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Unformatted text preview: ECE329 Summer 2009 Homework 6 — Solution July 13, 2009 1. Infinite current sheets. a) Magnetic field intensity at the origin due to sheet 1 ( J s 1 = 2ˆ z A m ) H 1 = 1 2 J s 1 × ˆ x = 1ˆ y A m . Magnetic field intensity at the origin due to sheet 2 ( J s 2 = 2ˆ z A m ) H 2 = 1 2 J s 2 × ( ˆ x ) = 1ˆ y A m . Resultant displacement vector due to both sheets H = H 1 + H 2 = A m . b) Magnetic field intensity at the origin due to sheet 1 ( J s 1 = 2ˆ z A m ) H 1 = 1 2 J s 1 × ˆ x = 1ˆ y A m . Magnetic field intensity at the origin due to sheet 2 ( J s 2 = 2ˆ z A m ) H 2 = 1 2 J s 2 × ( ˆ x ) = 1ˆ y A m . Resultant displacement vector due to both sheets H = H 1 + H 2 = 2ˆ y A m . 2. Let us compute the magnetic field B generated by a current flowing in a infinite slab of width W = 2 m parallel to the x y plane. The current density in the slab is J = 2ˆ y A m 2 . a) BiotSavart law tells us that the direction of the magnetic field generated by an infinitesimal current element is parallel to the cross product between the direction of the current and the vector joining the current element and the point under consideration. In our case, the current is parallel to ˆ y , then, using the righthand rule, we can easily verify that, above the slab ( z > ), B should be parallel to ˆ x , B k 2ˆ y × ˆ z = ⇒ B k ˆ x, while, below the slab ( z < ), B should be parallel to ˆ x , B k 2ˆ y ×  ˆ z = ⇒ B k  ˆ x....
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This note was uploaded on 09/16/2011 for the course ECE 329 taught by Professor Kim during the Fall '08 term at University of Illinois, Urbana Champaign.
 Fall '08
 Kim
 Electromagnet

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