1 of 7 RIII-1. R C R A capacitor C is charged up to a voltage V. The switch is closed at time t=0 connecting the capacitor to two identical resistors as shown. Immediately after the switch is closed, how does the current through the second resistor (on far right) compare with the current through the capacitor? A) IR = Ic B) IR = (1/2) Ic C) IR = 2 Ic E) IR = 0 Answer: IR = Ic The circuit elements are in series. Immediately after the switch is closed, how does the voltage across the second resistor (on far right) compare with the voltage across the capacitor? A) VR = Vc B) VR = (1/2) Vc C) VR = 2 Vc D) VR = 0 Answer: VR = (1/2) Vc By Kirchhoff's voltage law, we have Vc = VR1 + VR2 = 2VR. Phys1120 M.Dubson ©University of Colorado at Boulder2 of 7 RevIII-2. The circuit shown has two identical capacitors, both with capacitance C, and two identical resistors, both with resistance R. Initially, the switch is open and both capacitors are completely uncharged. What is the initial current I through the battery, immediately after the switch is closed? R C (A) zero (B) VR2 (C) VR (D) VR2 VRC(E) None of these. Answer: initVIR=2. When a capacitor is uncharged, it acts like a short (a wire). RevIII-3. A metal bar (not attached to any outside circuitry) is moving through a uniform magnetic field as shown. The electric field E within the bar is .. A) non-zero and downward ↓ V B(in) B) non-zero and upward ↑ C) zero / don't know Answer: Non-zero and downward. Although there is no current, there is a charge separation. (+) charges piles up at the top of the bar, (–) charges at the bottom. The E-field in the bar is such that the force from the E-field (F=qE) and the force from the B-field (F=qvB) cancel everywhere in the bar, and charges move neither up nor down. RevIII-4. A charged particle is orbiting in a uniform B-field in the sense shown. Is the particle positive or negative? B(in) A) positive B) negative C) impossible to tell Phys1120 M.Dubson ©University of Colorado at Boulder3 of 7 Answer: The direction of acceleration and net force must be toward the center of the circle. For CW rotation and B(in) and a negative particle, the force on particle is toward the center of the circle. RevIII-5.Two particles with the same charge but different masses are moving in circular orbits in a magnetic field. They have the same speed. Which one will have the larger radius orbit? A) Neither, the orbits have the same size. B) The larger mass particle will have a larger orbit. C) The smaller mass particle will have a larger orbit. Answer: The larger mass particle will have a larger orbit. You can see this from a physical argument or from the formula for the radius of an orbit) R = mv/(qB). The physical argument is this) both particles feel the same size (sideways) force F = qvB. The more massive particle has more inertia and is therefore harder to turn, so it turns through a larger orbit. RevIII-6. A coil of wire carrying current I can rotate freely about an axis in a magnetic field. If released from rest in the position shown, which way does it rotate? A) right side will move out of page B) left side will move out of page. B axis I C) loop will not rotate at all Answers: Right side will move out of page. Use Fwire = I L B Phys1120 M.Dubson ©University of Colorado at Boulder4 of 7 RevIII-7. Two bar magnets are brought near each other as shown. The magnets... A) attract B) repel C) exert no net force on each other. N S S N Hint: Answer: Parallel currents attract. Think about the current directions on the sides of the magnets. The two magnets will attract. RevIII-8. A long U-shaped wire carries a current I in the sense shown. Consider the magnetic field B at the point A, which is equi-distant from the two corners of the U and in the same plane as the U. Which one of the following statements is true? A) B has a z-component only which is due entirely to the bottom segment of the U. B) B has a z-component only which has contributions from the bottom and sides of the U. C) B has a non-zero x-component, as well as a z-component. D) B has a non-zero y-component, as well as a z-component. atomic current on rim, like solenoid N S B-field enters B-field comes "South" end out of "North" end I A x y z(out) Phys1120 M.Dubson ©University of Colorado at Boulder5 of 7 Answer: B has a z-component only which has contributions from the bottom and sides of the U. Use the Biot-Savart Law to see this. RIII-9. A long U-shaped wire carries a current I in the sense shown. Consider the (imaginary) loop of radius r centered on the right side of the U as shown. True or False? For the loop shown, GGBdl I⋅=zµ0. A) True B) False. r I A Answer: True. Ampere's Law is always true (for steady currents and no changing E-fields). RIII-10. True or False? At point A, the magnitude of the magnetic field is BIro=µπ2. A) True B) False. False: That formula is for an infinitely long straight wire, with no other wires nearby. It doesn't apply here because the other side of the U breaks the symmetry of the situation. In this messy situation, with a U-shaped wire, Ampere's Law is true, but not useful since the integral is very messy. To get the field at A, one would use Biot-Savart. RIII-11. Consider the following configuration of B-field lines (solid lines) and the imaginary rectangular loop (dotted line). Which one of the following statements is true? A) There must be a non-zero current thru the imaginary loop going into the page. B) There must be a non-zero current thru the loop coming out of the page. C) There is no current going thru the loop . Phys1120 M.Dubson ©University of Colorado at Boulder6 of 7 Answer: There must be a non-zero current thru the loop going into the page. Apply Ampere's Law to the loop. Note carefully) The B-field lines show the total field due to all currents present. For instance, the field lines on the right are not due just to a current in the center of those lines. RevIII-12. A coil of wire attached to a resistor is in an oscillating magnetic field, as shown. The B-field vs. time is shown. At what point in time is the magnitude of the current through the resistor a maximum? B 10Ω (oscilating) time 0 B-field (T) (B) (C) (D) (E) Answer: At point D, the rate of change of B is maximum. So the rate of change of flux is maximum, so maximum emf by Faraday's law.
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