PRACTICE EXERCISE 7.7 A current distribution gives rise to the vector magnetic potential A = x'ya¸ + yxa, - 4xyza, Wb/m. Calculate the following: (a) Bat (−1, 2, 5) (b) The flux through the surface defined by z = 1, 0 ≤ x ≤ 1, -1 ≤ y ≤ 4 Answer: (a) 20a, + 40a, + 3a₂ Wb/m², (b) 20 Wb.

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Electromagnet, picture 2 is an example of doing Practice Exercise 7.7, thank you in advance.
PRACTICE EXERCISE 7.7
A current distribution gives rise to the vector magnetic potential A = x'ya¸ +
yxa, - 4xyza, Wb/m. Calculate the following:
(a) B at (−1, 2, 5)
(b) The flux through the surface defined by z = 1, 0 ≤ x ≤ 1, -1 ≤ y ≤ 4
Answer: (a) 20a, + 40a, + 3a Wb/m², (b) 20 Wb.
2
Transcribed Image Text:PRACTICE EXERCISE 7.7 A current distribution gives rise to the vector magnetic potential A = x'ya¸ + yxa, - 4xyza, Wb/m. Calculate the following: (a) B at (−1, 2, 5) (b) The flux through the surface defined by z = 1, 0 ≤ x ≤ 1, -1 ≤ y ≤ 4 Answer: (a) 20a, + 40a, + 3a Wb/m², (b) 20 Wb. 2
EXAMPLE 7.7
Given the magnetic vector potential A = -p²/4 a₂ Wb/m, calculate the total magnetic flux
crossing the surface = 7/2, 1 ≤ p ≤ 2 m, 0≤z ≤ 5 m.
Solution:
We can solve this problem in two different ways: using eq. (7.32) or eq. (7.51).
Method 1:
Hence,
Method 2:
We use
B = VX A =
JA,
др
Y = 3.75 Wb
Y =
JL
ab
1
15
* - [* - - - - _ ___ ²² - | ||(0) - "
= B.ds =
p dp dz
=
4
S
z=0p=1
a ds =
A dl = ₁ + ₂ +
dp dz a
Y₁ = 0 = V₂
3 + V4
where L is the path bounding surface S; ¥₁, Y₂, ¥3, and Y4 are, respectively, the evalua-
tions of ₁A dl along the segments of L labeled 1 to 4 in Figure 7.20. Since A has only a
z-component,
.
Transcribed Image Text:EXAMPLE 7.7 Given the magnetic vector potential A = -p²/4 a₂ Wb/m, calculate the total magnetic flux crossing the surface = 7/2, 1 ≤ p ≤ 2 m, 0≤z ≤ 5 m. Solution: We can solve this problem in two different ways: using eq. (7.32) or eq. (7.51). Method 1: Hence, Method 2: We use B = VX A = JA, др Y = 3.75 Wb Y = JL ab 1 15 * - [* - - - - _ ___ ²² - | ||(0) - " = B.ds = p dp dz = 4 S z=0p=1 a ds = A dl = ₁ + ₂ + dp dz a Y₁ = 0 = V₂ 3 + V4 where L is the path bounding surface S; ¥₁, Y₂, ¥3, and Y4 are, respectively, the evalua- tions of ₁A dl along the segments of L labeled 1 to 4 in Figure 7.20. Since A has only a z-component, .
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