EE2111_A03_SOL6

EE2111_A03_SOL6 - EE2111 Homework #6 A03 Assigned 02...

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Unformatted text preview: EE2111 Homework #6 A03 Assigned 02 October 03 Due 08 October 03 Problem 1 . Hambly 15.6 Solution: Equation 15.4 in the text states: sin ilB f = Since the wire is perpendicular to the field, 90 = . Solving for the flux: T il f B 6 . 5 . 10 3 = = = Problem 2. Hambly 15.7 Solution: Each wire produces a field in the region surrounding it given by: T r I B 4 7 10 2 01 . 2 10 10 4 2 = = = The field in turn produces a force on the other wire given by ( ) ( ) N Bli f 3 4 10 10 5 . 10 2 = = = Using the right hand rule, we determine the direction of the field at the second wire due to the current in the first wire. Then using B il f = we can determine the direction of the force on the second wire. The direction of the forces is such that the wires attract one another. Problem 3. Hambly 15.9 Solution: The flux linking coil is the product of the coil area and the flux density: Wb r B BA 0314 . 1 . 1 2 2 = = = = The flux linkages are given by...
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This note was uploaded on 10/13/2009 for the course ECE 2111 taught by Professor Labonte during the Spring '09 term at WPI.

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EE2111_A03_SOL6 - EE2111 Homework #6 A03 Assigned 02...

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