Problem#1
In a particular region there is a uniform current density of 15 A/m2 in the positive z direction. What is the value ∮B. ds of when that line integral is calculated along a closed path consisting of the three straight-line segments from (x, y, z) coordinates (4d, 0, 0) to (4d, 3d, 0) to (0, 0, 0) to (4d, 0, 0), where d = 20 cm?Answer:
 |
| Fig.1 |
The path of integration is a right triangle with base length 4d and height 3d, so its area is
A = (1/2)(4d)(3d) = 6d².
Since d = 20 cm = 0.20 m, the area is
A = 6 x (0.20 m)² = 0.24 m².
Since the loop is orthogonal to the current density (and only because of that), we are able to get away with saying
I = JA = (15 A/m²)(0.24 m²) = 3.60 A.
So that, applying Ampere’s law, we get
∮B. ds = μ0ienc = (4π x 10─7 Tm/A)(3.60 A) = 4.52 x 10─6 T.m
Problem#2
Figure 2 shows a cross section across a diameter of a long cylindrical conductor of radius a = 2.00 cm carrying uniform current 170 A.What is the magnitude of the current’s magnetic field at radial distance (a) 0, (b) 1.00 cm, (c) 2.00 cm (wire’s surface), and (d) 4.00 cm?
Answer:
we use the equation
B = μ0ir/2πa2 for the B─field inside the wire (r < a) and
B = μ0i/2πr for that outside the wire (r > a)
(a) at r = 0, B = 0
(b) at r = 1.00 cm = 0.0010 m,
B = μ0ir/2πa2 = (4π x 10─7 Tm/A)(170 A)(0.0010 m)/[2π x (0.0020 m)2] = 8.50 x 10─4 T
(c) at r = 2.00 cm = 0.0020 m,
B = μ0ir/2πa2 = (4π x 10─7 Tm/A)(170 A)(0.0020 m)/[2π x (0.0020 m)2] = 1.70 x 10─3 T
(d) at r = 4.00 cm = 0.0040 m,
B = μ0i/2πr = (4π x 10─7 Tm/A)(170 A)/[2π x 0.0040 m] = 8.50 x 10─4 T
Problem#3
Figure 3 shows two closed paths wrapped around two conducting loops carrying currents i1 = 5.0 A and i2 = 3.0 A.What is the value of the integral ∮B. ds for (a) path 1 and (b) path 2?
Answer:
Applying Ampere’s law, we get
(a) For Path 1
∮B. ds = μ0ienc
= (4π x 10─7 Tm/A)(─3 A + 5 A)
∮B. ds = +2.51 x 10─6 T.m
A = (1/2)(4d)(3d) = 6d².
Since d = 20 cm = 0.20 m, the area is
A = 6 x (0.20 m)² = 0.24 m².
Since the loop is orthogonal to the current density (and only because of that), we are able to get away with saying
I = JA = (15 A/m²)(0.24 m²) = 3.60 A.
So that, applying Ampere’s law, we get
∮B. ds = μ0ienc = (4π x 10─7 Tm/A)(3.60 A) = 4.52 x 10─6 T.m
Problem#2
Figure 2 shows a cross section across a diameter of a long cylindrical conductor of radius a = 2.00 cm carrying uniform current 170 A.What is the magnitude of the current’s magnetic field at radial distance (a) 0, (b) 1.00 cm, (c) 2.00 cm (wire’s surface), and (d) 4.00 cm?
