Chapter 31 Solutions

Transcription

1 Chapter 31 Solutions 31.1 ε = Φ B 31.2 ε = Φ B = ( NBA) = 500 mv = ( B A) ε 1.60 mv = 1.60 mv and I loop = = = ma 2.00 Ω 31.3 ε = N BA cos θ = 25.0( T)π( 0.500m) E = mv 2 cos 180 cos s = NB π r 2 cos θ f cos θ i 31.4 (a) ε = dφ B = A db = AB max τ e t/τ ε = (0.160 m2 )(0.350 T) 2.00 s e 4.00/2.00 = 3.79 mv (c) At t = 0, ε = 28.0 mv 31.5 ε = N dφ B = (NBA) = 3.20 kv so I = ε = 160 A

2 Chapter 31 Solutions 219 Goal Solution A strong electromagnet produces a uniform field of 1.60 T over a cross-sectional area of m 2. A coil having 200 turns and a total resistance of 20.0 Ω is placed around the electromagnet. The current in the electromagnet is then smoothly decreased until it reaches zero in 20.0 ms. What is the current induced in the coil? G : O : A strong magnetic field turned off in a short time ( 20.0 ms) will produce a large emf, maybe on the order of 1 kv. With only 20.0 Ω of resistance in the coil, the induced current produced by this emf will probably be larger than 10 A but less than 1000 A. According to Faraday s law, if the magnetic field is reduced uniformly, then a constant emf will be produced. The definition of resistance can be applied to find the induced current from the emf. A : Noting unit conversions from F = qv B and U = qv, the induced voltage is L : ε = N d(b A) I = ε = 3200 V 20.0 Ω = 160 A ( ) cos 0 ( ) m2 ( ) = N 0 B i Acosθ T = s 1 N s/c m T 1 V C N m = 3200 V This is a large current, as we expected. The positive sign is indicative that the induced electric field is in the positive direction around the loop (as defined by the area vector for the loop) ε = N dφ B N(BA 0) = = NBA ε = NB(πr 2 ) = 500(0.200)π ( ) 2 ε = s 31.7 ε = d(ba) = µ 0 na di = V (a) I ring = ε = = 1.60 A B ring = µ 0 I 2r ring = 20.1 µt (c) Coil's field points downward, and is increasing, so B ring points upward 31.8 ε = d(ba) = µ 0 na di = µ 0 nπ r 2 2 I

3 220 Chapter 31 Solutions (a) I ring = ε = µ 0nπ r I B = µ 0I 2r 1 = µ 2 0nπ r 2 2 4r 1 I (c) The coil's field points downward, and is increasing, so B ring points upward (a) dφ B = B da = µ 0 I 2πx Ldx: Φ B = h+w µ 0 IL x=h 2π dx x = µ 0 IL 2π ln h + w h ε = dφ B = d µ 0 IL 2π ln h + w h = µ 0 L 2π ln h + w h di ( 4π 10 7 T ma) ( 1.00 m) ε = ln 2π A = 4.80 µv s The long wire produces magnetic flux into the page through the rectangle (first figure, above). As it increases, the rectangle wants to produce its own magnetic field out of the page, which it does by carrying counterclockwise current (second figure, above).

4 Chapter 31 Solutions Φ B = (µ 0 ni)a solenoid ε = N dφ B 2 = Nµ 0 n( πr solenoid ) di ( ) 600 A s = Nµ 0 n πr 2 solenoid ( )cos(120t) ε = 15.0( 4π 10 7 T ma) ( m)π m E = 14.2 cos(120t) mv ( ) A s ( )cos(120t) For a counterclockwise trip around the left-hand loop, with B = At d [ At(2a2 )cos 0 ] I 1 (5) I PQ =0 and for the right-hand loop, d [ At a2] + I PQ I 2 (3)=0 where I PQ = I 1 I 2 is the upward current in QP Thus, 2Aa 2 5(I PQ + I 2 ) I PQ =0 and Aa 2 + I PQ = I 2 (3) 2Aa 2 6I PQ 5 3 (Aa2 + I PQ )=0 I PQ = Aa2 23 upward, and since = ( Ω /m)( m) = Ω I PQ = ( T / s)(0.650 m) 2 23( Ω) = 283 µa upward ε = Φ B At t = 5.00 s, = N db A = N t ( )A ε = 30.0(0.410 T) [ π( m) 2 ] = 61.8 mv

