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EE2111_A03_SOL6

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

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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: θ sin ilB f = Since the wire is perpendicular to the field, 90 = θ . 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 − − × = × × × = = π π µ 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 = × × = = = π π φ The flux linkages are given by...
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EE2111_A03_SOL6 - EE2111 Homework#6 A03 Assigned 02 October...

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