Physics 2212 GJ Quiz #4 Solutions Fall 2015

Transcription

1 Physics 2212 GJ Quiz #4 Solutions Fall 215 I. (17 points) The magnetic fiel at point P ue to a current through the wire is 5. µt into the page. The curve portion of the wire is a semicircle of raius 2. cm, an the two straight segments are each 3. cm. Calculate the magnitue of the current. Show your work, starting from a funamental physical principle. In what irection oes it flow? Use the Biot-Savart Law. B = µ I 4π l ˆr r 2 B = µ I l sin ϕ 4π r 2 In each of the straight segments, l is parallel or anti-parallel to ˆr (ϕ = or π) so sin ϕ =. These segments make no contribution to the magnetic fiel at point P. At every location in the curve portion, l is perpenicular to ˆr (ϕ = π/2) so sin ϕ = 1. Every location in the curve portion is also the same istance r from the point P, an the current through every element of length l prouces magnetic fiel in the same irection. The magnitue of the fiel is B = B = µ I 4π l (1) r 2 = µ I 4πr 2 πr l = µ I 4πr 2 l πr = µ I 4πr 2 (πr ) = µ I 4r Solve for I. I = 4rB ( 4 (.2 m) T ) = µ 4π 1 7 =.32 A T m/a Consier the element of at the very top of the semicircle. The irection of ˆr from that point to P is own the page. With the magnetic fiel at point P being into the page, the Right-Han-Rule shows that the current must be flowing to the right, or clockwise. Quiz #4 Solutions Page 1 of 7

2 II. (17 points) A long, straight conucting wire of raius R has a nonuniform current ensity J(r) whose magnitue epens on the istance r from the center accoring to [ ] J(r) = J 1 r3 R 3 where J is a positive constant. Fin an expression for the magnetic fiel magnitue insie the wire, at raius r < R, in terms of parameters efine in the problem an physical or mathematical constants. Use Ampere s Law. B s = µ I thru = µ J A From symmetry, the fiel lines must be circular. To fin the fiel at an arbitrary istance r < R from the center of the wire, choose an Amperian Loop that follows a fiel line of raius r. Then B is parallel to s an has the same magnitue B at every point on the loop. Only current at a istance r or less from the center of the wire passes through this loop. Since the current varies with r, the area element A shoul be small in this imension. Choose A to be a thin ring, with area 2πr r. r B s = µ [ ] J 1 r3 R 3 2πr r B 2πr = µ J 2π r r r r r 4 R 3 r [ ] r 2 r [ ] = µ J 2π 2 r5 r 2 5R 3 = µ J 2π 2 r5 5R 3 Solve for B. [ ] r B = µ J 2 r4 5R 3 1. (5 points) Outsie the wire in the problem above (r > R), how oes the magnitue of the magnetic fiel B epen on the istance r from the center of the wire? Use Ampere s Law. B s = µ I thru From symmetry, the fiel lines must be circular. Choose an Amperian Loop that follows a circular fiel line of raius r. Then B is parallel to s an has the same magnitue B at every point on the loop. So Outsie the wire, B is proportional to 1/r. B s = B2πr = µ I thru B = µ I thru 2πr Quiz #4 Solutions Page 2 of 7

3 III. (16 points) A 12 cm 25 cm rectangular loop, which has a total resistance of 1. mω, is place in a uniform 1. mt magnetic fiel, as shown in Figure (a). The applie fiel graually flips irection from into the page to out of the page with the change sweeping slowly from left to right. The time elapse between figures (a) an (c) is.5 s. What is the inuce current in the loop at time t =.5 s, when the fiel flip is halfway along the loop? Use Faraay s Law to fin the magnitue of the emf inuce in the loop. E = Φ B t = [ t ] B A E = Φ B t = Φ B t = Φ Bf Φ Bi t Note that the final flux through the loop Φ Bf is zero at time t =.5 s, while the magnitue of the initial flux through the loop Φ Bi at time t =.5 s is just BA. E = BA t = BA t The current can be foun from the efinition of resistance. V = IR I = V R = E R = BA R t = ( T ) (.12 m.25 m) ( =.6 A Ω) (.5 s) 2. (5 points) What is the irection of the current, if any, in the loop in figure (c), as seen from above (the vantage point of the figures): Use Lenz Law. The flux through the loop is initially into the page. It ecreases with time. The inuce current must flow to create an inuce fiel that opposes this change. Therefore, the inuce fiel must be into the page. Using the Right-Han-Rule, the current must be Clockwise. Quiz #4 Solutions Page 3 of 7

