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

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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 points A conducting loop (octgonl) round mgnetic field (circulr) contins three lightuls (leled A,, nd C). The wires connecting the uls re idel, with no resistnce. The loop is ground t one point s shown in the figure. The mgnetic field is incresing rpidly. A nk order the rightness of the three uls, from rightest to lest right. 1. A > > C 2. A = C > 3. A > C > 4. A > = C 5. None of these correct 6. > C > A Thecurrentintheloopisthesmethroughout the loop. Consequently, ll the lightuls urn t the sme rightness A = = C. ince the ground mintins reltive potentil, it provides nothing of consequence to this rgument. C Flt Coil of Wire points A flt coil of wire consisting of 32 turns, ech with n re of 53 cm 2, is positioned perpendiculrly to uniform mgnetic field tht increses its mgnitude t constnt rte from 1.9 T to 14 T in 1.9 s. If the coil hs totl resistnce of 0.79 Ω, wht is the mgnitude of the induced current? Correct nswer: A. keywords: E = dφ dt Φ = N d A = N A E = N A( 2 1 ) t I = E = N A( 2 1 ) t = A. Holt F 22ev points A 53-turn coil with n re of m 2 is dropped from position where = 0.0 T to new position where = 0.67 T. The displcement occurs in 0.37 s nd the re of the coil is perpendiculr to the mgnetic field lines. Wht is the resulting verge emf induced in the coil? Correct nswer: V. sic Concept: E = N A(cosθ) = NA(cosθ)

2 Version 001 HW#6 - Electromgnetism rts (00224) 2 Given: olution: so N = 53 A = m 2 i = 0 T f = 0.67 T = 0.37 s θ = 0 = f i = 0.67 T 0 T = 0.67 T E = NA(cosθ) = (53)( m 2) (cos0 )(0.67 T) (0.37 s) = V so the mgnetic field is = E N πr 2 ( s)(0.215 V) = (363)π(0.074 m) µt T = µt. ectngulr Loop points A rectngulr conducting loop of wire is pproximtely hlf-wy into mgnetic field (into the pge) nd is free to move. uppose the mgnetic field egins to increse rpidly in strength. c erwy CP points A 363-turn circulr-loop coil 14.8 cm in dimeter is initilly ligned so tht its xis is prllel to Erth s mgnetic field. In 3.64 ms thecoil isflipped so tht itsxisisperpendiculr to Erth s mgnetic field. If n verge voltge of V is therey induced in the coil, wht is the vlue of Erth s mgnetic field t tht loction? Correct nswer: µt. Given : N = 363 turns, r = 7.4 cm = m, t = 3.64 ms = s, nd E = V. The chnge of the mgnetic flux is nd Φ = Φ f Φ i = πr 2 0 = πr 2 E = N Φ = N πr2, 0 x 0 Wht hppens to the loop? 1. The loop will rotte. 2. The loop doesn t move. 3. The loop is pushed upwrd, towrds the top of the pge. 4. The loop is pushed downwrd, towrds the ottom of the pge. 5. The loop is pushed to the right, wy from the mgnetic field. correct 6. The loop is pushed to the left, towrd the mgnetic field. Mgnetic flux is defined s: Φ = A d