5 222 Chapter 31 Solutions B = µ 0 ni = µ 0 n( 30.0 A) ( 1 e 1.60t ) Φ B = BdA = µ 0 n( 30.0 A) ( 1 e 1.60t ) da Φ B = µ 0 n( 30.0 A) ( 1 e 1.60t )π 2 ε = N dφ B = Nµ 0 n( 30.0 A)π2 ( 1.60)e 1.60t ε = (250)(4π 10 7 NA 2 )(400 m 1 )(30.0 A) [ π( m) 2 ] 1.60 s 1 e 1.60t ε = ( 68.2 mv)e 1.60t counterclockwise B = µ 0 ni = µ 0 ni max (1 e α t ) Φ B = BdA = µ 0 ni max (1 e α t ) da Φ B = µ 0 ni max (1 e α t )π 2 ε = N dφ B = Nµ 0 ni max π 2 αe α t ε = Nµ 0 ni max π 2 αe α t counterclockwise ε = d (NBl2 cos θ) = Nl2 B cos θ l = ε N B cosθ = ( V)(0.400 s) (50)( T T)cos(30.0 ) = 1.36 m Length = 4 ln = 4(1.36 m)(50) = 272 m

6 Chapter 31 Solutions 223 Goal Solution A coil formed by wrapping 50.0 turns of wire in the shape of a square is positioned in a magnetic field so that the normal to the plane of the coil makes an angle of 30.0 with the direction of the field. When the magnetic field is increased uniformly from 200 µ T to 600 µ T in s, an emf of 80.0 mv is induced in the coil. What is the total length of the wire? G : O : If we assume that this square coil is some reasonable size between 1 cm and 1 m across, then the total length of wire would be between 2 m and 200 m. The changing magnetic field will produce an emf in the coil according to Faraday s law of induction. The constant area of the coil can be found from the change in flux required to produce the emf. A : By Faraday s law, For magnitudes, ε = N dφ B = N d db (BAcosθ) = NAcosθ B ε = NA cosθ ε V and the area is A = = B N cosθ 50(cos 30.0 ) T T s = 1.85 m 2 Each side of the coil has length d = A, so the total length of the wire is L = N( 4d)= 4N A = (4)(50) 1.85 m 2 = 272 m L : The total length of wire is slightly longer than we predicted. With d = 1.36 m, a normal person could easily step through this large coil! As a bit of foreshadowing to a future chapter on AC circuits, an even bigger coil with more turns could be hidden in the ground below high-power transmission lines so that a significant amount of power could be stolen from the electric utility. There is a story of one man who did this and was arrested when investigators finally found the reason for a large power loss in the transmission lines! The average induced emf is given by ε = N Φ B Here N = 1, and Φ B =B(A square A circle ) with A circle = πr 2 = π(0.500 m) 2 = m 2 Also, the circumference of the circle is 2π r = 2π (0.500 m) = 3.14 m Thus, each side of the square has a length L = 3.14 m = m, 4 and A square = L 2 = m 2 So Φ B = (0.400 T)(0.617 m m 2 ) = T m 2 The average induced emf is therefore: ε = T m s = V

7 224 Chapter 31 Solutions In a toroid, all the flux is confined to the inside of the toroid. B = µ 0NI 2πr = 500 µ 0 I 2πr Φ B = BdA = 500 µ 0 I max 2π Φ B = 500 µ 0 I max 2π ε = N dφ B sinωt asinωtln b + dzdr r = µ 0 I max 2π ω aln b + cosωt ε = 104 2π 4π N A A 2 ( ) 377 rad s ( ) cm ( m )ln 4.00 cm cosωt = (0.422 V) cos ω t The field inside the solenoid is: N B = µ 0 ni = µ 0 l I Thus, through the single-turn loop N Φ B = BA solenoid = µ 0 l ( πr2 )I and the induced emf in the loop is ε = Φ B N = µ 0 l ( ) πr2 I = µ 0 Nπr2 l I 2 I ε = N dφ B I = N dφ B I = N dφ B Q = N Φ B = N AB ( f B i ) I = N dφ B 200 Q = 5.00 Ω ( m 2 )( ) T = C I = ε = Blv v = 1.00 m/s

