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# 14 - Electromagnetics Workshop Week 14 Solutions...

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Electromagnetics Workshop – Week 14 Solutions Magnetostatics Problem 1: A long solenoid of length L is wound with many layers of thin wire. A cross section of the solenoid is shown in the figure. There are a total of N turns, each carrying a current I . The inner radius of the solenoid is a and the outer radius is b . Neglect fringing of the fields near the end of the solenoid. a) What is the magnetic field H in the region b ρ ? b) Find the magnetic field H in the region a ρ < . c) Find the magnetic field H in the region a b ρ < < . Problem 1 Solution: . encl d I × = H l Ñ (a) b ρ : . 0 0 encl I = = H (b) a ρ < : . encl z z N I I N I H L N I H L = = = [A/m] (c) a b ρ < < : At b ρ = : . 0 0 encl I = = H At a ρ = : . encl z z N I I N I H L N I H L = = = [A/m] The field should vary smoothly between these two points. . encl z z b b N I b I N I H L N I H a b a b L a b ρ ρ ρ - - - = = = ÷ ÷ ÷ - - - [A/m] 1

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Electromagnetics Workshop – Week 14 Solutions Magnetostatics Problem 2: Determine the magnetic field H at the point ( x , 0, 0) due to the infinitely long wire with a right- angle bend, as shown below (the wire goes to infinity in the – x and – y directions). Helpful integral: ( 29 3 2 2 2 2 2 2 1 du u a u a u a = + + Problem 2 Solution: This is the superposition the magnetic field of two semi-infinite wires. “Wire 1” is the wire on the x -axis. “Wire 2” is the wire on the y -axis. Wire 1: ( 29 ˆ x x = - 1 R x ( 29 ( 29 2 ˆ ˆ 0 4 C I dx x x π × = = - 1 x x H Wire 2: 2 ˆ ˆ x y = - R x y ( 29 ( 29 ( 29 ( 29 ( 29 0 0 0 2 3 2 3 2 2 2 2 2 2 2 ˆ ˆ ˆ ˆ ˆ ˆ 4 4 4 4 I x y I x dy I y I dy x x x y x y x y π π π π -∞ -∞ -∞ - × - = = = = + + + y x y H z z z [A/m] 2 I I x y
Electromagnetics Workshop – Week 14 Solutions Magnetostatics Problem 3: The operation of a magnetic compass depends on the presence of the earth’s magnetic field.

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