The value of the line integral of B around the closed path in figure is 3.77×10-6 Tm. What is I3? If I₁ 3.0 A and I2 = 5.0 A. 1₁ = Ø 13

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The value of the line integral of B around the closed path in figure is 3.77×10-6 Tm.
What is I3? If I₁ = 3.0 A and I₂ = 5.0 A.
1₁
0 13
Transcribed Image Text:The value of the line integral of B around the closed path in figure is 3.77×10-6 Tm. What is I3? If I₁ = 3.0 A and I₂ = 5.0 A. 1₁ 0 13
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Step 1
Given here the value of the line integral of B⇒ around the closed path in figure is 3.77 × 10-
6 Tm.here we have to find the value of 13.
Step 2
Given here the value of the line integral of B⇒ around the closed path in figure is 3.77 × 10-
6 Tm.we know that by ampere's law § B→ dl→ = µ0 lenclosedwhere μ0 = permeability of free space = 4T
x 10-7 H/mgiven § B→ dl⇒ = 3.77 × 10-
6 Tmfrom above given diagram § B→ dl⇒ = µ0 lenclosed
11 + 12 - 13
3.77 x 10-6 = 4 x 10-7 -3 +5-13
64π x 10-7
2-13 = 3
13 = 2-
3
13 = -1 A if you take only magnitude then value of 13 is 1A.
§ B→ dl⇒ = μ0 -
-3 +5-13 = 3.77 × 10-
Transcribed Image Text:Step 1 Given here the value of the line integral of B⇒ around the closed path in figure is 3.77 × 10- 6 Tm.here we have to find the value of 13. Step 2 Given here the value of the line integral of B⇒ around the closed path in figure is 3.77 × 10- 6 Tm.we know that by ampere's law § B→ dl→ = µ0 lenclosedwhere μ0 = permeability of free space = 4T x 10-7 H/mgiven § B→ dl⇒ = 3.77 × 10- 6 Tmfrom above given diagram § B→ dl⇒ = µ0 lenclosed 11 + 12 - 13 3.77 x 10-6 = 4 x 10-7 -3 +5-13 64π x 10-7 2-13 = 3 13 = 2- 3 13 = -1 A if you take only magnitude then value of 13 is 1A. § B→ dl⇒ = μ0 - -3 +5-13 = 3.77 × 10-
The value of the line integral of B around the closed path in figure is 3.77x10-6 Tm.
What is I3? If I₁ = 3.0 A and I₂ = 5.0 A.
2
1₁
13
12
Transcribed Image Text:The value of the line integral of B around the closed path in figure is 3.77x10-6 Tm. What is I3? If I₁ = 3.0 A and I₂ = 5.0 A. 2 1₁ 13 12
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