8 Chapter 31 Solutions (a) F B = I 1 B = I lb. When I = E / and ε = Blv, we get F B = Blv (lb) = B2 l 2 v = (2.50)2 (1.20) 2 (2.00) = 3.00 N 6.00 The applied force is 3.00 N to the right P = I 2 = B2 l 2 v 2 = 6.00 W or P = Fv = 6.00 W *31.22 F B = IlB and E = Blv I = E = Blv so B = I lv (a) F B = I 2 l lv and I = F B v = A I 2 = 2.00 W (c) For constant force, P = F v = ( 1.00 N) ( 2.00 m / s)= 2.00 W The downward component of B, perpendicular to v, is ( T) sin 58.0 = T E = Blv = ( T) ( 60.0 m) ( 300 m / s)= V The left wing tip is positive relative to the right ε = N d BA cos θ = NB cos θ A ε = 1(0.100 T) cos 0 I = 1.21 V 10.0 Ω = A (3.00 m 3.00 m sin 60.0 ) (3.00 m) s = 1.21 V The flux is into the page and decreasing. The loop makes its own magnetic field into the page by carrying clockwise current ω = (2.00 rev/s)(2π rad/rev) = (4.00)π rad/s E = 1 2 Bω l2 = 2.83 mv

9 226 Chapter 31 Solutions (a) B ext = B ext i and B ext decreases; therefore, the induced field is B 0 = B 0 i (to the right). Therefore, the current is to the right in the resistor. B ext = B ext ( i) increases; therefore, the induced field B 0 = B 0 (+ i) is to the right, and the current is to the right in the resistor. (a) (c) B ext = B ext ( k) into the paper and B ext decreases; therefore, the induced field is B 0 = B 0 ( k) into the paper. Therefore, the current is to the right in the resistor. (d) By the Lorentz force law, F B = q (v B). Therefore, a positive charge will move to the top of the bar if B is into the paper. (c) (d) (a) The force on the side of the coil entering the field (consisting of N wires) is F = N( ILB)= N( IwB) The induced emf in the coil is ε = N dφ B = N d( Bwx) = NBwv, so the current is I = ε = NBwv counterclockwise. The force on the leading side of the coil is then: F = N NBwv wb = N 2 B 2 w 2 v to the left Once the coil is entirely inside the field, Φ B = NBA = constant, so ε = 0, I = 0, and F = 0. (c) As the coil starts to leave the field, the flux decreases at the rate Bwv, so the magnitude of the current is the same as in part (a), but now the current flows clockwise. Thus, the force exerted on the trailing side of the coil is:

10 Chapter 31 Solutions 227 F = N 2 B 2 w 2 v to the left again (a) Motional emf ε = Bwv appears in the conducting water. Its resistance, if the plates are submerged, is ρl A = ρw ab Kirchhoff's loop theorem says Bwv I Iρw ab = 0 I = Bwv + ρw ab = abvb ρ + ab w I sc = (100 m)(5.00 m)(3.00 m/s)( T) 100 Ω m = ma Look in the direction of ba. The bar magnet creates a field into the page, and the field increases. The loop will create a field out of the page by carrying a counterclockwise current. Therefore, current must flow from b to a through the resistor. Hence, Va Vb will be negative E = 1 2 Bω l2 = mv Name the currents as shown in the diagram: Left loop: + Bdv 2 I 2 2 I 1 1 =0 ight loop: + Bdv 3 I I 1 1 =0 At the junction: I 2 = I 1 + I 3 Then, Bdv 2 I 1 2 I 3 2 I 1 1 =0 I 3 = Bdv I So, Bdv 2 I 1 ( ) Bdv I =0

