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# 329lect13 - 13 Current sheet solenoid vector potential and...

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Unformatted text preview: 13 Current sheet, solenoid, vector potential and current loops In the following examples we will calculate the magnetic fields B = μ o H established by some simple current configurations by using the integral form of static Ampere’s law. z J s = ˆ zJ s x y L W B = ˆ yB ( x ) B B ( x ) = μ o J s 2 C As shown in Example 1 mag- netic field of a current sheet is independent of distance | x | from the current sheet. Also H changes discontinu- ously across the current sheet by an amount J s . Example 1: Consider a uniform surface current density J s = J s ˆ z A/m flowing on x = 0 plane (see figure in the margin) — the current sheet extends infinitely in y and z directions. Determine B and H . Solution: Since the current sheet extends infinitely in y and z directions we expect B to depend only on coordinate x . Also, the field should be the superposition of the fields of an infinite number of current filaments, which suggests, by right-hand- rule, B = ˆ yB ( x ) , where B ( x ) is an odd function of x . To determine B ( x ) , such that B (- x ) =- B ( x ) , we apply Ampere’s law by computing the circulation of B around the rectangular path C shown in the figure in the margin. We expand C B · d l = μ o I C as B ( x ) L + 0- B (- x ) L + 0 = μ o J s L, from which we obtain B ( x ) = μ o J s 2 ⇒ B = ˆ y μ o J s 2 sgn ( x ) and H = ˆ y J s 2 sgn ( x ) . 1 Example 2: Consider a slab of thickness W over- W 2 < x < W 2 which extends in- finitely in y and z directions and conducts a uniform current density of J = ˆ zJ o A/m 2 . Determine H if the current density is zero outside the slab. Solution: Given the geometric similarities between this problem and Example 1, we postulate that B = ˆ yB ( x ) , where B ( x ) is an odd function of x , that is B (- x ) =- B ( x ) . To determine B ( x ) we apply Ampere’s law by computing the circulation of B around the rectangular path C shown in the figure in the margin. For x < W 2 , we expand C B · d l = μ o I C as B ( x ) L + 0- B (- x ) L + 0 = μ o J o 2 xL, from which we obtain B ( x ) = μ o J o x . For x < W 2 , the expansion gives B ( x ) L + 0- B (- x ) L + 0 = μ o J o WL, leading to B ( x ) = μ o J o W 2 . Hence, we find that H = ˆ yJ o x sgn ( x ) , | x | < W 2 ˆ yJ o W 2 sgn ( x ) , otherwise....
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329lect13 - 13 Current sheet solenoid vector potential and...

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