Problem Solving 7: Faraday s Law Solution

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Aldous Gallagher
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1 MASSACHUSETTS NSTTUTE OF TECHNOLOGY Deprtment of Physics: 8.02 Prolem Solving 7: Frdy s Lw Solution Ojectives 1. To explore prticulr sitution tht cn led to chnging mgnetic flux through the open surfce ounded y n electric circuit. 2. To clculte the rte of chnge of mgnetic flux through the open surfce ounded y tht circuit in this sitution. 3. To determine the sense of the induced current in the circuit from Lenz s Lw. 4. To look t the forces on the current crrying wires in our circuit nd determine the effects of these forces on the dynmics of the circuit. Reference: Sections , 8.02 Course Notes. Prolem-Solving Strtegy for Frdy s Lw n Chpter 10 of the 8.02 Course Notes, we hve seen tht chnging mgnetic flux induces n emf: dφ ε = dt ccording to Frdy s lw of induction. For conductor tht forms closed loop, the emf sets up n induced current = ε / R, where R is the resistnce of the loop. To compute the induced current nd its direction, we follow the procedure elow: (1) For the closed loop of re A, define n re vector A nd choose positive direction. For convenience of pplying the right-hnd rule, let A point in the direction of your thum. Compute the mgnetic flux through the loop using Φ A ( is uniform) = da ( is non-uniform) (2) Evlute the rte of chnge of mgnetic flux dφ / dt. Keep in mind tht the chnge could e cused y Solving 7-1

2 (i) chnging the mgnetic field d / dt 0, (ii) chnging the loop re if the conductor is moving ( da/ dt 0 ), or (iii) chnging the orienttion of the loop with respect to the mgnetic field ( dθ / dt 0 ). Determine the sign of dφ / dt (is the flux incresing or decresing) (3) The sign of the induced emf is opposite the sign of dφ / dt. The direction of the induced current cn e found y using the Lenz s lw discussed in Section Prolem 1 Flling Loop g = 0 ˆk î ĵ = 0 l w z(t) v(t) = v z (t) ˆk z = 0 A rectngulr loop of wire with mss mg, width w, verticl length l, nd resistnce R flls out of mgnetic field under the influence of grvity. The mgnetic field is uniform nd out of the pper (! = î, > 0 within the re shown (see sketch) nd zero outside of tht re. At the time shown in the sketch, the loop is exiting the mgnetic field t! speed v(t) = v z (t) ˆk, where v z (t) < 0 (mening the loop is moving downwrd, not upwrd). Suppose t time t the distnce from the top of the loop to the point where the mgnetic field goes to zero is z(t) (see sketch). Question 1 Wht is the reltionship etween v z (t) nd z( t )? e creful of you signs here, rememer tht z( t ) is positive nd decresing with time, so dz( t) / dt < 0. v z (t) = dz(t) dt Solving 7-2

3 Prolem Solving Strtegy Step (1): Define A nd Compute Φ Question 2: f we define the re vector A to e out of the pge, wht is the mgnetic flux Φ through our circuit t time t (in terms of z( t ), not v z (t) ). Φ = wz(t) Prolem Solving Strtegy Step (2): Compute dφ / dt Question 3: Wht is dφ / dt? s this positive or negtive t time t? e creful here, your nswer should involve v z (t) (not z( t ) ), nd rememer tht v z (t) < 0. d dt Φ = d dt (wz(t)) = wv z (t) ecuse v z (t) < 0, the chnge in mgnetic flux is negtive. Prolem Solving Strtegy Step (3): Determine the sign of the induced emf (the sme s the direction of the induced current) Question 4: f dφ / dt < 0 then your induced emf (nd current) will e right-hnded with respect to A, nd vice vers. Wht is the direction of your induced current given your nswer to Question 3, clockwise or counterclockwise? Counterclockwise. Question 5: Lenz s Lw sys tht the induced current should e such s to crete selfmgnetic field (due to the induced current lone), which tries to keep things from chnging. Wht is the direction of the self-mgnetic field due to your induced current inside the circuit loop, into or out of the plne of the figure? s it in direction so s to keep the flux through the loop from chnging? The mgnetic field generted y the induced current is out of the pge, hence positive mgnetic flux since we choose out of the pge s positive. This opposes the decrese in externl mgnetic flux s more of the loop enters the region of zero externl mgnetic field. Solving 7-3

4 Question 6: Wht is the mgnitude of the current flowing in the circuit t the time shown (use = ε / R )? The mgnitude of the induced current is induced = wv z (t) R, Question 7: esides grvity, wht other force cts on the loop in the ± ˆk -direction? Give the mgnitude nd direction of this force in terms of the quntities given (hint: d F mg = d s ). There is mgnetic force directed upwrds (opposing the motion) on the top leg of the loop given y F mg = top leg d s = induced L = wv z (t) R ( wĵ) (î) = 2 w 2 R v z (t) ˆk. The force on the two side legs cncel nd there is no force on the ottom leg. Question 8: Assume tht the loop hs reched "terminl velocity"--tht is, tht it is no longer ccelerting. Wht is the mgnitude of tht terminl velocity in terms of given quntities? When the loop hs reched terminl speed, the sum of the forces on the loop re zero, hence F mg + F grv = 2 w 2 R v ˆk mg ˆk = 0. z,term So the terminl speed is v z,term = mgr 2 w 2 Question 9: Show tht t terminl velocity, the rte t which grvity is doing work on the loop is equl to the rte t which energy is eing dissipted in the loop through Joule heting (the rte t which force does work is F v )? ecuse the loops now moving t terminl speed, the rte tht grvity does work is given y F grv v = ( mg ˆk ) v y,term ˆk = mgv y,term = m2 g 2 R 2 w 2 Recll tht the terminl speed is v y,term = mgr nd the terminl induced current is 2 2 w Solving 7-4

