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
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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
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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
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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
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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
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=
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
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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
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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
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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:
ee08.uet@gmail.com
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:
ee08.uet@gmail.com
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
(c) (∇ × H) =
=
354
6
= 59 mA2
ax
ay
az
∂
∂x
∂
∂y
∂
∂z
Hx
Hy
Hz
=
ax
ay
az
∂
∂x
∂
∂y
∂
∂z
3z
0
−2x
= (0 − 0)ax −(−6x2 − 3)ay +(0 − 0)az = (6x2 + 3)ay
3
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
ay
az
ax
ay
az
∂
∂x
∂
∂y
∂
∂z
∂
∂x
Hx
Hy
Hz
∂
∂y
2
∂
∂z
2
(a) J = ∇ × H =
At P: J = −16ax +9ay +16az
=
x z
0
= (−2yx − x2 )ax −(−y 2 − 0)ay +(2xz − 0)az
−y x
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
1 ∂Hρ
ρ ∂φ
az
A
m2
(c) Using equation (26):
1
1
= r sin
θ (0 − 0) ar + r (0 − 0) aθ +
1
r
∂
r
∂r sinθ
− 0 aφ =
1
r sin θ aφ
A
m2
At P: J =aφ
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
(b) Now,
´
S
(∇ × H) =
(∇ × H).dS :
ax
ay
az
∂
∂x
∂
∂y
∂
∂z
Hx
Hy
Hz
=
ax
ay
∂
∂x
∂
∂y
6xy
−3y
az
2
∂
∂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)(0.8m)
2π(1m)2 = 3.2 mT
2
ee08.net.tc
CHAPTER 8 DRILLS
(c) φ =
φ
d
´
B.dS =
S
´d´a
Iρ
dρdz
0 2πa2
0
=
IµO
ρ2
2πa2 d 2
a
, so
0
Wb
= 4π × 10−7 (20)
4π = 2 µ m
(d) φ =
´
B.dS =
S
´ d ´ 0.5m
0
0
Iρ
2πa2 dρdz
=
Iµo
ρ2
2π(1m)2 d 2
0.5m
, so
0
2
b
φ = 4π × 10−7 (20)d(0.5m)
= 0.5d µ W
4π(1m)2
m
(e) φ =
´
S
B.dS =
´d´∞
0
0
Iρ
2πa2 dρdz
=
Iµo d
ρ2
2π(1m)2 2
∞
, so
0
2
Wb
φ = 4π × 10−7 (20)d(∞)
4π(1m)2 = ∞ m
D8.8
I
(a) H = 2πρ
aφ , so first we find I:
´
´ 2π
I = KdN = 0 (2.4)(ρdφ) = 18.09 A
H=
18.09
2πρ aφ
=
2.88
ρ aφ
(b) H = −∇Vm , so
2.88
∇Vm = − (1.5)
aφ = −1.92aφ
1 ∂Vm
ρ ∂φ
= −1.92⇒ Vm = −2.88
´ 0.6π
0
dφ = −5.43V
(c) Proceeding similar as above:
´ 0.6π
⇒ Vm = −2.88 2π dφ = −2.88(0.6 − 2)π = 12.66V
(d) Similarly,
´ 0.6π
⇒ Vm = −2.88 π dφ = −2.88(0.6 − 1)π = 3.62V