11 228 Chapter 31 Solutions v 2 3 v 3 2 I 1 = Bd upward 3 (4.00 m/s)(15.0 Ω) (2.00 m/s)(10.0 Ω) I 1 = ( T)(0.100 m) = 145 µa (5.00 Ω)(10.0 Ω)+(5.00Ω)(15.0 Ω)+(10.0Ω)(15.0 Ω) upward (a) db = 6.00t t ε = dφ B At t = 2.00 s, E = π 2 (db/) 2π r 2 = 8.00π (0.0250)2 2π (0.0500) F = qe = N clockwise for electron When 6.00t t = 0, t = 1.33 s db = t ε = dφ B At t = 3.00 s, E = π r1 2 db 2π r 1 = N/C perpendicular to r 1 and counterclockwise *31.34 ε = dφ B = πr 2 db = E d1 E(2π) = πr 2 db πr2, or E = 2π db B = µ 0 ni I = 3.00 e 0.200t db = µ 0 n di di = 0.600e0.200t At t = 10.0 s, E = πr2 ( 2π µ 0 n)(0.600e0.200t ) becomes E = ( m)2 2( m) (4π 10 7 N/A 2 )(1000 turns / m)(0.600)e 2.00 = N/C

12 Chapter 31 Solutions (a) E d1 = dφ B 2πrE = (πr 2 ) db so E = (9.87 mv/m) cos (100 π t) The E field is always opposite to increasing B. clockwise

13 230 Chapter 31 Solutions For the alternator, ω = 3000 rev min 2π rad 1 rev 1 min 60 s = 314 rad/s ε = N dφ B = 250 d [ ( T m 2 )cos(314 t /s)] = +250( T m 2 )(314/s) sin(314t) (a) ε = (19.6 V) sin(314t) ε max = 19.6 V (a) ε max = NABω = (1000)(0.100)(0.200)(120π) = 7.54 kv ε(t) = NBAω sin ωt = NBAω sin θ ε is maximal when sin θ = 1, or θ = ± π 2, so the plane of coil is parallel to B Let θ represent the angle through which the coil turns, starting from θ = 0 at an instant when the horizontal component of the Earth's field is perpendicular to the area. Then, ε = N d BA cos θ = NBA d cos ωt =+ NBAω sin ωt Here sin ω t oscillates between +1 and 1, so the spinning coil generates an alternating voltage with amplitude ε max = NBAω = NBA2π f = 100( T)(0.200 m) 2 2π rad (1500) = 12.6 mv 60.0 s B = µ 0 ni = ( 4π 10 7 T ma) ( 200 m 1 )( 15.0 A)= T For the small coil, Φ B = NB A = NBA cosωt = NB( πr 2 )cosωt Thus, ε = dφ B = NBπr 2 ω sinωt ε = ( 30.0) ( T)π( m) 2 ( 4.00π s 1 )sin 4.00πt ( )= mv sin 4. 00πt ( ) ( )

14 Chapter 31 Solutions As the magnet rotates, the flux through the coil varies sinusoidally in time with Φ B = 0 at t = 0. Choosing the flux as positive when the field passes from left to right through the area of the coil, the flux at any time may be written as Φ B = Φ max sinωt so the induced emf is given by I/I max ε = dφ B = ωφ max cosωt. -1 t/t = (ω t/2 π ) The current in the coil is then I = ε = ωφ max cosωt = I max cosωt (a) F = NI lb τ max = 2Fr = NI lwb = N m P = τω = (0.640 N m)(120π rad/s) P max = 241 W (about 1 3 hp) (a) ε max = BAω = B( 2 1 π2 )ω ε ε = ε max = ( 1.30 T) π m ε max = 1.60 V 2π 0 ε ( )2 4.00π rad s dθ = BAω 2π sinθ dθ = 2π 2π 0 0 ε Figure 1 (c) (d) (e) The maximum and average ε would remain unchanged. See Figure 1 at the right. See Figure 2 at the right. Figure (a) Φ B = BA cos θ = BA cos ωt = (0.800 T)( m 2 ) cos 2π (60.0)t = (8.00 mt m 2 ) cos (377t)

