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Unformatted text preview: Version 038/AACBC – midterm 03 – Turner – (59130) 1 This printout should have 20 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Three very long wires are strung parallel to each other as shown in the figure below. Each wire is at the perpendicular distance 28 cm from the other two, and each wire carries a current of magnitude 7 . 5 A in the directions shown in the figure. I I I 3 2 1 z x y 28 cm 28 cm 28 cm × 3 2 1 y x z Crosssectional View The permeability of free space is 1 . 2566 × 10 − 6 T · m / A. Find the magnitude of the net force per unit length exerted on the upper wire (wire 3) by the other two wires. 1. 2.7614e05 2. 2.04628e05 3. 5.97669e06 4. 2.12879e05 5. 6.33417e06 6. 1.50892e05 7. 6.95893e05 8. 1.73267e05 9. 4.44932e05 10. 4.1568e06 Correct answer: 6 . 95893 × 10 − 5 N / m. Explanation: Let : r = 28 cm = 0 . 28 m and I = 7 . 5 A . Magnetic field due to a long straight wire is B = μ I 2 π r , and the force per unit length between two parallel wires is F ℓ = μ I 1 I 2 2 π r . There are two ways to solve this problem which are essentially the same. The first way is to find the net magnetic field at the upper wire from the two wires below ( vector B net = vector B 1 + vector B 2 ) and then find the force from vector F = I vector L × vector B . The crucial step here will be to add the magnetic fields as vectors . The second way would be to use vector F = I vector L × vector B to find the net force on the upper wire from the two lower wires vector F net = vector F 1 + vector F 2 , where we must be sure to add the forces as vectors. You should recognize that the two methods are formally identical. Let’s do it the first way. The magnitude magnetic field from wire 1 is found from Ampere’s law to be B 1 = μ I 2 π r . Using the right hand rule the direction points up and to the left of wire as shown in figure 2. 3 ◦ 3 ◦ 60 ◦ 6 ◦ 6 ◦ B 1 B B 2 × 3 2 1 ˆ ˆ ı ˆ k Adding Magnetic Fields Version 038/AACBC – midterm 03 – Turner – (59130) 2 3 ◦ 3 ◦ 60 ◦ 6 ◦ 6 ◦ F 1 3 F F 2 3 × 3 2 1 ˆ ˆ ı ˆ k Adding Forces Its components will then be vector B 1 = B [sin(30 ◦ )ˆ − cos (30 ◦ )ˆ ı ] . Similarly, vector B 2 points down and to the left of wire 3; its components are given by vector B 2 = B [ − sin(30 ◦ )ˆ − cos (30 ◦ )ˆ ı ] . Notice that by symmetry the ˆ ( y ) compo nent of the magnetic field vanishes. The net magnetic field is therefore vector B net = − 2 μ I cos (30 ◦ ) 2 π r ˆ ı . The force is then vector F = I vector L × vector B = I L parenleftbigg − 2 μ I cos (30 ◦ ) 2 π r parenrightbigg ( − ˆ k × ˆ ı ) = I L parenleftbigg − 2 μ I cos (30 ◦ ) 2 π r parenrightbigg ( − ˆ ) bardbl vector F bardbl bardbl vector L bardbl = μ I 2 cos(30 ◦ ) π r = (1 . 2566 × 10 − 6 T · m / A) π (0 . 28 m) × (7 . 5 A) 2 cos(30 ◦...
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This note was uploaded on 03/26/2012 for the course PHY 303L taught by Professor Turner during the Spring '08 term at University of Texas at Austin.
 Spring '08
 Turner

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