Chapter In Fig , the magnetic flux through the loop increases according to the relation Φ B. =12.0t

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1 Chapter In Fig , the magnetic lux through the loop increases according to the relation = 6.0t t where the lux is in milliwebers and t is in seconds. (a) What is the magnitude o the em induced in the loop at t=2.0s? (b) In what direction will the current low through the resistor. The magnitude o the induced em is given by = 6.0t t ε = d =12.0t ε(2) = = 31mV The lux is pointing out o the paper and rising. The coil reacts to oppose this rising outward ield by generating a ield that points into the paper. To do this, the current must low to the let through the resistor A small loop o area 6.8 mm 2 is placed inside a long solenoid that has 854 turns/cm and carries a sinusoidally varying current i o amplitude 1.28 A and angular requency 212 rad/s. The central axis o the loop and the solenoid coincide. What is the amplitude o the em induced in the loop The em generated can be calculated by inding the rate o change o magnetic lux through the inner loop that results rom the changing ield due to the solenoid B = µ 0 ni = µ 0 ni 0 sinω t = BA loop = µ 0 ni 0 sinω t A loop d == ω µ 0 n i 0 A loop cosω t ω µ 0 n i 0 A loop = 212 4π turns / m 1.28A m 2 = V 30.7 In Fig , a wire orms a closed circular loop, with radius R=2.0 m and resistance 4.0 X. The circle is centered on a long straight wire; at time t=0, the current in the long straing wire is 5.0A rightward. Thereater, the current changes according to i = 5A - (2A/s 2 )t 2. The straight wire is insulated; so there is no electrical contact between the current induced and the wire o the loop. What is the magnitude o the current induced in the loop at times t>0?

2 Since the ield lines encircle the straight wire, every ield line that passes through the circle upward also passes through the circle downward, so the net lux through the circle is always zero, no matter what the current is. The lux never changes, so the em induced is A rectangular coil o N turns and o length a and wih b is rotated at requency in a uniorm magnetic ield B as indicated. The coil is connected to co rotating cylinders against which metal brushes slide to make contact. (a) Show that the em matches the given expression. (b) What value o Nab gives an em such that its maximum value is 150V. Take the angular speed to be 60 rev/s and the ield to be 0.5 T. This is a classic induced em problem. You are being asked to calculate the expression that describes how a generator works. We begin by writing out the lux or the loop in the coniguration below (shown edge on). q B = BAcosθ = B Nabcosθ θ = ω t = 2π t = BAcos2π t Note that since the loop is rotating, we have substituted in or the angle in terms o the angular velocity and then requency. We can now ind the induced em = B Nabcos2π t ε = d d t = 2π B Nabsin(2π t) The maximum value o the em occurs when the sine is 1. We can now solve or the loop parameters. ε 0 = 2π B Nab Nab = ε 0 2π B = 150V 2π T = 0.796

3 30.15 In Fig , a sti wire bent into a semicircle o radius a = 2.0cm is rotated at constant angular speed 40 rev/s in a uniorm 20 mt magnetic ield. What are the (a) requency and (b) amplitude o the em in the loop. This problem is really the same as 30.13, except the area is a hal-circle. = BAcosθ = B Nabcosθ θ = ω t = 2π t = BAcos2π t = B π r2 2 cos2π t ε = d d t = 2π B π r2 2 sin(2π t) The requency is 40 Hz. 2π B π r2 2 = V I 50.0 cm o copper wire (diameter 1.00mm) is ormed into a circular loop and placed perpendicular to a uniorm magnetic ield that is increasing at the constant rate o 10mT/s, at what rate i thermal energy generated in the loop. We ind the area o the wire loop irst. 0.5m = 2rr 0.5m r = = 7.96 # 10-2 m 2r A loop = rr 2 = 1.99 # 10-2 m 2 Now that we know the area, we can compute the change in lux and induced em. = d{ B d db = (BAloop) = A loop = 1.99 # 10-4 Tm 2 The power will depend on the resistance o the wire. L R = t Cu = 1.69 # m Xm $ Awire r(0.5 # 10-3 = 1.08 # 10-2 X m) 2 P = R 2 = 3.67 # 10-6 W In Fig , a metal rod is orced to move with constant velocity v along two parallel metal rails connected with a strip o metal at one end. A magnetic ield o magnitude B=0.350T points out o the page. (a) I the rails are separated by L=25.0 cm and the speed o the rod is 55.0 cm/s, what em is generated? (b)i the rod has a resistance o 18 Ohms and the rails and