 |
| Fig.2 |
we use the equation
B = μ0ir/2πa2 for the B─field inside the wire (r < a) and
B = μ0i/2πr for that outside the wire (r > a)
(a) at r = 0, B = 0
(b) at r = 1.00 cm = 0.0010 m,
B = μ0ir/2πa2 = (4π x 10─7 Tm/A)(170 A)(0.0010 m)/[2π x (0.0020 m)2] = 8.50 x 10─4 T
(c) at r = 2.00 cm = 0.0020 m,
B = μ0ir/2πa2 = (4π x 10─7 Tm/A)(170 A)(0.0020 m)/[2π x (0.0020 m)2] = 1.70 x 10─3 T
(d) at r = 4.00 cm = 0.0040 m,
B = μ0i/2πr = (4π x 10─7 Tm/A)(170 A)/[2π x 0.0040 m] = 8.50 x 10─4 T
Problem#3
Figure 3 shows two closed paths wrapped around two conducting loops carrying currents i1 = 5.0 A and i2 = 3.0 A.What is the value of the integral ∮B. ds for (a) path 1 and (b) path 2?
 |
| Fig.3 |
Applying Ampere’s law, we get
(a) For Path 1
∮B. ds = μ0ienc
= (4π x 10─7 Tm/A)(─3 A + 5 A)
∮B. ds = +2.51 x 10─6 T.m
 |
| Fig.4 |
(b) For Path 2
∮B. ds = μ0ienc
= (4π x 10─7 Tm/A)(─5 A + 5 A ─ 3 A)
∮B. ds = ─3.77 x 10─6 T.m
Problem#4
Each of the eight conductors in Fig. 5 carries 2.0 A of current into or out of the page.Two paths are indicated for the line integral ∮B. ds. What is the value of the integral for (a) path 1 and (b) path 2?
= (4π x 10─7 Tm/A)(─5 A + 5 A ─ 3 A)
∮B. ds = ─3.77 x 10─6 T.m
Problem#4
Each of the eight conductors in Fig. 5 carries 2.0 A of current into or out of the page.Two paths are indicated for the line integral ∮B. ds. What is the value of the integral for (a) path 1 and (b) path 2?
 |
| Fig.5 |
Answer:
Applying Ampere’s law, we get
(a) For Path 1
∮B. ds = μ0ienc
= (4π x 10─7 Tm/A)(─4 A + 2 A)
∮B. ds = ─2.5 x 10─6 T.m
(b) For Path 2
∮B. ds = μ0ienc
= (4π x 10─7 Tm/A)(─4 A + 4 A)
∮B. ds = 0
Problem#5
Eight wires cut the page perpendicularly at the points shown in Fig. 6. A wire labeled with the integer k (k = 1, 2, . . . , 8) carries the current ki, where i = 4.50 mA. For those wires with odd k, the current is out of the page; for those with even k, it is into the page. Evaluate ∮B. ds along the closed path indicated and in the direction shown.
Answer:
Observe that only i1, i3, i6, i7 are enclosed.
So ienc = i1 + i3 ─ i6 + i7 = i + 3i ─ 6i + 7i = 5i = 5 x 4.5 mA = 22.5 mA
Applying Ampere’s law, we get
∮B. ds = μ0ienc = (4π x 10─7 Tm/A)(22.5 mA) = 2.83 x 10─8 T.m
Applying Ampere’s law, we get
(a) For Path 1
∮B. ds = μ0ienc
= (4π x 10─7 Tm/A)(─4 A + 2 A)
∮B. ds = ─2.5 x 10─6 T.m
(b) For Path 2
∮B. ds = μ0ienc
= (4π x 10─7 Tm/A)(─4 A + 4 A)
∮B. ds = 0
Problem#5
Eight wires cut the page perpendicularly at the points shown in Fig. 6. A wire labeled with the integer k (k = 1, 2, . . . , 8) carries the current ki, where i = 4.50 mA. For those wires with odd k, the current is out of the page; for those with even k, it is into the page. Evaluate ∮B. ds along the closed path indicated and in the direction shown.
 |
| Fig.6 |
Observe that only i1, i3, i6, i7 are enclosed.
So ienc = i1 + i3 ─ i6 + i7 = i + 3i ─ 6i + 7i = 5i = 5 x 4.5 mA = 22.5 mA
Applying Ampere’s law, we get
∮B. ds = μ0ienc = (4π x 10─7 Tm/A)(22.5 mA) = 2.83 x 10─8 T.m
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| Fig.7 |
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