HW_11 Solutions - Physics 2213 28.20. HW #11 Solutions Fall...

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Physics 2213 HW #11 Solutions Fall 2009 28.20. IDENTIFY and SET UP: The magnitude of B is given by Eq.(28.9) and the direction is given by the right-hand rule. (a) EXECUTE: Viewed from above, the current is in the direction shown in Figure 28.20. Directly below the wire the direction of the magnetic field due to the current in the wire is east. Figure 28.20 75 0 800 A (2 10 T m/A) 2.91 10 T 2 5.50 m I B r µ π −−  = = ×⋅ = ×   (b) EVALUATE: B from the current is nearly equal in magnitude to the earth's field, so, yes, the current really is a problem. 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 . 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. EVALUATE: This force is too small to have an appreciable effect for an ordinary cord. 28.42. IDENTIFY and SET UP: At the center of a long solenoid 00 . N B nI I L µµ = = EXECUTE: 7 0 (0.150 T)(1.40 m) 41.8 A (4 10 T m/A)(4000) BL I N µπ = = = EVALUATE: The magnetic field inside the solenoid is independent of the radius of the solenoid, if the radius is much less than the length, as is the case here.
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This note was uploaded on 12/25/2009 for the course PHYS 213 at Cornell University (Engineering School).

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HW_11 Solutions - Physics 2213 28.20. HW #11 Solutions Fall...

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