4 connectors have negligible resistance, what is the current in the rod? (c) At what rate is energy being transerred to the thermal energy The em generated is given by calculating the rate at which the lux is changing. In this problem, the lux changes because the area changes. { = BA = Blvt d{ = - = -Blv = $ 0.25m $ 0.55m/s = 0.048V Now that we know the em, we can compute the current. We can also compute the power. i = R = 0.048V = A 18X P = i 2 R = ( A) 2 $ 18X = 1.29 # 10-4 W In Fig , a long rectangular conducting loop o wih L, resistance R and mass m is hung in a horizontal uniorm magnetic ield B that is directed into the page and that exists only above line aa. The loop is then dropped during its all, it accelerates until it reaches a certain terminal speed vt. Ignoring air drag, ind the an expression or vt. The terminal velocity is reached when the upward orce on the top wire equals the weight o the loop. This orce arises because o the current running through this wire. We begin by writing an equation the sets the orces equal mg = ilb mg i = LB Now we need to ind the current. We do this by irst writing the em and then inding the current. = BLv t BLv i = R = t R The inal step is to set the current expressions equal and solve or the velocity BLv t mg R = LB mgr v t = L2 B 2

5 30.40 The inductance o a closely packed coil o 400 turns is 8.0 mh. Calculate the magnetic lux through the coil when the current is 5 ma. N{ L = i il 5 # 10 { = N = -3 A $ 8 # 10-3 H 400 = 1 # 10-7 Tm A 12 H inductor carries a current o 2A. At what rate must the current be changed to produce a 60V em in the inductor. di = L di 60V = L = 12H = 5A/s The switch in Fig is closed on a at time t=0. What is the ratio L/ o the inductor s sel-induced em to the battery s em (a) just ater t=0 and (b) at t = 2.00x L? (c) At what multiple o x L will L/ = 0.5? The equation or rising current in an RL circuit is i = R (1 - e -t/ ) From this expression, we can ind the voltage across the inductor. di 1 L = L = L(R $ $ e -t/ ) x L = R L L = e -t/ (a) At t=0 L = e -t/ L = 1 (b) At t = 2.00x L (c) The time or L/ = 0.5 is L = e -2/ = e -2

6 L 1 = 2 = e -t/ 1 t ln( 2 ) =- t =-x L ln( 2 1 ) A solenoid having an inductance o 6.30 nh is connected in series with a 1.2 kx resistor. (a) I a 14.0 V battery is connected across the pair, how long will it take or the current through the resistor to reach 80% o its inal value? b) What is the current through the resistor at time t = 1.0x L It s convenient to compute the time constant irst. x L = R L = 6.30 # 10-6 H = 5.25 # 10-9 s 1.2 # 103 X We can write the current in the RL circuit and ind the time or part (a) i = R (1 - e -t/ ) = is(1 - e -t/ ) i i = 0.8 = 1 - e -t/ s e -t/ = = 0.2 We can ind the current at one time constant. t - = ln(0.2) t =-x L ln(0.2) = 8.45 # 10-9 s i = R (1 - e -t/ 14V ) = 1.2 # 103 X (1 - e -/ ) = 7.37 # 10-3 A At t=0, a battery is connect to a series arrangement o a resistor and an inductor. I the inductive time constant is 37 ms, at what time is the rate at which the energy is dissipated in the resistor equal to the rate at which energy is stored in the inductors s magnetic ield.

7 P R = i 2 2 R = a R (1 - e -t/ ) k R du L du L d 1 di = (2 Li2 ) = Li = L(R (1 - e -t/ ))(R = P R L( R (1 - e -t/ e ))(R -t/ 2 ) = a R (1 - e -t/ ) k R 2 R (1 - e -t/ ) $ e -t/ = 2 R (1 - e -t/ ) 2 e -t/ - e -2t/ = 1-2e -t/ + e -2t/ 3e -t/ - 2e -2t/ = 1 2e -2t/ - 3e -t/ + 1 = 0 x = e -t/ 2x 2-3x - 1 = x = 2-4 $ 2 $ $ 2 = 2 1 e -t/ = 2 t =-x L ln( 2 1 ) e -t/ ) Note: The book has this as the answer or It s actually the answer to What must be the magnitude o a uniorm electric ield i it is to have the same energy density as that possessed by a 0.50 T magnetic ield? E 2 = 2n0 B 2 1 E = B = 1.5 # 10 8 V/m 0n 0

Exam 2 Solutions. PHY2054 Spring Prof. Paul Avery Prof. Pradeep Kumar Mar. 18, 2014

Exam 2 Solutions. PHY2054 Spring Prof. Paul Avery Prof. Pradeep Kumar Mar. 18, 2014 Exam 2 Solutions Prof. Paul Avery Prof. Pradeep Kumar Mar. 18, 2014 1. A series circuit consists of an open switch, a 6.0 Ω resistor, an uncharged 4.0 µf capacitor and a battery with emf 15.0 V and internal