ps11soln

# ps11soln - Physics 217 Homework 11 Solutions Ruxandra...

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Unformatted text preview: Physics 217 - Homework 11 Solutions Ruxandra Bondarescu May 8, 2005 Problem 1. Purcell 11.4 B = 2 m R 3 = 0 . 62gauss (1) The radius of the Earth is about R = 6 × 10 8 cm. So, from the equation above we get m = 6 . 7 × 10 25 erg / gauss = 6 . 7 × 10 22 J / T. If we have a current ring of radius b = 3 × 10 8 cm then . 62gauss = 2 π b 2 I c(b 2 + z 2 ) 3 / 2 = 2 π (3 × 10 8 ) 2 c[(3 × 10 8 ) 2 + (6 × 10 8 ) 2 ] 3 / 2 (2) So, the intensity of the current in the loop is I = 9 . 9 × 10 18 esu / sec = 3 . 3 × 10 9 Amp (3) Problem 2. Purcell 11.11 We know that Maxwell’s equations are ∇× E =- 1 c ∂ B ∂t (4) ∇× B = 1 c ∂ E ∂t + 4 π c J (5) ∇· E = 4 πρ (6) ∇· B = 0 (7) If magnetic charge existed Eq. (7) becomes ∇· B = 4 π K (8) 1 where K is the density of magnetic charge. Moving magnetic charge would generate magnetic current. Let the magnetic current density N = K v in analogy to the current density J = ρ v . In analogy with the Continuity equation that results from the conservation of electric charge ∇· J = ∂ρ/∂t we get an equation for the conservation of magnetic charge ∇· N =- ∂ K ∂t (9) A magnetic current would be the source of an electric field. So, we need to add a term to Eq. (4) ∇× E =- 1 c ∂ B ∂t- 4 π c N (10)...
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ps11soln - Physics 217 Homework 11 Solutions Ruxandra...

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