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Unformatted text preview: 6 WINTER QUARTER 2010 MARCH 1, 2010 I(t) The long wire in the vertical direction carries current I(t)=I0t/t0, with I0=20A, t0=1s. The circular loop of wire shown has its center at the long wire and is on the plane perpendicular to the long wire, has radius a=1m and resistance 1. The current induced in the circular loop of wire at time t=1s is, in A (10-6A): (a) ; (b) 2 ; (c) 4 ; (d) 0 ; (e) 0.5 Problem 6 solenoid a P 2a The solenoid shown is oriented perpendicular to the paper. The time-dependent magnetic field inside the solenoid points into the paper and is given by B(t) = B0e"t / # with =2s and B0=10T. The radius of the solenoid is a=0.3m. Point P is at distance 2a=0.6m to the right from the center of the solenoid in the horizontal direction along a line that goes through the center of the solenoid. ! The induced electric field at point P at time t=1s points (a) up ; (b) down ; (c) right ; (d) left ; (e) into the paper Problem 7 For the situation of problem 6, the magnitude of the induced electric field at point P at time t=1s is (a) 0.13V/m ; (b) 0.23V/m ; (c) 0.33V/m ; (d) 0.43V/m ; (e) 0.53V/m Problem 8 v For the triangular conducting loop shown in the figure that is being pushed into a region of uniform magnetic field at constant speed v, the power dissipated when half the horizontal side is in the region of the magnetic field is 100W. Right before the entire loop is in the region of uniform magnetic field the power dissipated is (a) 200W ; (b) 400W ; (c) 100W ; (d) 800W ; (e) 50W...
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This note was uploaded on 03/03/2010 for the course PHYS 2b taught by Professor Schuller during the Winter '08 term at UCSD.

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