(e) Similarly,
h
i
´π
⇒ Vm = −2.88 −0.6π dφ + 5 = −14.47 + 5 = −9.47V
D8.9
We know that,
B = µo H
For solid conductor along z-axis:
H=
Iρ
2πa2 aφ ,
so
Iρ
B = µo 2πa
2 aφ
Also,
B=∇×A
Comparing both sides’ aφ components:
´
Iρ
Iρ
∂Az
µo 2πa
µo 2πa
2 = − ∂ρ ⇒ Az = −
2 dρ + C
From given conditions, at ρ = a for Az :
3
ee08.net.tc
CHAPTER 8 DRILLS
C=
´
2
Iρ
Iρ
Iln 5
µo 2πa
2 dρ + µo 2πa2 = µo 4πa2
a
0
Iln 5
−7
+ µo 2πa
2 = 4.218 × 10
Now,
h
i
2
Iρ2
−7
) = −10−7 ρa2 + (4.218 × 10−7 ) I
Az = −µo 4πa
2 + (4.39 × 10
(a) At ρ = 0
A = 0.4218I az µ Wb
m
(b) At ρ = 0.25a
A = 0.415I az µ Wb
m
(c) At ρ = 0.75a
A = 0.365I az µ Wb
m
(d) At ρ = a
A = 0.3217I az µ Wb
m
D8.10
Using Eq. (66) with b = 4cm:
Az =
µo I
2π
ln ρb
For red conductor:
Wb
Azr = 2.4 ln 4cm
ρ1 µ m
For black conductor:
Wb
Azb = −2.4 ln 4cm
ρ2 µ m
Also,
A = Azr + Azb
(a) ρ1 = 4cm = ρ2 , so Azr = −Azb
b
A = 0W
m
b
Wb
(b) ρ1 = 4cm and ρ2 = 12cm , so Azb = 2.63µ W
m and Azr = 0µ m
4
ee08.net.tc
CHAPTER 8 DRILLS
b
A = 2.63µ W
m
(c) ρ1 = 4cm and ρ2 =
√
b
Wb
82 + 42 = 8.94cm , so Azr = 0µ W
m and Azb = 1.93µ m
b
A = 1.93µ W
m
(d) ρ1 = 2cm and ρ2 =
√
b
Wb
82 + 22 = 8.24cm , so Azr = 1.66µ W
m and Azb = 1.734µ m
b
A = 3.39µ W
m
5
ee08.net.tc
CHAPTER 8 DRILLS
Please report to the following e-mail, if you find any mistake:
ee08.uet@gmail.com
6
ee08.net.tc
CHAPTER 9 DRILLS
Solved by Zaeem A. Varaich
www.ee08.net.tc
D9.1
(a) F = qv × B = qvav × B = 9 × 10−5 (3.3ax − 4.5ay + 4.65az )
so, F = 654µN
(b) F = qE = 18n × 103 (−3ax + 4ay + 6az ) = 140.58µN
(c)F = q(E + v × B)
= (2.97 × 10−4 ax − 4.05 × 10−4 ay + 4.185 × 10−4 az ) + (−5.4 × 10−5 ax + 7.2 × 10−5 ay + 1.08 × 10−4 az )
= 2.43 × 10−4 ax − 3.3 × 10−4 ay + 5.265 × 10−4 az
so, F = 664µN
D9.2
(a) aAB = ax , so
¸
¸2
´2
F = −I B×dL = −(12) 1 (−2ax +3ay +4az )×ax dx.10−3 = −12×10−3 1 −3az +4ay dx = −48ay + 36az mN
√x +5ax , so
(b) aAB = 2ax +4a
3 5
¸
F = −I B × dL
i
h´
´5
´6
3
= −(12×10−3 ) 1 (−2ax + 3ay + 4az ) × ax dx + 1 (−2ax + 3ay + 4az ) × ay dy + 1 (−2ax + 3ay + 4az ) × az dz
= −(12 × 10−3 )
h´
3
(−3az
1
+ 4ay )dx +
´5
1
(−2az − 4ax )dy +
´6
1
i
(2ay + 3ax )dz
= −(12 × 10−3 ) [2(−3az + 4ay ) + 4(−2az − 4ax ) + 5(2ay + 3ax )]
= 12ax − 216ay + 168az mN
1
ee08.net.tc
D9.3
(a) V = E × l = 800 × 1.3c = 10.4 V
(b) vd = µe E = 0.13 × 800 = 104 m
s
(c) Ft = qvB = (1)(104)(0.07) = 7.28 N
C
(d) Et =
F
q
V
= 7.28 m
(e) VH = Et × b = 7.28 × 1.1c = 80.1 mV
D9.4
−6
−6
I1 I2
3×10 ×3×10
(a) d(dF2 ) = µo 4πR
2 dL2 × (dL1 × aR12 ) = µo 4π(12 +22 +22 ) (−0.5ax + 0.4ay + 0.3az ) × (ay ×
12
ax +2ay +2az
3
)
