831_PartUniversity Physics Solution

831_PartUniversity Physics Solution - Sources of Magnetic...

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Sources of Magnetic Field 28-9 EVALUATE: In part (c), if all four currents are reversed in direction, the net field at the center of the square would be to the right. Figure 28.23 28.24. IDENTIFY: Use Eq.(28.9) and the right-hand rule to determine the field due to each wire. Set the sum of the four fields equal to zero and use that equation to solve for the field and the current of the fourth wire. SET UP: The three known currents are shown in Figure 28.24. 123 , ⊗⊗ BBB ! !! " 0 ; 0.200 m 2 I Br r μ π == for each wire Figure 28.24 EXECUTE: Let " be the positive z -direction. 12 3 10.0 A, 8.0 A, 20.0 A. II I = Then 5 1 1.00 10 T, B 5 2 0.80 10 T, B and 5 3 2.00 10 T. B 555 1z 2z 3z 1.00 10 T, 0.80 10 T, 2.00 10 T −−− =− × × =+ × 1234 0 zzzz BBBB +++= 6 41 2 3 () 2 . 0 1 0 T zz z z BB B B + + × To give 4 in the B ! direction the current in wire 4 must be toward the bottom of the page. 6 04 44 7 0 (0.200 m)(2.0 10 T) so 2.0 A 2( / 2 ) ( 2 1 0 T m / A ) Ir B BI r πμ × = = ×⋅ EVALUATE: The fields of wires #2 and #3 are in opposite directions and their net field is the same as due to a current 20.0 A & 8.0 A = 12.0 A in one wire. The field of wire #4 must be in the same direction as that of wire #1, and 4 10.0 A 12.0 A. I += 28.25. IDENTIFY: Apply Eq.(28.11). SET UP: Two parallel conductors carrying current in the same direction attract each other. Parallel conductors carrying currents in opposite directions repel each other. EXECUTE: (a) 6 012 0 (5.00 A)(2.00 A)(1.20 m) 6.00 10 N, 22 ( 0 . 4 0 0 m ) IIL F r μμ ππ = × and the force is repulsive since the currents are in opposite directions. (b) Doubling the currents makes the force increase by a factor of four to 5 2.40 10 N. F EVALUATE: Doubling the current in a wire doubles the magnetic field of that wire. For fixed magnetic field, doubling the current in a wire doubles the force that the magnetic field exerts on the wire. 28.26. IDENTIFY: Apply Eq.(28.11). SET UP: Two parallel conductors carrying current in the same direction attract each other. Parallel conductors carrying currents in opposite directions repel each other. EXECUTE: (a) 2 FI I L r = gives 5 2 01 0 2 2 (0.0250 m) (4.0 10 N m) 8.33 A. (0.60 A) Fr I LI × = (b) The two wires repel so the currents are in opposite directions. EVALUATE: The force between the two wires is proportional to the product of the currents in the wires.
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28-10 Chapter 28 28.27. IDENTIFY: The lamp cord wires are two parallel current-carrying wires, so they must exert a magnetic force on each other. SET UP: First find the current in the cord. Since it is connected to a light bulb, the power consumed by the bulb is P = IV . Then find the force per unit length using F / L = 0 2 II r μ π . EXECUTE: For the light bulb, 100 W = I (120 V) gives I = 0.833 A. The force per unit length is F / L = 72 5 4 π 10 T m/A (0.833 A) 4.6 10 N/m 2 π 0.003 m ×⋅ Since the currents are in opposite directions, the force is repulsive.
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This note was uploaded on 07/16/2011 for the course PHY 2053 taught by Professor Buchler during the Spring '06 term at University of Florida.

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831_PartUniversity Physics Solution - Sources of Magnetic...

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