Lecture13 - 1 Lecture 13 Chapter 18. Magnetic Field * End...

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Unformatted text preview: 1 Lecture 13 Chapter 18. Magnetic Field * End of Chapter Special case: r<<L , , y x r − = From Lecture 12: A Long Straight Wire B = μ o LI 4 π r r 2 + L /2 ( ) 2 B = μ o 2 I 4 π r Step 1: Cut up the distribution into pieces Δ l = R cos θ + d θ ( ) , R sin θ + d θ ( ) ,0 − − R cos θ , R sin θ ,0 , sin , cos , , θ θ R R z r − = Make use of symmetry! Need to consider only B z due to one l Magnetic Field of a Wire Loop Step 2: B due to one piece Origin: center of loop Vector r: source loc obs r − = . z R R z r , , , , , , − = − = Magnitude of r : 2 2 z R r + = Unit vector : 2 2 , , ˆ z R z R r + − = l: , , θ Δ − = Δ R l 2 ˆ 4 r r l I B × Δ = Δ π μ Δ B = μ 4 π I − R Δ θ ,0,0 × 0, − R , z R 2 + z 2 ( ) 3/2 Magnetic field due to one piece: Magnetic Field of a Wire Loop 2 B z = μ 4 π 2 π R 2 I R 2 + z 2 ( ) 3/2 Magnetic Field of a Wire Loop B z = μ 4 π 2 π I R At the Center of the Loop ( z= 0) : B z = μ 4 π 2 π R 2 I z 3 for z >> R : DEMO 2/22/11 6 Magnetic Field Lines of a Current Loop DEMO What if we had a coil of wire?What if we had a coil of wire?...
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This note was uploaded on 04/23/2011 for the course PHYS 272 taught by Professor K during the Spring '07 term at Purdue University.

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Lecture13 - 1 Lecture 13 Chapter 18. Magnetic Field * End...

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