MIT6_641s09_sol_exam2006

MIT6_641s09_sol_exam - MIT OpenCourseWare http/ocw.mit.edu 6.641 Electromagnetic Fields Forces and Motion Spring 2009 For information about citing

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Unformatted text preview: 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 . 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 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 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 H x µH x = = x =0 + 1 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 ) < 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 ) < x < s = 0 ⇒ H x B.C. III ⇒ satisfied x = s ⇒ − µ C 1 k = − µA 1 k sinh( − ks ) ⇒ C 1 = − µ B.C. IV A 1 sinh ks ⇒ µ H x µH x = x =0 + x =0 − µ ⇒ H y B.C. I − H y = K cos ky ⇒ − A 1 k cosh( − ks ) + C 1 k = K x =0 + x =0 − K µ K ⇒ − A 1 cosh ks + C 1 = ⇒ − A 1 cosh ks − A 1 sinh ks = k µ k µ K ⇒ A 1 ( µ cosh ks + µ sinh ks ) = − k µ K ⇒ A 1 = − k ( µ cosh ks + µ sinh ks µK sinh ks ⇒ C 1 = k ( µ cosh ks + µ sinh ks ) 2 C Spring 2006 Final Exam 6.641, Spring 2005 µ sinh kse kx sin ky x < K ⇒ χ = k ( µ cosh ks + µ sinh ks ) − µ cosh k ( x − s ) sin ky 0 < x < s − µ sinh kse kx (sin kyi x + cos kyi y ) x < K H = ( µ cosh ks + µ sinh ks ) + µ (sin ky sinh k ( x − s ) i x + cos ky cosh k ( x − s ) i y ) < x < s Question: What is the surface current distribution on the x...
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This note was uploaded on 12/02/2010 for the course ECE 6.641 taught by Professor Zahn during the Spring '09 term at MIT.

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MIT6_641s09_sol_exam - MIT OpenCourseWare http/ocw.mit.edu 6.641 Electromagnetic Fields Forces and Motion Spring 2009 For information about citing

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