3 Version 001 HW#6 - Electromgnetism rts (00224) 3 Φ totl = N Φ = N [ A] The flux is incresing in the loop. To oppose this increse, the induced mgnetic field needs to point out of the pge. This requires counter-clockwise induced current ( down to ). Using the right-hnd rule, the mgnetic force on the current in the left edge of the loop is to the right, wy from the field. The mgnetic forces on the top nd ottom segments of the loop re in opposite directions nd cncel ech other. erwy CP points A solenoid 3.93 cm in dimeter nd 25.6 cm long hs 301 turns nd crries current of 18.6 A. Find the mgnetic flux through the circulr cross-sectionl re of the solenoid. Correct nswer: T m 2. Let : µ 0 = T m/a, r = cm = m, l = 25.6 cm = m, N s = 301 turns, nd I = 18.6 A. The mgnetic field strength is = µ 0 ni = µ 0N s I l s nd since θ = 0, the mgnetic flux is Φ = A cosθ = µ 0N s I(πr 2 ) l s = ( T m/a) (301 turns)(18.6 A)π( m) m = T m 2. keywords: AP EM 1993 MC points Inthe figure shown, the mgnet is first moved downwrd towrd the loop of wire, then withdrwn upwrd from the loop of wire. Clockwise induced current I N down then up Counterclockwise I induced current As viewed from ove, the induced current in the loop is 1. for oth cses clockwise with incresing 2. for oth cses counterclockwise with decresing 3. first clockwise, then counter-clockwise. correct 4. first counter-clockwise, then clockwise. 5. for oth cses counterclockwise with incresing 6. for oth cses clockwise with decresing From Ohm s lw nd Frdy s lw, the current in mgnitude is I = V = 1 dφ dt, whereφisthemgneticfluxthroughtheloop. We know the sign of the rte of chnge of the mgnetic flux is chnged when the mgnet is withdrwn upwrd, s is the current direction ccording to the ove eqution. Using the right-hnd-rule nd from Lenz s lw, we know tht when the mgnet is first moved downwrd towrd the loop of wire, then withdrwn upwrd from the loop of wire, the current in the loop

4 Version 001 HW#6 - Electromgnetism rts (00224) 4 is first clockwise, then counter-clockwise, s viewed from ove. Induced Current Direction (prt 1 of 2) 10.0 points ε Introduce the following nottions: A) After is closed, the direction of the mgnetic field in the cylinder is A1 : to the left A2 : to the right A3 : upwrd A4 : downwrd A5 : into the pper ) The direction of the current induced in is 1 : from through to 2 : from through to Wht is correct immeditely fter the switch in the figure is closed? 1. A4, 1 2. A3, 1 3. A1, 1 4. A5, 2 5. A2, 1 6. A3, 2 7. A4, 2 8. A1, 2 9. A5, A2, 2 correct Whentheswitchisopen,thereisnomgnetic field. At the moment the switch is closed, the mgnetic flux through the circuit increses from zero, nd the mgnetic field is directed fromlefttoright,sotheinducedcurrent must produce mgnetic field pointing to the left. 009 (prt 2 of 2) 10.0 points I Introduce the nottions elow: A) The direction of the induced mgnetic field within the rectngulr circuit is A1 : into the pper A2 : out of the pper A3 : prllel to I A4 : ntiprllel to I A5 : rdilly outwrd nd I ) The direction of the current induced in is 1 : from through to 2 : from through to Wht is correct when the current I in the figure decreses? 1. A3, 1 2. A4, 1 3. A1, 1 correct 4. A4, 2 5. A2, 2 6. A2, 1 7. A5, 2

5 Version 001 HW#6 - Electromgnetism rts (00224) 5 8. A1, 2 9. A5, A3, 2 The mgnetic field due to the current is directed into the plne through the circuit. When the current decreses, the mgnetic flux lso decreses, so the induced current must produce mgnetic field to oppose tht decrese; i.e., into the plne. erwy CP (prt 1 of 2) 10.0 points A coil is suspended round n xis which is coliner with the xis of r mgnet. The coil is connected to resistor with ends leled nd. The r mgnet moves from left to right with North nd outh poles leled s in the figure. right ( induced = ) to resist ny chnge of mgnetic flux in the coil (Lenz s Lw). 011 (prt 2 of 2) 10.0 points Wht is the direction of the induced current in resistor when the r mgnet is moving from left to right? 1. from å through to (I ) correct 2. from through to å ( I) 3. the induced current is zero mperes The helicl coil when viewed from the r mgnet winds round the solenoid from terminl counter-clockwise. As the induced field is left to right ( induced = ), the induced current must flow counter-clockwise nd therefore it goes from å through to (I ). N v Wht is the direction of the induced mgnetic field in the coil when the r mgnet is moving from left to right? 1. left to right ( induced = ) correct 2. the induced field is zero tesl 3. right to left ( = induced ) The induced mgnetic field depends on whether the flux is incresing or decresing. The mgnetic flux through the coil is from left to right. When the mgnet moves from left to right, the mgnetic flux through the coils decreses. The induced current in the coil must produce n induced mgnetic field from left to