15 232 Chapter 31 Solutions ε = dφ B = (3.02 V) sin (377t) (c) I = ε = (3.02 A) sin (377t) (d) P = I 2 = (9.10 W) sin 2 (377t) (e) P = Fv = τω so τ = P ω = (24.1 mn m) sin2 (377t) At terminal speed, the upward magnetic force exerted on the lower edge of the loop must equal the weight of the loop. That is, ε Mg = F B = IwB = wb = Bwv t wb = B2 w 2 v t Thus, B = Mg w 2 v t = ( ) Ω ( kg) 9.80 m s 2 ( ) ( 1.00 m) 2 ( 2.00 m s) = T See the figure above with Problem ε (a) At terminal speed, Mg = F B = IwB = wb = Bwv t wb = B2 w 2 v t or v t = Mg B 2 w 2 (c) The emf is directly proportional to v t, but the current is inversely proportional to. A large means a small current at a given speed, so the loop must travel faster to get F m = mg. At given speed, the current is directly proportional to the magnetic field. But the force is proportional to the product of the current and the field. For a small B, the speed must increase to compensate for both the small B and also the current, so v t B 2. *31.46 The current in the magnet creates an upward magnetic field, so the N and S poles on the solenoid core are shown correctly. On the rail in front of the brake, the upward flux of B increases as the coil approaches, so a current is induced here to create a downward magnetic field. This is clockwise current, so the S pole on the rail is shown correctly. On the rail behind the brake, the upward magnetic flux is decreasing. The induced current in the rail will produce upward magnetic field by being counterclockwise as the picture correctly shows.

16 Chapter 31 Solutions F = ma = qe + qv B a = e i j k m [E + v B] where v B = = 200(0. 400)j + 200(0. 300)k a = [50.0j 80.0j k]= [ 30.0j k] a = [ j + 2k]m s 2 = ( j k)m s F = ma = qe + qv B so a = e [E + v B] where v B = m a = ( ) [2.50i j 4.00j] = ( )[2.50i j] i j k = 4.00j a = ( i j ) ms 2 *31.49 ε = N d ( BAcosθ )= N( πr2)cos0 db ε = ( 30.0)π( m) 2 1 ε = ( 30.0)π( m) 2 ( T) 2π ε = ( V)cos( 2π 523t s) () d [ 50.0 mt + ( 3.20 mt)sin ( 2π 523t s )] ( )cos 2π 523t s ( ) 523 s ( ) *31.50 (a) Doubling the number of turns. Amplitude doubles: period unchanged (c) Doubling the angular velocity. doubles the amplitude: cuts the period in half Doubling the angular velocity while reducing the number of turns to one half the original value. Amplitude unchanged: cuts the period in half

17 234 Chapter 31 Solutions *31.51 ε = N ( BAcosθ )= N( πr2)cos0 B = 1( m 2 )() T 5.00 T s = V (a) I = ε = V Ω = 43.8 A P = EI = ( V) ( 43.8 A)= 38.3 W In the loop on the left, the induced emf is ε = dφ B = A db = π( m)2 ( 100 T s)= π V and it attempts to produce a counterclockwise current in this loop. In the loop on the right, the induced emf is ε = dφ B = π( m) 2 ( 100 T s)= 2.25π V and it attempts to produce a clockwise current. Assume that I 1 flows down through the 6.00-Ω resistor, I 2 flows down through the 5.00-Ω resistor, and that I 3 flows up through the 3.00-Ω resistor. From Kirchhoff s point rule: I 3 = I 1 + I 2 (1) Using the loop rule on the left loop: 6.00 I I 3 = π (2) Using the loop rule on the right loop: 5.00 I I 3 = 2.25π (3) Solving these three equations simultaneously, I 1 = A, I 2 = A, and I 3 = A *31.53 The emf induced between the ends of the moving bar is ε = Blv = ( 2.50 T) ( m) ( 8.00 m s)= 7.00 V The left-hand loop contains decreasing flux away from you, so the induced current in it will be clockwise, to produce its own field directed away from you. Let I 1 represent the current flowing upward through the 2.00-Ω resistor. The right-hand loop will carry counterclockwise current. Let I 3 be the upward current in the 5.00-Ω resistor.

