb_m

(c) At ρ = 0.75a A = 0.365I a

_z

µ

^Wb_m

(d) At ρ = a A = 0.3217I a

_z

µ

^Wb_m

D8.10

Using Eq. (66) with b = 4cm:

A

_z

=

^µ_2π^o^I

ln

^b_ρ

For red conductor:

A

_zr

= 2.4 ln

^4cm_ρ

1

µ

^{W b}_m

For black conductor:

A

_zb

=

−2.4 ln^4cm_ρ

2

µ

^{W b}_m

Also,

A = A

_zr

+ A

_zb

(a) ρ

1

= 4cm = ρ

2

, so A

zr

=

−Azb

A = 0

^{W b}_m

(b) ρ

1

= 4cm and ρ

2

= 12cm , so A

zb

= 2.63µ

^{W b}_m

and A

zr

= 0µ

^{W b}_m

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CHAPTER 8 DRILLS

A = 2.63µ

^{W b}_m

(c) ρ

1

= 4cm and ρ

2

=

√

8

²

+ 4

²

= 8.94cm , so A

zr

= 0µ

^{W b}_m

and A

zb

= 1.93µ

^{W b}_m

A = 1.93µ

^{W b}_m

(d) ρ

1

= 2cm and ρ

2

=

√

8

²

+ 2

²

= 8.24cm , so A

zr

= 1.66µ

^{W b}_m

and A

zb

= 1.734µ

^{W b}_m

A = 3.39µ

^{W b}_m

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CHAPTER 8 DRILLS

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