= 3.33 × 10−20 (−0.5ax + 0.4ay + 0.3az ) × (−az + 2ax ) = 3.33 × 10−20 (−0.4ax + 0.1ay − 0.8az )
= (−1.33ax + 0.33ay − 2.66az ) × 10−20 N
−6
−6
I1 I2
3×10 ×3×10
(b) d(dF1 ) = µo 4πR
2 dL1 × (dL2 × aR21 ) = µo 4π(12 +22 +22 ) (ay ) × (−0.5ax + 0.4ay + 0.3az ×
12
−ax −2ay −2az
3
)
= 3.33 × 10−20 (ay ) × (−0.2ax − 1.3ay + 1.4az ) = 3.33 × 10−20 (0.2az + 1.4ax )
= (4.67ax + 0.66az ) × 10−20 N
D9.5
(a) FA = IL × B = (0.2)(−4ax − 2ay + 2az ) × (0.2ax − 0.1ay + 0.3az ) = −0.08ax − 0.32ay + 0.16az N
(b) FA = −0.08ax + 0.32ay + 0.16az N
FB = −0.12ax − 0.24ay + 0az N
FC = 0.2ax − 0.08ay − 0.16az N
F = FA + FB + FC = 0 N
(c) FA = −0.08ax + 0.32ay + 0.16az N
RA = ay + az
TA = RA × FA = −0.16ax − 0.08ay + 0.08az N.m
(d) It was proved in part (b) that the sum of forces is zero, so:
TC = TA = −0.16ax − 0.08ay + 0.08az N.m
D9.6
(a) M =
B−µo H
µo
=
µH−µo H
µo
= 1598.87 A/m
(b) M =(No. of atoms / volume)×(Dipole Moment of each atom)= 8.3 × 1028 × 4.5 × 10−27 = 373.5 A/m
(c) M = χm H = χm B
µ = 15 ×
300µ
(1+χm )µo
= 223.8 A/m
2
ee08.net.tc
D9.7
(a) M = χm H = χm B
µ ⇒B=M×
JT = ∇ ×
B
µo
(1+χm )µo
χm
= 2.12 × 10−4 z 2 ax T
= ∇ × 168.75z 2 ax = 2 × 168.75zay
so, JT |z=0.04 = 13.5 A/m2
(b) J = ∇ ×
M
χm
= ∇ × 18.75z 2 ax = 2 × 18.75zay
so, J |z=0.04 = 1.5 A/m2
(c) Jb = ∇ × M = ∇ × 150z 2 ax = 2 × 150zay
so, Jb |z=0.04 = 12 A/m2
D9.8
(a) |HtA | = |−400ay + 500az | = 640.3 A/m
(b) |HN A | = |300ax | = 300 A/m
(c) Ht1 − Ht2 = aN12 × K⇒ Ht2 = Ht1 − (aN12 × K) = (−400ay + 500az ) − (200ay + 150az )
⇒ Ht1 = 694.62 A/m
(d) HN 2 =
µ1
µ2 HN 1
=
5
20
× 300 = 75 A/m
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CHAPTER 10 DRILLS
Solved by Zaeem A. Varaich
www.ee08.net.tc
D10.1
Given:
F
= 10−11 m
−5 H
µ = 10 m
B = 2 × 10−4 cos105 t sin10−3 y ax T
(a) B = µH, so
∇×
B
µ
= ∂E
∂t and
∇×B=
⇒E=
(b) φ =
∂Bx
∂z ay
1
µ
´
´
s
−
∂Bx
∂y az
=0−
∂Bx
∂y az
= 2 × 10−7 cos105 t cos10−3 y
∇ × Bdt = −2 × 104 sin105 t cos10−3 y ax
B.dS =
´
s
Bx dydz =
´ 40 ´ 2
0
0
V
m
−7
Bx dydz = 2 × 10
5
−6
cos(10 × 10
)
2
z|0
−cos(10−3 y)|
10−3
40
0
= 0.318 mW b
(Remember to use radian mode in calculator while solving this problem)
¸
´2
´
× 109 sin105 t cos10−3 y)dz − 20 (−2 × 109 sin105 t cos10−3 y)dz
2
= −2 × 104 sin(105 10−6 ) z|0 cos(10−3 × 0) − cos(10−3 × 40) = −3.19V
(c)
E.dL =
0 (−2
(Remember to use radian mode in calculator while solving this problem)
D10.2
Given:
d = 7 cm
B = 0.3 az T
v = 0.1 ´ay e20y m
s´
so, y = vdt = 0.1 e20y dt = 0.1 e20y t + C
As, y = 0 at t = 0
y = 0.1 e20y t ay m
(a) v(t = 0) = 0.1 e20(0)
m
s
= 0.1 m
s
(b) y(t = 0.1) = 0.1 e20y (0.1)
m
s
⇒
y
e20y
= 0.01 m
s
(Solve it using TABLE mode in calculator!)