5 therefore ind, term = wv z,term R = mg w. Therefore t terminl velocity, Prolem 2: Pie Shped Genertor F grv v = m2 g 2 R 2 w 2 = 2 ind, term R. A pie-shped circuit is mde from stright verticl conducting rod of length welded to conducting rod ent into the shpe of semi-circle with rdius (see sketch). The circuit is completed y conducting rod of length pivoted t the center of the semi-circle, (t point P ), nd free to rotte out tht point. This moving rod mkes electricl contct with the verticl rod t one end nd the semi-circulr rod t the other end. The ngle θ is the ngle etween the verticl rod nd the moving rod, s shown. The circuit sits in constnt mgnetic field! ext pointing out of the pge. ext P. ) f the ngle θ is incresing with time, wht is the direction of the resultnt current flow round the pie-shped circuit? Wht is the direction of the current in the circuit? The flux out of the pge is incresing, so we wnt to generte field into the pge (Lenz Lw). This requires clockwise current (see rrows eside pie shped wedge). For the next two prts, ssume tht the ngle θ is incresing t constnt positive rte, ω = dθ(t) / dt. ) Wht is the mgnitude of the rte of chnge of the mgnetic flux through the pie-shped circuit due to ext only (you my ignore the mgnetic field ssocited with ny induced current in the circuit)? Solving 7-5

6 d dt Φ = d dt d ( ext A) = ext π 2 θ dt 2π = ext 2 2 dθ dt = ext 2 ω 2 c) f the pie-shped circuit hs constnt resistnce R, wht is the mgnitude nd direction of the mgnetic force due to the externl field on the moving rod in terms of the given quntities. The mgnetic force is determined y the current, which is determined y the EMF, which is determined y Frdy s Lw: ε = d dt Φ = ext 2 ω = ε 2 R = ext 2 ω 2R F mg = ext = 2 3 ω ext 2R The force opposes the motion, which mens it is currently down nd to the left (the cross product of rdilly outwrd current with mgnetic field out of the pge). Solving 7-6

7 Prolem 3: Self-nductor Coxil Cle An inductor consists of two very thin conducting cylindricl shells, one of rdius nd one of rdius, oth of length h. Assume tht the inner shell crries current out of the pge, nd tht the outer shell crries current into the pge, distriuted uniformly round the circumference in oth cses. The z -xis is out of the plne of the figure elow right. long the common xis of the cylinders nd the r -xis is the rdil xis perpendiculr to the z -xis. h ˆk ˆr ) Use Ampere s Lw to find mgnitude of the mgnetic field (r) etween the cylindricl shells, s function of r for < r <. ndicte the direction of the mgnetic field on the sketch. The enclosed current enc in the Ampere s loop with rdius r is given y Applying Ampere s lw, d s counterclockwise direction. Then we otin (r) = 0, r < enc =, < r < 0, r > = (2π r) = µ 0 enc. Choose ˆ 0, r < µ 0 ˆθ, < r < (counterclockwise in the figure) 2πr 0, r > θ to point in the Solving 7-7

8 ) Wht is the mgnetic energy density u (r) s function of r for < r <? The mgnetic energy density for < r < is t is zero elsewhere. u = 2 = 1 µ 0 2µ 0 2µ 0 2πr 2 = µ 0 2 8π 2 r 2 c) Where is the energy density highest? The field exists only for < r <, nd u 1/ r 2. So, the energy density is highest right outside the inner shell (i.e. r = + ε with ε << ). d) Clculte the inductnce of this long inductor reclling tht U = 1 2 L 2 nd using your results for the mgnetic energy density in prt ). The volume element in this cse is 2πrhdr. The mgnetic energy is : U = u dvol = V µ 0 2 8π 2 r 2 2π h rdr = µ 2 h 0 4π Setting U = (1/ 2)L 2 yields for the self-inductnce is L = µ h 0 2π ln. ln. Note tht the self-inductnce hs the dimensions of µ 0 times length s it must ecuse flux divided y current hs the dimensions of mgnetic field time re divided y current, nd the mgnetic field hs the dimensions of µ 0 times current divided y length. Solving 7-8

9 e) Clculte the inductnce of this long inductor y using the formul Φ = L = da nd your results for the mgnetic field in prt ). To do open surfce this you must choose n pproprite open surfce over which to evlute the mgnetic flux. Does your result clculted in this wy gree with your result in prt c)? The mgnetic field is perpendiculr to rectngulr surfce shown in the ove figure. The mgnetic flux through thin strip of re da = ldr is Thus, the totl mgnetic flux is dφ = da = µ 0 2πr hdr ( ) = µ 0 h 2πr dr Φ = dφ = µ 0 h 2πr dr = µ 0 h 2π dr r = µ h 0 2π ln Thus, the inductnce is L = Φ = µ h 0 2π ln, which grees with tht otined in prt d). Solving 7-9

Version 001 HW#6 - Electromagnetism arts (00224) 1

Version 001 HW#6 - Electromagnetism arts (00224) 1 Version 001 HW#6 - Electromgnetism rts (00224) 1 This print-out should hve 11 questions. Multiple-choice questions my continue on the next column or pge find ll choices efore nswering. rightest Light ul