PHYS 1101 Practice problem set 12, Chapter 32: 21, 22, 24, 57, 61, 83 Chapter 33: 7, 12, 32, 38, 44, 49, 76

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1 PHYS 1101 Practce problem set 1, Chapter 3: 1,, 4, 57, 61, 83 Chapter 33: 7, 1, 3, 38, 44, 49, Vsualze: Please reer to Fgure Ex3.1. Solve: Because B s n the same drecton as the ntegraton path s rom to, the dot product o B and ds s smply Bds. Hence the lne ntegral B ds = Bds = B ds = B 0.50 m ( ) + ( 0.50 m ) = 0.10 T ( ) ( 0.50 m) = T m 3.. Model: The magnetc eld s that o a current lowng nto the plane o the paper. The current carryng wre s very long. Vsualze: Please reer to Fgure Ex3.. Solve: Dvde the lne ntegral nto three parts: B d s = B d s + let lne B d s + semcrcle rght lne B d s B s everywhere The magnetc eld o the current-carryng wre s tangent to clockwse crcles around the wre. perpendcular to the let lne and to the rght lne, thus the rst and thrd parts o the lne ntegral are zero. Along the semcrcle, B s tangent to the path and has the same magntude B = 0 I/πd at every pont. Thus B ds = 0 + BL + 0 = I 0 π d (πd) = I 7 (4π 10 T m / A) (.0 A) 0 = = T m where L = πd s the length o the semcrcle, whch s hal the crcumerence o a crcle o radus d Model: Assume that the solenod s an deal solenod. Solve: We can use Equaton 3.16 to nd the current that wll generate a 3.0 mt eld nsde the solenod: B solenod = 0 NI L Usng L = 0.15 m and N = 0.15 m m = 150, ( T ) 0.15 m I = 4π I = B solenod L 0 N ( ) ( ) 10 7 T m / A ( ) =.39 A ( ) 150 Assess: Ths s a reasonable current to pass through a good conductng wre o dameter 1 mm Model: The cols are dentcal, parallel, and carry equal currents n the same drecton. The magnetc eld s that o the currents n these cols. Solve: (a) The on-axs magnetc eld o an N-turn current loop at a dstance z rom the loop center s B loop = 0 NI ( z + ) 3/ 1

2 I the spacng between the loops s, then the mdpont between them s z = /. Snce the currents are n the same drecton, the eld o each loop s n the same drecton and the net eld s smply twce the eld due to a sngle loop. Thus (b) The magnetc eld s B = 0 NI = 1.5 ( 4 + ) 3/ ( ) 3/ 0 NI B = ( 0.716) 4π ( 10 7 T m / A)10( 1.0 A ) = T 0.05 m Model: The magnetc eld s that o the current whch s dstrbuted unormly n the hollow wre. Vsualze: Ampere s ntegraton paths are shown n the gure or the regons 0 m < r < 1, 1 < r <, and < r. Solve: For the regon 0 m < r < 1, B ds = 0 I through. Because the current nsde the ntegraton path s zero, B = 0 T. To nd I through n the regon 1 < r <, we multply the current densty by the area nsde the ntegraton path that carres the current. Thus, I I through = π ( ) 1 ( ) π r 1 where the current densty s the rst term. Because the magnetc eld has the same magntude at every pont on the crcular path o ntegraton, Ampere s law smples to B ds = B ds I r 1 ( ) = 0 = B πr ( ) ( 1 ) B = I 0 πr r 1 1 For the regon < r, I through s smply I because the loop encompasses the entre current. Thus, B ds = B ds = Bπr = 0 I B = I 0 πr Assess: The results obtaned or the regons r > and 1 < r < yeld the same result at r =. Also note that a hollow wre and a regular wre have the same magnetc eld outsde the wre Model: The magnetc eld s that o a coaxal cable consstng o a sold nner conductor surrounded by a hollow outer conductor o essentally zero thckness. Vsualze: Please reer to Fgure CP3.83. The sold nner conductor and the hollow outer conductor carry equal currents but n opposte drectons. The coaxal cable has perect cylndrcal symmetry. So the magnetc eld must be tangent to crcles that are concentrc wth the wre. Solve: (a) Ampere s law s B ds = 0 I through = B ds = B( πr) B = I 0 through πr For r < 1,