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so, y(t = 0.1) ≈ 0.012 m
Alternatively, using MAPLE:
y = −0.05000000000 LambertW (−2.0 t)
y = 0.01295855509 m = 1.29 cm
(c) v(t = 0.1) = 0.1 e20(0.0129)
m
s
= 0.129 m
s
(d) V12 = −Bvd = −0.3 × 0.129 × 0.07 = −2.718 mV
D10.3
(a) When J = 0:
∇ × H = Jd
8
8
x
∇ × H = − ∂H
∂y = −0.15 cos 3.12 × 3 × 10 t − 3.12y × −3.12 = 0.468 cos 3.12 × 3 × 10 t − 3.12y
|Jd | = |∇ × H| = 0.468 mA2
(b) When J = 0:
Jd = ∇ × H = ∇ ×
B
µ
∂Hy
∂x
= −0.8 sin 1.257 × 10−6 × 3 × 108 t − 1.257 × 10−6 x × −1.257 × 10−6
= 1.0056 × 10−6 sin 1.257 × 10−6 × 3 × 108 t − 1.257 × 10−6 x
∇×B =
|Jd | = ∇ ×
B
µ
= 0.80023 mA2
(c) D = r o E
∂D
∂t
= r o ∂E
∂t
√ = −r o 0.9 sin 1.257 × 10−6 × 3 × 108 t − 1.257 × 10−6 z 5 × −1.257 × 10−6 × 3 × 108 M = 0.015025 mA2
Jd =
(d) E =
|Jd | =
J
σ
∂D
∂t
and D = o E, so
pA
= o 377M
= 57.55 m
2
σ
D10.4
(a) ∇.B = 0 ⇒ ∇.µH = 0 ⇒ ∇.H = 0
⇒ ∇.(kxax + 10yay − 25zaz ) = 0
⇒ k + 10 − 25 = 0
⇒ k = 15A/m2
(b) As ρv = 0, so J = ρv v = 0:
∇×H =
∂D
∂t
⇒ ax = (−4k × 10−9 )ax ⇒ k =
1
−4×10−9
2
= −2.5 × 108 V /(m.s)
ee08.net.tc
D10.5
Let u = 0.64ax + 0.6ay − 0.48az
(a) BN 1 = B1 .u = 2T
(b) Bt1 = |B1 × u| = 3.16T
(c) BN 2 = BN 1 = 2T
(d) B2 = BN 2 + Bt1 = 5.16T
D10.6
(a) ρs = DN 1 = |o r E|t=6ns,z=0.3 = 20o r cos(2 × 108 t − 2.58z) = 0.806 nC/m2
´
∂µH
1
∇ × E dt
(b) ∇ × E = − ∂B
∂t = − ∂t ⇒ H = − µ
= 13.68 × 106 ×
− cos(2×108 t−2.58z)
ax
2×108
t=6ns, z=0.3
= −62.3ax mA/m
(c) Ht1 = K × aN ⇒ −62.3ax = K × ay , so K = −62.3az mA/m
(Cyclic rule of cross product)
D10.7
(a) [ρv1 ] = 4cos(108 π(t −
[ρv1 ] = 4cos(108 π(t −
8
[ρv2 ] = −4cos(10 π(t
R
v )),
R
v )) =
−R
v )),
where t = 15ns, R = 450 − 1.5, v = 3 × 108 m/s
4µ
where t = 15ns, R = 450 + 1.5, v = 3 × 108 m/s
[ρv2 ] = −4(−1) = 4µ
[ρv ] = [ρv1 ] + [ρv2 ] = 8µ
Now, considering a unit volume:
[ρv ]
V = 4π
= 159.77V
oR
(b) [ρv1 ] = 4cos(108 π(t −
[ρv1 ] = 4cos(108 π(t −
8
[ρv2 ] = −4cos(10 π(t
R
v )),
R
v )) =
−R
v )),
where t = 15ns, R =
−0.01047µ
where t = 15ns, R =
√
√
4502 + 1.52 = 450.0025, v = 3 × 108 m/s
4502 + 1.52 = 450.0025, v = 3 × 108 m/s
[ρv2 ] = −4(−1) = 0.01047µ
[ρv ] = [ρv1 ] + [ρv2 ] = 0µ
Now, considering a unit volume:
V =
[ρv ]
4πo R
= 0V
(c) [ρv1 ] = 4cos(108 π(t −
[ρv1 ] = 4cos(108 π(t −
[ρv2 ] = −4cos(108 π(t
R
v )),
R
v )) =
−R
v )),
where t = 15ns, R =
3.46µ
where t = 15ns, R =
√
√
316.72 + 318.22 = 449, v = 3 × 108 m/s
319.72 + 318.22 = 451.06, v = 3 × 108 m/s
[ρv2 ] = −4(−1) = 3.58µ
[ρv ] = [ρv1 ] + [ρv2 ] = 7.04µ
Now, considering a unit volume:
[ρv ]
V = 4π
= 140.65V
oR
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