Exam III review.CT3-1An electron and a proton have the same momentum (p=mv) and enter a region where there is a uniform magnetic field. Which particle is bent more?A: electron B: proton C: both are bent the same amountAnswer: Both the electron and proton are bent into circles with the same radius of curvature. R=(mv)/qB. Same p=mv, same q, same B, so same R. The electron has larger acceleration and speed, however. CT3-2A charged particle is orbiting in a uniform B-field in the sense shown.Is the particle positive or negative?A: positiveB: negativeC: impossible to tell Answer: The direction of acceleration and net force must be toward the center of the circle. For a negative particle, the force on particle is toward the center of the circle.CT3-3A 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 pageB: left side will move out of page.C: loop will not rotate at allB(in) B axis IAnswer: Right side will move out of page. Use Fwire = I L BCT3-4A 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.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.I A x y z(out)CT33-5.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 loopshown, B dl I z0.A: True B: False.Answer: True. Ampere's Law is always true (for steady currents and no changing E-fields). True or False? At point A, the magnitude of the magnetic field is BIro2.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.CT3-6Consider 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 .r I AAnswer: 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.CT3-7. A bar magnet with the ends properly labeled N and S is placed near an unlabeledbar magnet as shown. The magnets attract. The left side of the unlabeled bar magnet must be a A: South pole B: North poleAnswer: South. Opposite poles attract. Recall that a bar magnet is like a current-carrying solenoid, with the B-field coming out of the N end. Parallel currents attract, so the orientation must be such as to cause the currents in the near ends to be parallel.S N ?CT3-8A circuit with a battery and a variable resistor is near a loop of wire as shown. When theresistance R is decreased, the induced current in the loop is ..A: CWB: CCWAnswer: CW. When R decreases, the current I = V/R in the circuit increases, causing an increasing B-field out of the page at the position of the loop. A CW current in the loop will fight the increase in B. RCT3-9 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 B: non-zero and upward C: zero / don't knowAnswer: 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-fieldin 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.CT3-10A solenoid has an increasing current which creates an increasing B-field in its interior. The electric field at point c isA: up B: down C: left D: right E: zeroAnswer: down. Apply Faraday's Law and Lenz's Law.At which of the three points (a, b, or c) is the E-field maximum?A: a B: b C: cD: the E-fields at b and c have the same magnitude and are larger than the E-field at a.Answer: point bV B(in) a c b I, increasing B, increasingCT3-11 A uniform solid sphere of copper rotates about a stationary axis in a uniform magnetic field B. Are there eddy currents?A: Yes, there are eddy currents.B: No, no eddy currents.ANSWER: NO EDDY CURRENTS. The flux through any loop rotating with the sphere does not change. Therefore no emf, no eddy currents. CT3-12 The switch in the circuit below has been closed for a long time and then is opened at t=0. What is the current in the inductor at t=0+?A: zeroB: 1AC: 0.5AD: some other answer Answer: 1A. Just before the switch was opened the current thru the inductor was 1A = (V/R1). The current thru an inductor cannot change instantly.At t=0+ , what is the current through resistor R2?A: zero B: 1A C: 0.5A D: some other answer Answer: 1A. The inductor, the capacitor, and the resistor are in series, so they must all have the same current. The current thru R2 is the same as thru the inductor.At t=0+, what is the voltage across the capacitor?A: zero B: not zero.B L = 10H V = 10V R2 = 10 R1 = 10 C= 1FAnswer: zero! This is a tricky one! Initially, there is no charge q on the capacitor so no VC = q/C. Prior to opening the switch, the capacitor is discharged, because the inductor acts like a short (a wire) which discharges the capacitor. When the switch is opened, current begins flowing through the capacitor instantly, but it takes time for the charge to build up.At t=0+, energy
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