{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

2000 Exam 3

# 2000 Exam 3 - Spring 2000 Physics 2220 Exam 3 Version 1 l A...

This preview shows pages 1–3. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Spring 2000, Physics 2220, Exam 3 - Version 1 l. A conducting circular loop is placed in a magnetic ﬁeld that is uniform and perpendicular to the loop plane at any time, but its magnitude B increases at the constant rate of 0.60 T/s. The radius of the loop is r = 2.0 m and the total resistance of the loop circuit is R = 80.0 9. Find the induced emf and the induced current. If the loop is replaced by one of much larger resistance, what effect does this have on the induced emf and induced current? ’1 541/ , 00342 A 2. A rectangular loop with width a = 25.0 cm and a slide wire with mass m = 15.0 g are placed in a uniform magnetic ﬁeld of magnitude B = 0.015 T and perpendicular to the loop as shown in the ﬁgure. The slide wire is given an initial speed v0 = 2.0 m/s and then released. Assume that there is no friction between the slide wire and the rest of the loop. Moreover, the resistance of immobile parts of the loop is considered negligible in comparison to the resistance R = 20.0 9 of the slide wire. Obtain an expression for the magnitude of the force exerted on the Wire while it is moving at speed v. Write an equation of motion for the sliding wire using Newton’s 2“Cl law, and solve this equation to ﬁnd the slide’s velocity as a function of time. To simplify the solution, you may assume that v changes negligibly from vO during the time period of interest. Determine the _ wire velocity after t = lnS s. . Lomgg'Ma (Au : 1,511/0’41413) 3. A 2.5 Q resistor and an inductor are connected in series across a 12 V battery at time t=0 5. At time F22 s, the current ﬂowing in the circuit is 3.1 A. Find the value of the inductor in H. 3.37 b. 4.35 © 5.30 d. 6.50 e 8.26 4. In the circuit described in Problem 3, the inductor is replaced with a 14 H inductor. Calculate how much energy in J 15 stored 1n the magnetic ﬁeld of the inductor after the current in the circuit reaches its maximum value. a. 489 b. 273 © 161 d. 61.7 e 26.7 5. A uniformly wound, air-ﬁlled, cylindrical solenoid has 250 turns, is 0.65 m long, and has a radius of 0.04 m. Assume that current in the solenoid produces a uniform magnetic ﬁeld in its interior. If the current in the solenoid is increasing at the rate of 25 NS, determine the self— induced emf in mV in the solenoid. a. 52.7 b. 78.4 c. 93.1 152 e. 32.0 Version 1 6. A series RLC circuit has a 120 Hz voltage source with an rms voltage of 180 V, a resistance of 150 9, an inductance of 200 mH, and a capacitance of 6 uF. Determine the maximum current in the circuit in A. a. 1.02 b. 0.78 Cc) 1.54 d. 2.67 e. 2.04 7. Calculate the average power in W delivered to the series RLC circuit in problem 6. a. 124 ® 177 c. 59.7 d. 231 e. 415 8. A generator produces an rms current of 15 A at 500 V. The voltage is stepped up to 6, 400 V by an ideal transformer and is transmitted a long distance through a power line of total resistance Of 150 (2. Determine the ratio of the power lost when the voltage 1S not stepped up to the power lost when it is stepped up; ‘ b 131 c 223 d' 278 c 76.6 9. The sun delivers ~1,000 W/m2 of electromagnetic ﬂux to the earth’s surface. The radiation is incident normal to the roof of a house and 60% is absorbed and 40% reﬂected. If the house roof dimensions are 24 m x 13 m, determine the radiation force on the roof in mN. a. 3.20 b. 3.68 c. 2.63 @ 1.46 e. 2.15 10. At What distance in m from a 200 W electromagnetic wave point source does Emax = 14 V/m? a. 11.2 b. 9.42 111354“ 2220 1533 _ 61 —~ a , L ‘€l~ 3% Wk?“ (1)“: KM ”SA \$18,“??? I L1 11!: “if H R‘mcreaSes, 8‘45 (MAW, 3'73“an ,, R 1 deems“ 2- F=UKB= IaB Mmh [fit éﬁf—Y’ Hm ¢5= {MM BA= 15m m 95%M1306k ”Bali I From NW5 ZMLMC ma ‘ F15 \$ W a? ' ” ”f U (‘4qu 5"3" "5 WW 43m ”Wm" mﬁm) or 3435:..A1r wave (2:?va '9‘ dV=—30(t =>ﬂmr Wm '27: cit =>- \\1 [Han Wm sm’fbgwﬁmw ”—525 I 112% él‘é 9712); \$ air-Ira ﬂ-lvgt 3;- Uﬁma'ﬂtm Whewt‘ﬂmu. WWW ' ﬂ ’t/lf '§‘ :1:— -- , “Rt 3.1400 e )Wem If}? MM: 12 ,> L'M.-Bg_) L ’8 Z- 4UL=~§LI “AMI-E \$- 5 151=Nd¢5 when: (b3: (i112 =3an MAB=}L0%I \$ RPMIZTW‘FES (1’- Ivmuc: V’gﬁ‘x We“. VW:EVW 04W! z=JKZ+O(L’)(c)z WW )(L:wL=2n¥L debzi‘zjfa Ji‘vm 1+(2ﬁ¥L'§n"I—FE)L FBM=I¢212 I or ”Raﬁ ‘Vms'l MS'COSLP Welue wap‘m XER Xe. :> 1%: 8. ﬁM=PM\$ V3.11: V,I 53>- 12: V L 'Phss‘ : IQZR E3? Ems, : (YE); Plow). ‘ I: R ‘ PIOSSZ‘ V. 9. ﬁ’wﬁé Pram: Z“ 1‘; «\$490125!!va [30M mﬂwd)’ 1M F’A'(¥Fnﬁl+("9P“h’) :: ﬂwﬁ .gﬁz-S—nyﬂ ~03 )3]: (W 0"”? ,— wﬁtere \$35M31: [000W W» 17- . 10. P=S-A= Em 4w ~=> V“ 313% Zyoc ...
View Full Document

{[ snackBarMessage ]}