3 I through = I π 1 πr = Ir B = Ir 0 1 π 1 For 1 < r <, I through = I. Hence, B = 0 I πr. For r >, I through = 0 A and B = 0 T. (b) In the regon r < 1, B s lnearly proportonal to r; n the regon 1 < r <, B s nversely proportonal to r; and n the regon or r >, B = 0 T. A B versus r graph n the range r = 0 m to r = s shown below Model: Assume that the magnet s a bar magnet wth eld lnes pontng away rom the north end. Vsualze: As the magnets move, they create a change n the lux through the solenod, there wll be an nduced current and correspondng eld. Accordng to Lenz s law, the nduced current creates an nduced eld that opposes the change n lux. Solve: (a) When magnet 1 s close to the solenod there s lux to the let through the solenod. As magnet 1 moves away there s less lux to the let. The nduced current wll oppose ths change and produce an nduced current and a correspondng lux to the let. By the rght-hand rule, ths corresponds to a current n the wre nto the page at the top o the solenod and out o the page at the bottom o the solenod. So, the current wll be rght to let n the resstor. (b) When magnet s close to the solenod the dvergng eld lnes o the bar magnet produce a lux to the let n the let hal o the solenod and a lux to the rght n the rght hal. Snce the lux depends on the orentaton o the loop, the lux on the two halves have opposte sgn and the net lux s zero. Movng the magnet away changes the strength o the eld and lux, but the total lux s stll zero. Thus there s no nduced current Model: Assume the eld s unorm. Vsualze: Please reer to Fgure Ex33.1. The moton o the loop changes the lux through t. Ths results n an nduced em and current. Solve: The nduced em s E = dφ/ and the nduced current s I = E /. The area A s changng, but the eld B s not. Take A as beng out o the page and parallel to B, so Φ = AB and dφ/ = B(dA/). The lux s through that porton o the loop where there s a eld, that s, A = lx. The em and current are E = B da = B d( lx) = Blv = ( 0.0 T )( m) ( 50 m / s) = 0.50 V I = E = 0.50 V 0.10Ω = 5.0 A The eld s out o the page. As the loop moves the lux ncreases because more o the loop area has eld through t. To prevent the ncrease, the nduced eld needs to pont nto the page. Thus, the nduced current lows clockwse. Assess: Ths seems reasonable snce there s rapd moton o the loop. 3

4 33.3. Model: Snce the solenod s arly long compared to ts dameter and the loop s located near the center, assume the solenod eld s unorm nsde and zero outsde. Vsualze: Please reer to Fgure P33.3. The solenod s magnetc eld s perpendcular to the loop and creates a lux through the loop. Ths lux changes as the solenod s current changes, causng an nduced em and correspondng nduced current. Solve: Usng Faraday s law, the nduced current s I loop = E loop = 1 dφ = A sol db sol = πr sol d 0 NI l = π r sol 0 N l We have used the act that the eld s approxmately zero outsde the solenod, so the lux s conned to the area A sol = πr sol o the solenod, not the larger area o the loop tsel. The current s constant rom 0 s t 1 s and s t 3 s, so di/ = 0 A/s and I loop = 0 A durng these ntervals. The current s changng at the rate di/ = 40 A/s durng the nterval 1 s t s, so the current durng ths nterval s I loop = π( m ( ) 4π 10 7 T m / A) ( 100) ( 40 A / s) = A = 158 A ( 0.10 Ω )( 0.10 m ) Durng the rst hal o ths nterval, rom 1.0 s to 1.5 s, the eld s to the rght and decreasng. To oppose ths decrease, the nduced eld must pont to the rght. Ths requres an nduced current comng out o the page at the top o the loop a postve current wth the sgn denton gven n the problem. Durng the second hal o ths nterval, rom 1.5 s to.0 s, the eld s to the rght and ncreasng. To oppose ths ncrease, the nduced eld must pont to the rght. Ths also requres a postve nduced current comng out o the page at the top o the loop. di Assess: The nduced current s proportonal to the negatve dervatve o the solenod current Model: Assume the eld due to the solenod s unorm nsde and vanshes outsde. Vsualze: The changng current n the solenod produces a changng eld and lux through the col. Ths changng lux creates an nduced em n the col. Solve: The lux s only nonzero wthn the area o the solenod, not the entre area o the col. The nduced current n the col s I col = E col = 1 N col = N πr col s 0 N s l s dφ di s = 1 N col A s db s = 1 N colπ r s d = N π r col s 0 N s ( π I 0 ) cos( πt) l s 0 N s I s l s The maxmum col current occurs when the cosne s +1. The maxmum current n the solenod s I col, max l s I 0 = N col π r s 0 N s ( π ) = ( 0.0 A )( 0.40Ω) ( 0.0 m ) 40π( m ) 4π 10 7 T m / A ( )π ( 60 Hz ) = 5.97 A Assess: Ths s a large current, but the nduced current s large as well. ( ) Model: Assume there s no resstance n the rals. I there s any resstance, t s accounted or by the resstor. Vsualze: Please reer to Fgure P The movng wre wll have a motonal em that produces a current n the loop. Solve: (a) At constant velocty the external pushng orce s balanced by the magnetc orce, so F push = F mag = IlB = E Blv lb = lb = B l v = 0.50 T ( ) ( 0.10 m ) ( 0.50 m / s).0 Ω = N 4