18 Chapter 31 Solutions 235 (a) Kirchhoff s loop rule then gives: V I 1 ( 2.00 Ω)= 0 I 1 = 3.50 A and V I 3 ( 5.00 Ω)= 0 I 3 = 1.40 A The total power dissipated in the resistors of the circuit is P = EI 1 + EI 3 = E( I 1 + I 3 )= ( 7.00 V) ( 3.50 A A)= 34.3 W (c) Method 1 : The current in the sliding conductor is downward with value I 2 = 3.50 A A = 4.90 A. The magnetic field exerts a force of F m = IlB = ( 4.90 A) ( m) ( 2.50 T)= 4.29 N directed toward the right on this conductor. A n outside agent must then exert a force of 4.29 N to the left to keep the bar moving. Method 2 : The agent moving the bar must supply the power according to P = F v = Fvcos0. The force required is then: F = P v = 34.3 W 8.00 m s = 4.29 N *31.54 Suppose we wrap twenty turns of wire into a flat compact circular coil of diameter 3 cm. Suppose we use a bar magnet to produce field 10 3 T through the coil in one direction along its axis. Suppose we then flip the magnet to reverse the flux in 10 1 s. The average induced emf is then ε = N Φ B = N [ BAcosθ] = NB πr 2 ε = ( 20) ( 10 3 T)π( m) 2 2 ( ) 10 1 s ~10 4 V cos180 cos I = ε + ε Induced and ε Induced = d (BA) F = m dv = IBd dv = IBd m = Bd m (ε + ε Induced) = Bd m (ε Bvd) To solve the differential equation, let u = (ε Bvd), du = Bd dv. 1 du Bd = Bd m u so u u 0 du u = t t=0 Integrating from t = 0 to t = t, ln u u 0 = (Bd)2 m e B2 d 2 t/m (Bd) 2 m Since v = 0 when t = 0, u 0 = ε and u = ε Bvd t or u u 0 =

19 236 Chapter 31 Solutions ε Bvd = εe B2 d 2 t/m and v = ε Bd (1 e B2 d 2 t/m )

20 Chapter 31 Solutions (a) For maximum induced emf, with positive charge at the top of the antenna, F + = q + (v B), so the auto must move east ε = Blv = ( T)(1.20 m) m 3600 s cos 65.0 = V I = ε = B A so q = I = (15.0 µt)(0.200 m) Ω = 1.20 µc Goal Solution The plane of a square loop of wire with edge length a = m is perpendicular to the Earth's magnetic field at a point where B = 15.0 µt, as shown in Figure P The total resistance of the loop and the wires connecting it to the galvanometer is Ω. If the loop is suddenly collapsed by horizontal forces as shown, what total charge passes through the galvanometer? G : For the situation described, the maximum current is probably less than 1 ma. So if the loop is closed in 0.1 s, then the total charge would be Q = I = ( 1 ma) ( 0.1 s)= 100 µc O : We do not know how quickly the loop is collapsed, but we can find the total charge by integrating the change in magnetic flux due to the change in area of the loop ( a 2 0). A : Q = I = ε = 1 dφ B = 1 dφ B = 1 d(ba) = B A 2 =0 A 1 =a 2 da A 2 =0 Q = B A A 1 =a 2 = Ba2 = ( T)(0.200 m) 2 = C Ω L : The total charge is less than the maximum charge we predicted, so the answer seems reasonable. It is interesting that this charge can be calculated without knowing either the current or the time to collapse the loop. Note: We ignored the internal resistance of the galvanometer. D Arsonval galvanometers typically have an internal resistance of 50 to 100 Ω, significantly more than the resistance of the wires given in the problem. A proper solution that includes G would reduce the total charge by about 2 orders of magnitude ( Q ~0.01 µc).

