{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Solutions7

# Solutions7 - Physics 208 Assignment 7 Solutions RT Problem...

This preview shows pages 1–3. Sign up to view the full content.

Physics 208, Assignment 7, Solutions, 3/16/09, RT Problem 1. V =10 V F μ f =10 kHz U C U L U C U L + U C U L U C U L + C =0.1 L t (ms) q i V E q i V E ohm (e) R = 10 (d) R = 0 (a) f = ω 2 π = 1 2 πLC ; L = parenleftbigg 1 2 π × 10 4 parenrightbigg 2 1 10 - 7 = 2 . 5 mH . (b) 1 2 CV 2 0 = 1 2 Li 2 max ; i max = radicalbigg C L V 0 = radicalbigg 10 - 7 2 . 5 × 10 - 3 10 = 0 . 063 A . (c) U max L = 1 2 Li 2 max = 1 2 CV 2 0 = 1 2 × 10 - 7 × 10 2 = 0 . 5 × 10 - 5 J . 1

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Problem 2. (a) The only non-vanishing electic field component of an EM wave is given by E ( z, t ) = E m sin( kz ωt ) ˆy . The “phase” kz ωt , is constant, for example vanishing when z = ( ω/k ) t = ct . So the wave is moving in the positive z direction. (b) The wave is plane-polarized in the y -plane. i.e. E bardbl ˆy . (c) For E × B to be parallel to ˆ z , with B E = 0 requires B = B m ˆx , where B m remains to be determined. In a plane-polarized plane wave c | E | = | B | . Therefore we have B ( z, t ) = E m c sin( kz ωt ) ˆx . A priori the sign here should have been ± but the minus sign has been selected so that E × B bardbl ˆ z , (d) Given λ = 550 nm, f = c λ = 3 × 10 8 0 . 55 × 10 - 6 = 5 . 4 × 10 1
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern