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Let the parallel current sheets be as in Figure 7.14. Also let H = H₂ + H₂ where H, and H, are the contributions due to the current sheets z = 0 and z = 4, respec- tively. We make use of eq. (7.23).

Let the parallel current sheets be as in Figure 7.14. Also let H = H₂ + H₂ where H, and H, are the contributions due to the current sheets z = 0 and z = 4, respec- tively. We make use of eq. (7.23).

Delmar's Standard Textbook Of Electricity
Delmar's Standard Textbook Of Electricity
7th Edition
ISBN:9781337900348
Author:Stephen L. Herman
Publisher:Stephen L. Herman
Chapter25: Surge, Spike, And Lightning Protection
Section: Chapter Questions
Problem 10RQ: In reference to the load, how are capacitors connected to help prevent voltage spikes and surges?...
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hi can you explain the calculations 

Planes z = 0 and z = 4 carry current K = -10a, A/m and K = 10a, A/m, respectively. Determine Hat (a) (1, 1, 1) (b) (0, -3, 10) Solution: Let the parallel current sheets be as in Figure 7.14. Also let H = H₂ + H₂ 4, respec- where H, and H4 are the contributions due to the current sheets z = 0 and z = tively. We make use of eq. (7.23). (a) At (1, 1, 1), which is between the plates (0 < z = 1 < 4), H₂ = 1/2 K X a,, = 1/2 (-10a,) × a₂ = 5a, A/m H₁ = 1/2 K X a, = 1/2 (10a,) × (-a) = 5a, A/m Solved Examples x-1 (b) At (0, -3, 10), which is above the two sheets (z = 10 > 4> 0), H₂ = 1/2 (-10a,) × a₂ = 5a, A/m H₁ = 1/2 (10a,) × a₂ = -5a, A/m Hence, H = 0 A/m
Transcribed Image Text:Planes z = 0 and z = 4 carry current K = -10a, A/m and K = 10a, A/m, respectively. Determine Hat (a) (1, 1, 1) (b) (0, -3, 10) Solution: Let the parallel current sheets be as in Figure 7.14. Also let H = H₂ + H₂ 4, respec- where H, and H4 are the contributions due to the current sheets z = 0 and z = tively. We make use of eq. (7.23). (a) At (1, 1, 1), which is between the plates (0 < z = 1 < 4), H₂ = 1/2 K X a,, = 1/2 (-10a,) × a₂ = 5a, A/m H₁ = 1/2 K X a, = 1/2 (10a,) × (-a) = 5a, A/m Solved Examples x-1 (b) At (0, -3, 10), which is above the two sheets (z = 10 > 4> 0), H₂ = 1/2 (-10a,) × a₂ = 5a, A/m H₁ = 1/2 (10a,) × a₂ = -5a, A/m Hence, H = 0 A/m
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