21 238 Chapter 31 Solutions *31.58 (a) I = dq = ε where E = N dφ B so Φ 2 dq = N dφ B Φ 1 and the charge through the circuit will be Q = N (Φ 2 Φ 1 ) Q = N BAcos0 BAcos π 2 = BAN so B = Q NA = (200 Ω)( C) (100)( m 2 ) = T (a) ε = B lv = V I = ε = A (c) (d) F B = IlB = N Since the magnetic flux B A is in effect decreasing, the induced current flow through is from b to a. Point b is at higher potential. N o. Magnetic flux will increase through a loop to the left of ab. Here counterclockwise current will flow to produce upward magnetic field. The in is still from b to a ε = Blv at a distance r from wire ε = µ 0 I 2πr lv (a) At time t, the flux through the loop is Φ B = BAcosθ = ( a + bt) ( πr 2 )cos0 = π( a + bt)r 2 At t = 0, Φ B = π ar 2 ε = dφ B = πr 2 da+ ( bt) = πbr 2 (c) I = ε = πbr2 (d) P = ε I = πbr2 ( πbr2 )= π 2 b 2 r 4

22 Chapter 31 Solutions ε = d db (NBA) = 1 π a2 = π a 2 K (a) Q = C ε = C π a 2 K B into the paper is decreasing; therefore, current will attempt to counteract this. Positive charge will go to upper plate. (c) The changing magnetic field through the enclosed area induces an electric field, surrounding the B-field, and this pushes on charges in the wire The flux through the coil is Φ B = B A = BAcosθ = BAcosωt. The induced emf is (a) ε = N dφ B = NBA d( cosωt) = NBAω sinωt. ( ) 30.0 rad s ε max = NBAω = 60.0( 1.00 T) m 2 dφ B = ε N, thus dφ B max (c) At t = s, t = rad = ε max N = 36.0 V 60.0 ( )= 36.0 V = V = Wb/s ω and ε = ε max sin 1.50 rad (d) The torque on the coil at any time is τ = µ B = NIA B = NAB When ε = ε max, sin ωt = and ε 2 τ = max ω = ( )= ( 36.0 V)sin( 1.50 rad)= 35.9 V ( 36.0 V) rad s ( )I sinωt = ε max ω ( )( 10.0 Ω) = 4.32 N m ε sinωt (a) We use ε = N Φ B, with N = 1. Taking a = m to be the radius of the washer, and h = m, Φ B = B 2 A B 1 A = A(B 2 B 1 )= π a 2 µ 0 I 2π(h + a) µ 0I 2π a = a2 µ 0 I h + a a = µ 0 ahi 2(h + a) The time for the washer to drop a distance h (from rest) is: = 2h g Therefore, ε = µ 0 ahi 2(h + a) = µ 0 ahi 2(h + a) g 2h = µ 0 ai 2(h + a) gh 2 and ε = ( 4π 10 T m/a) ( m) ( A) 2( m m) ( m/s )( m) = 97.4 nv Since the magnetic flux going through the washer (into the plane of the paper) is decreasing in time, a current will form in the washer so as to oppose that decrease. Therefore, the current will flow in a clockwise direction.

23 240 Chapter 31 Solutions ε = N dφ B ε = NBcosθ = N d ( BAcosθ) A ( ) cos 62.0 = T ( ) m s = 10.2 µv Find an expression for the flux through a rectangular area "swept out" by the bar in time t. The magnetic field at a distance x from wire is B = µ 0 I 2πx Φ B = µ 0Ivt 2π and Φ B = BdA. Therefore, r+l dx where vt is the distance the bar has moved in time t. x r Then, ε = dφ B = µ 0 Iv 2π ln 1 + l r The magnetic field at a distance x from a long wire is B = µ 0 I 2πx. flux through the loop. dφ B = µ 0 I 2πx (ldx) so Φ B = µ 0Il 2π r+w r dx x = µ 0 Il 2π ln 1 + w r Find an expression for the Therefore, ε = dφ B = µ 0 Ilv 2πr w (r + w) and I = ε = µ 0 Ilv 2πr w (r + w) As the wire falls through the magnetic field, a motional emf ε = Blv is induced in it. Thus, a counterclockwise induced current of I = ε = Blv flows in the circuit. The falling wire is carrying a current toward the left through the magnetic field. Therefore, it experiences an upward magnetic force given by F B = IlB = B 2 l 2 v. The wire will have attained terminal speed when the magnitude of this magnetic force equals the weight of the wire. Thus, B2 l 2 v t = mg, or the terminal speed is v t = mg B 2 l Φ B = (6.00t t 2 ) T m 2 and ε = dφ B Maximum ε occurs when dε Therefore, the maximum current (at t = 1.00 s) is = 18.0t t = 36.0t = 0, which gives t = 1.00 s. I = ε = ( )V 3.00 Ω = 6.00 A

