Exam2_2009Spr_Solutions

# Exam2_2009Spr_Solutions - Physics 8.02 Exam Two Solutions...

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Physics 8.02 Exam Two Solutions Spring 2009 Please Remove this Tear Sheet from Your Exam Some (possibly useful) Relations: inside closedsurface o d κ Q ε ⋅= ∫∫ EA w G G points from inside to outside d A G b d G G moving from to ab b a a VV V Δ= = Es 0 d −⋅ = G G v many point charges 1 4 i i oi V πε = = rr GG N q V C Δ = Q C V = Δ Q parallelplate o C d A = U = 1 2 C Δ V 2 = 2 Q 2 C series 1 2 CC =+ 11 1 C parallel 1 2 where ρρ = EJ C is the resistivity Δ V = i R L R A ρ = 1 R parallel = 1 R 1 + 1 R 2 R series = R 1 + R 2 2 2 ohmic heating V Pi V i R R Δ =Δ = = 2 ˆ 4 o q c r μ π × = << vr Bv G G G 2 ˆ 4 o I d d r × = sr B G G ˆ points source observer rf r o mt o through contour o dI Bs G G v 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 right-handed with respect to d A sgl loop d d N dt dt Φ = Ed s BdA v ext q Fv B G G G ext d Fs B G 2 cent. Fm v r = = perpendicular to loop, right-handed with respect to IA I μ G nn ± ± = × τ μ B G G G z zz dB F dz = ˆˆ ˆˆˆˆ 0 Cross-products of unit vectors: ×=×= × = ii jjkk ˆˆ 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 Electron charge e = 1.6 x 10 -19 C Avogadro’s number N A = 6.02 x 10 23 mol -1

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MIT PHYSICS DEPARTMENT Problem 1: Five Short Questions. Circle your choice for the correct answer. Question A (4 points out of 20 points): Consider a charged circular loop of radius R and total charge q glued down to the loop. Suppose the loop is rotating about the axis normal to it and going through its center and that it makes N complete revolutions in a time T. During this time the loop acts as a magnetic dipole with magnetic dipole moment μ given by 1) 2 TqR N = 2) 4 TqR = 3) 2 qR T μπ = 4) 2 Nq R T = 5) 2 2 Nq R T = 6) 2 qR π = 2 2 q Nq R IA R TN T == = Question B (4 points out of 20 points): Consider two identical circular loops of radius a , separated by a distance a , with the second (right hand) coil rotated slightly clockwise relative to the first when looking from above: Top view Side view A large current is suddenly injected into the left hand loop. What does the right hand loop feel due to this? 1. Nothing 2. Force to the left, no torque 3. Force to the right, no torque 4. Force to the left, torque rotates clockwise (in top view) 5. Force to the left, torque rotates counterclockwise (in top view) 6. Force to the right, torque rotates clockwise (in top view) 7. Force to the right, torque rotates counterclockwise (in top view) 8. No force, torque rotates clockwise (in top view) 9. No force, torque rotates counterclockwise (in top view) 10. Can’t tell without knowing which direction current injected into right loop Lenz’s law says whatever the direction of the new flux, we want to get away from it, so move away (to right) and rotate out of alignment (clockwise) 8.02 Exam #2 Spring 2009
MIT PHYSICS DEPARTMENT 8.02 Exam #2 Spring 2009

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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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Exam2_2009Spr_Solutions - Physics 8.02 Exam Two Solutions...

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