MIT6_641s09_sol_exam2006

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MIT OpenCourseWare http://ocw.mit.edu 6.641 Electromagnetic Fields, Forces, and Motion Spring 2009 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms .
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6.641 — Electromagnetic Fields, Forces, and Motion Spring 2006 Final- Solutions Prof. Markus Zahn MIT OpenCourseWare Problem 1 A Question: What are the boundary conditions necessary to solve for the magnetic fields for x < 0 and for 0 < x < s ? Solution: y σ = 0, 0 σ z s x K z (x = 0, y ) = K 0 cos ky H = -� χ 2 χ = 0 σ =0, Figure 1: A sheet of surface current at x = 0. (Image by MIT OpenCourseWare.) B.C. I H II H I K H y ( x = 0 + , y ) H y ( x = 0 , y ) = K 0 cos ky × n n · µ II H II µ I H I = B.C. II χ 0 as x → −∞ n · µ II H II µ I H I B.C. III B.C. IV 0 H x ( x = s , y ) = 0 x =0 = 0 µ 0 H x µH x = = x =0 + 1
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Spring 2006 Final Exam 6.641, Spring 2005 B Question: What are the magnetic scalar potential and magnetic field distributions for x < 0 and 0 < x < s ? Hint: The algebra will be greatly reduced if you use one of the following forms of the potential for the region 0 < x < s I. sin( ky ) cosh k ( x s ) II. cos( ky ) cosh k ( x s ) III. sin( ky ) sinh k ( x s ) IV. cos( ky ) sinh k ( x s ) Solution: 2 χ 2 χ 2 χ = 0 + = 0 ∂x 2 ∂y 2 General solution is of the form χ ( x, y ) = e kx ( C 1 sin ky + C 2 cos ky ) + e kx ( C 3 sin ky + C 4 cos ky ) For x < 0 only e kx term is relevant from B.C. II lim χ 0; also from surface current boundary condition x →−∞ I only sin ky term is needed. Therefore, χ ( x, y ) = C 1 e kx sin ky for x < 0. For x > 0, we can work with the general solution in sinh, cosh, due to surface current condition in y , only sin ky term is needed; also due to boundary condition at x = 0, x = s we can use the form χ ( x, y ) = A 1 sin ky cosh( x s ) 0 < x < s C 1 sin kye kx x < 0 χ ( x, y ) = A 1 sin ky cosh k ( x s ) 0 < x < s C 1 ke kx (sin kyi x + cos kyi y ) x < 0 H = −� χ = A 1 k (sin ky sinh k ( x s ) i x + cos ky cosh k ( x s ) i y ) 0 < x < s = 0 H x B.C. III satisfied x = s ⇒ − µ 0 C 1 k = µA 1 k sinh( ks ) C 1 = µ B.C. IV A 1 sinh ks µ 0 H x µH x = x =0 + x =0 µ 0 H y B.C. I H y = K 0 cos ky ⇒ − A 1 k cosh( ks ) + C 1 k = K 0 x =0 + x =0 K 0 µ K 0 ⇒ − A 1 cosh ks + C 1 = ⇒ − A 1 cosh ks A 1 sinh ks = k µ 0 k µ 0 K 0 A 1 ( µ 0 cosh ks + µ sinh ks ) = k µ 0 K 0 A 1 = k ( µ 0 cosh ks + µ sinh ks µK 0 sinh ks C 1 = k ( µ 0 cosh ks + µ sinh ks ) 2
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C Spring 2006 Final Exam 6.641, Spring 2005 µ sinh kse kx sin ky x < 0 K 0 χ = k ( µ 0 cosh ks + µ sinh ks ) µ 0 cosh k ( x s ) sin ky 0 < x < s µ sinh kse kx (sin kyi x + cos kyi y ) x < 0 K 0 H = ( µ 0 cosh ks + µ sinh ks ) + µ 0 (sin ky sinh k ( x s ) i x + cos ky cosh k ( x s ) i y ) 0 < x < s Question: What is the surface current distribution on the x = s surface?
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