5 (b) The power s P = Fv = ( N )( 0.50 m / s) = W (c) The lux s out o the page and decreasng and the nduced current/eld wll oppose the change. The nduced eld must have a lux that s out o the page so the current wll be counterclockwse. The magntude o the current s (d) The power s I = Blv = ( 0.50 T )( 0.10 m )( 0.50 m / s) = A.0 Ω P = I = ( A ) (.0 Ω) = W Assess: From energy conservaton we see that the mechancal energy put n by the pushng orce shows up as electrcal energy n the resstor Model: Assume the magnetc eld s unorm over the loop. Vsualze: Please reer to Fgure As the wre alls, the lux nto the page wll ncrease. Ths wll nduce a current to oppose the ncrease, so the nduced current wll low counterclockwse. As ths current passes through the slde wre, t experences an upward magnetc orce. So there s an upward orce a retardng orce on the wre as t alls n the eld. As the wre speeds up the retardng orce wll become larger untl t balances the weght. Solve: (a) The orce on the current-carryng slde wre s F m = IlB. The nduced current s I = E = 1 dφ = 1 d AB = B d Blv lx = Consequently, the retardng magnetc orce s F m = l B v = l B v The mportant pont s that F m s proportonal to the speed v. As the wre begns to all and ts speed ncreases, so does the retardng orce. Wthn a very short tme, F m wll ncrease n sze to where t matches the weght w = mg. At that pont, there s no net orce on the loop, so t wll contnue to all at a constant speed. The condton that the magnetc orce equals the weght s (b) The termnal speed s l B v term = mg v term = mg l B ( ) 0.10Ω ( kg) 9.80 m / s ( ) v term = = 0.98 m / s ( 0.0 m ) ( 0.50 T ) Vsualze: Please reer to Fgure CP The area wthn n the loop s changng so the lux wll change. Ths wll produce an nduced em and correspondng current. The area o the loop s the area o a square. Solve: (a) The eld s out o the page, so the lux s ncreasng outward as the loop expands. Accordng to Lenz s law, the nduced current tres to prevent the ncrease o outward lux. It does so by generatng a eld nto the page. Ths requres a clockwse current low. (b) Take A parallel to B, so Φ = AB and E = dφ/ = B(dA/). The eld s constant but the loop area s changng. Let the length o each edge o the loop be x at tme t. Ths length s ncreasng lnearly wth tme as the corner o the loop moves outward wth speed v = 10 m/s. We have The loop s area at tme t s x = x 0 + v x t = 0 + [(10 m/s)cos45 ]t = 7.07t m 5

6 Consequently, the nduced em at tme t s A = x = (7.07t m) = 50t m da = 100t m / s E = B da = (0.1 T)(100t m / s) = 10t V The nduced current s I = E/, where s the resstance o the loop. The wre has resstance 0.01 Ω per meter, so = (0.01 Ω/m)l where l = 4x s the permeter o the square at tme t. That s, Thus, the nduced current s = 4(0.01 Ω/m)(7.07t m) = 0.8t Ω I = E = 10t V 0.8 t Ω = 35.5 A (c) At t = 0.1 s, E = 1.0 V and I = 35.5 A. Assess: The nduced em depends on tme, but so does the resstance. Ths means the nduced current s constant. 6