Exam3_2009Spr_Solutions - Physics 8.02 Exam Three Solutions...

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Physics 8.02 Exam Three Solutions Spring 2009 Please Remove this Tear Sheet from Your Exam Some (possibly useful) Relations: 2 ˆ 4 o d r πε = Er 1 dq G inside closedsurface o d Q ε ⋅= ∫∫ EA w G G points from inside to outside d A G total dt dt = ∫∫ Ed s BdA v dd Φ G GG G b d G G moving from to ab b a a VV V Δ= = Es 2 4 o c r ˆ q μ π =< × vr Bv G G < G 2 4 o d r ˆ I d × = B sr G G ˆ where points source observer rf r o mt o d BA 0 closed surface w G G through 0 contour E o dI dt με d Φ ⎛⎞ + ⎜⎟ ⎝⎠ v Bs G G where I through is the current flowing through any open surface bounded by the contour: through open surface I d =⋅ ∫∫ JA G G d s is right-handed with respect to d A 2 1 2 2 0 2 EB o B uE u == () GGG G G ext q =+ × FE v B ext d Fs B 2 cent. Fm v r = IA = μ n G ± τ μ B G U =− ⋅ μ B G G VI R L R A ρ = 2 2 ohmic heating V PI V I R R Δ =Δ = = Q C V = 2 2 1 2 2 Q UC V C = Δ B,self,sgl coil back N dI LL Φ I dt 2 1 2 L UL I = 22 2 fT k ω =π=π =πλ 1 2 00 cT f k =λ =ω = με 0l i g h t 0 ˆˆ ˆ Ev B ×= EB p = absorb reflect 0 12 ; SS PP cc = = μ SE B G G G 2 slit interference: sin Constructive 1 slit diffraction: sin Destructive dm am Interference θ λ θλ = = Far field: sin y L ˆˆ ˆˆˆˆ 0 Cross-products of unit vectors: ×=×= × = ii jjkk ˆ k ˆ × = i j × = ki ˆ j × = j Some potentially useful numbers 2 97 2 0 1N m T m 91 0 4 1 0 4C A eo k μπ × = × Breakdown of air E ~ 3 x 10 6 V/m Earth’s B Field B ~ 5 x 10 -5 T = 0.5 Gauss Speed of light c = 3 x 10 8 m/s Light (violet to red) λ = 400 nm to 700 nm Electron charge e = 1.6 x 10 -19 C Avogadro’s number N A = 6.02 x 10 23 mol -1 Calories 1 cal = 10 -3 Cal = 4.184 J
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MIT PHYSICS DEPARTMENT p. 2 of 11 Problem 1: Five Short Questions. Circle your choice for the correct answer. Question A (4 points out of 20 points): Consider a circular parallel plate capacitor, initially charged to ±Q, then discharged through a resistor connecting the centers of the two plates, as pictured at right. A standard analytic problem is to calculate the E & B fields and the Poynting vector at point P. Which of the following statements is true? -Q ˆ j ˆ i P +Q ( ) ˆ dir i ection 1) The Poynting vector at P points to the left ( ) ˆ dir + j 2) The Electric Field at P, calculated by Faraday’s law, points up 3) Using Ampere’s Law to calculate the B field at P, you would find that both the displacement current and physical current lead to B fields out of the page, so you need to sum their effects 4) None of the above The electric field at P is up, but by Gauss’s Law, not Faraday’s (there is no changing magnetic flux to generate an E field in that way). Current flows up the wire from the +Q to the –Q charge (discharging). This generates a magnetic field into the page at point P. The displacement current, being negative, points opposite the direction of the E field (down), meaning it generates a B field in the opposite direction (out of the page). But an Amperian loop through P won’t contain as
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This note was uploaded on 02/27/2011 for the course PHYSICS 8 taught by Professor Dourmashkin during the Fall '10 term at MIT.

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Exam3_2009Spr_Solutions - Physics 8.02 Exam Three Solutions...

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