Quiz8solutions

Quiz8solutions - Quiz #8 Solutions 10 am Section (Section...

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Quiz #8 Solutions 10 am Section (Section A) Version Version Version Version A B C D 1 E A B A 2 A C B C 3 A B A A 4 C E C A 5 C A A B 6 C A D B 11 am Section (Section B) Version Version Version Version A B C D 1 C A A B 2 A C A A 3 E A A B 4 B E A B 5 B C C B 6 C B D D Solutions for version A of each quiz are provided below. Any additional questions should be directed to: btoth@physics.ucsd.edu (Bryan) OR jleonard@physics.ucsd.edu (Jason)
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1.
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1. This is a difficult problem to explain, so I will try to be as clear as possible. The flux through the circuit area (two horizontal rails, resistor, and moving bar) is a function of the magnetic field due to the stationary wire and the velocity of the left- hand bar. Since the area of the circuit is changing, this flux is changing, and thus produces an emf, which drives current through the resistor. Call r the distance from the vertical wire. Then, from Ampere’s law, the B-field a distance r from a current- carrying wire is μ *I / (2*Pi*r). Let t=0 be at r = r0 = 0.2. Call the vertical height h. To find the flux, integrate B*dA over the area of the circuit, where dA = dr dh, and r is a function of time (since the left-hand side is moving). The limits of this integration are from r = r0 to r = r0 + v*t. This gives that the flux is equal to μ *I*h/(2*Pi) * ln(r / (v*t + r0)). Differentiating this result gives the change in flux with time, which is equal to the induced emf of 0.3 microVolts. (i.e., emf = μ *I*h*v/(2*Pi*(v*t + r0)). Solve this equation for t, to get t = 0.833 seconds.
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Quiz8solutions - Quiz #8 Solutions 10 am Section (Section...

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