Old exam answers

Exam 1, summer 2009, problem 4: this was also on the summer 2008 exam 2.

Exam 2, summer 2010: only the first three problems are material related to your upcoming exam.

1. Just walk around the middle loop - that gives 12-6I=0, and determines the current in the 6Ω resistor to be 2A.

2. Use the right-hand rule to determine directions. Two of the currents give fields pointing to the lower right corner, and two of them give fields going to the lower left corner. All fields have the same magnitude (same current, same distance; watch the geometries). All but the vertical components will cancel; find those and add together. The net field is in the -y direction, and has magnitude (field of 1 wire at center)*(sin 45)*(4 wires)~5e-6T.

3. (a) into the plane. (b) set r=mv/qB equal for both particles using the known electron and proton masses. q is the same for both, and will cancel, as will B. This gives (v_e/v_p)=(m_p/m_e)~1.8e3

4. (a,b) This filter passes speech - the inductors let low frequencies through, and the capacitor shorts out any remaining high frequencies. (c) Add all inductors together, they sum to 10.26mH. The cutoff frequency is when the inductive and capacitive reactances are equal, so f=1/(2*pi*sqrt(LC))~10.6kHz

5. This is motionally-induced voltage, like the conducting bar. (a) V=Blv~2.5mV (b) Right hand rule gives the force on a positive charge toward the drivers door, so it is the positive side.