24 Chapter 31 Solutions For the suspended mass, M: ΣF = Mg T = Ma For the sliding bar, m: ΣF = T I lb = ma, where I = ε = Blv Mg B2 l 2 v = (m + M)a or a = dv = Mg m + M B 2 l 2 v (M + m) v dv (α β v) = t where α = Mg 0 0 M + m and β = B 2 l 2 (M + m) Therefore, the velocity varies with time as v = α β (1 e β t ) = Mg B 2 l 2 1 e B 2 l 2 t/(m+m) *31.71 (a) ε = N dφ B = NA db = NA d (µ 0nI) where A = area of coil, N = number of turns in coil, and n = number of turns per unit length in solenoid. Therefore, ε = Nµ 0 An d 4 sin( 120πt) [ ] = Nµ0An(480π) cos (120π t) ε = 40(4π )[ π( m) ]( )(480π) cos(120π t) = (1.19 V) cos(120π t) I = V and P = VI = (1.19 V)2 cos 2 (120π t) (8.00 Ω) From cos 2 θ = cos 2θ, the average value of cos2 θ is 1 2, so P = 1 ( 1.19 V) Ω ( ) = 88.5 mw The induced emf is ε = Blv where B = µ 0I 2π y, v = v i + gt = ( 9.80 m s 2 )t, and y = y i 2 1 gt2 = m ( 4.90 m s 2 )t 2. ( ) 200 A ( ) ε = 4π 10 7 T ma 2π m ( 4.90 m s 2 )t 2 [ ] m ( )t = ( ) 9.80 m s2 ( )t [ t 2 ] V At t = s, ε = ( )( 0.300) [ ( 0.300) 2 V = 98.3 µv ]

25 242 Chapter 31 Solutions The magnetic field produced by the current in the straight wire is perpendicular to the plane of the coil at all points within the coil. The magnitude of the field is B = µ 0 I 2πr. Thus, the flux linkage is NΦ B = µ 0 NIL 2π h+w h dr r = µ 0 NI max L 2π ln h + w sin(ωt + φ) h Finally, the induced emf is ε = µ 0 NI max Lω 2π ln 1 + w h cos(ωt + φ) ( 4π 10 7 )(100)(50.0)(0.200 m)(200π s ε 1 ) 5.00 cm = ln 1 + cos(ωt + φ) 2π 5.00 cm ε = ( 87.1 mv)cos(200πt + φ) The term sin(ωt + φ)in the expression for the current in the straight wire does not change appreciably whenωt changes by rad or less. Thus, the current does not change appreciably during a time interval t < (200π s 1 ) = s. We define a critical length, ct = ( m/s)( s) = m equal to the distance to which field changes could be propagated during an interval of s. This length is so much larger than any dimension of the coilor its distance from the wire that, although we consider the straight wire to be infinitely long, we can also safely ignore the field propagation effects in the vicinity of the coil. Moreover, the phase angle can be considered to be constant along the wire in the vicinity of the coil. If the frequency ω were much larger, say, 200π 10 5 s 1, the corresponding critical length would be only 48.0 cm. In this situation propagation effects would be important and the above expression for ε would require modification. As a "rule of thumb" we can consider field propagation effects for circuits of laboratory size to be negligible for frequencies, f = ω 2π, that are less than about 10 6 Hz Φ B = BA cos θ I sin θ dφ B = ωba sinθ ; τ IB sin θ sin 2 θ The area of the tent that is effective in intercepting magnetic field lines is the area perpendicular to the direction of the magnetic field. This is the same as the base of the tent. In the initial configuration, this is A 1 = L(2L cos θ) = 2(1.50 m) 2 cos 60.0 = 2.25 m 2 After the tent is flattened, A 2 = L(2L) = 2L 2 = 2(1.50 m) 2 = 4.50 m 2 The average induced emf is: ε = Φ B = B( A) ( ) ( ) m2 = T s = 6.75 V