4 3. (5 points) The conucting slab moves own the page with velocity v, through a magnetic fiel B irecte out of the page. At what location, if any, in the conuctor is the electric potential highest? Consier a positively-charge particle free to move within the conuctor. It moves ownwar with the slab, so the magnetic force F = q v B is to the left. The left ege of the conuctor becomes positively charge, leaving the right ege to be negatively charge. Electric fiel points from positive to negative charges, or rightwar. Electric fiel also points from high to low potential, so The electric potential is highest on the left ege of the conuctor. 4. (5 points) Three ientical conucting loops enter a magnetic fiel at ientical velocities v. At the instant shown, rank the current magnitues in the loops from greatest to least The current is proportional to the inuce emf, I = E/R, where R is the resistance of the loops. Using Faraay s Law, the emf is relate to the time rate of magnetic flux change through a loop. E = Φ B t In a time t, each loop avances a istance v t. That increases the area of fiel within loops i an ii by their height at the ege of the fiel times v t. This is greater for loop ii than for loop i. But as loop iii is entirely within the fiel, there is no flux change as it avances. I ii > I i > I iii Quiz #4 Solutions Page 4 of 7

5 5. (5 points) Three long wires are aligne parallel to one another, with a uniform spacing between wires. Each wire carries the current inicate in the figure. Rank orer the magnitues of the net force exerte on each wire The force on a current-carrying wire F = I l B epens on the current I in the wire an the fiel B it is in. From Ampere s Law, the fiel from a current-carrying wire epens on the current in the wire an the istance r between them. So the force magnitue between two parallel current-carrying wires will epen on each of the currents, an the istance between them, F I 1 I 2 /r. Consier the force F L on the left wire. The center an right wires, having currents in opposite irections, prouce fiel (an therefore force) in opposite irections. F L I LI C r LC I LI R r LR = 2I2 2I2 2 = I2 Consier the force F C on the center wire. The left an right wires, being on opposite sies, prouce fiel (an therefore force) in opposite irections. F C I CI L r CL I CI R r CR = 2I2 I2 = I2 Consier the force F R on the right wire. The center an left wires, having currents in opposite irections, prouce fiel (an therefore force) in opposite irections. F R I RI C r RC I RI L r RL = I2 2I2 2 = so F left = F center > F right 6. (6 points) Consier a particle that is release from rest at the -5 V plate of a capacitor. It accelerates towar the +5 V plate, an passes through a small hole. Entering a uniform magnetic fiel irecte into the page, its path is eflecte up the page. Uner what circumstances, if any, is this possible? (Neglect the effects of gravity.) Electric fiel points from high potential to low potential. If the particle accelerates towar the high-potential plate, the force F E = q E on it must be in the irection of the electric fiel, so the particle must be negatively charge. If a negatively-charge particle enters the magnetic fiel as shown, the force F B = q v B woul be own the page. So, This is not possible. Quiz #4 Solutions Page 5 of 7

6 7. (5 points) Initially, an inuctor stores magnetic energy U with a current I running through it. Starting at time t =, the current changes linearly, as inicate by the graph. What is the final energy store in the inuctor? The energy store in an inuctor is U L = 1 2 LI2 At the final time, the current has half the magnitue it i at t =. Therefore, the energy is U /4 8. (5 points) A rectangular conucting loop is positione above a long wire that carries a current I(t). If I(t) is increasing with time, what force or torque oes the loop experience? From the Biot-Savart Law an the Right-Han-Rule (or a short cut) the fiel from the wire through the loop is out of the page. As the current in the wire increases, the flux out of the page through the loop increases. From Lenz Law, current must flow in the loop to create flux opposing that change, or into the page. Using the Right-Han-Rule again, the current in the loop must flow clockwise. The force on each sie of the loop can be consiere, using F = I L B, where B is the fiel ue to the wire. The force on the left sie of the loop is to the right, an the force on the right sie of the loop is to the left. Both are in the plane of the page an have the same magnitue, proucing no net torque or force. The force on the upper sie of the loop is own the page, an the force on the lower sie of the loop is up the page. Both are in the plane of the page, proucing no net torque. The upper sie of the loop, however, is farther from the wire, an so is in a weaker fiel. The magnitue of the force on the upper sie, then, is less than that on the lower sie. Therefore, there is Only a force up the page. Quiz #4 Solutions Page 6 of 7

7 9. (5 points) A circular loop of current rests on a table top. A large, uniform magnetic fiel is applie parallel to the table top, oriente from left to right, as shown. How oes the loop respon? The battery will prouce a counter-clockwise current in the loop. The magnetic moment µ of the loop, then, will be out of the page. Since the torque on current loop in a magnetic fiel is τ = µ B, the Right-Han- Rule inicates the torque is irecte up the page. With the torque in that irection, Its left ege lifts off the table. 1. (5 points) A negatively charge particle, moving with velocity v as shown, experiences a magnetic Lorentz force oriente irectly out of the page. What is the irection of the magnetic fiel at the particle s location? The magnetic force on a charge particle is F = q v B. Using the Right-Han-Rule for the vector prouct, the magnetic fiel must be in this irection: Quiz #4 Solutions Page 7 of 7