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Questions 1. Two blocks of masses m,=5 kg and m>=3 kg linked by a string lye on a horizontal frictionless table. Both blocks are connected by a string to the center of the table. my is at a distance r,=2 m from the center and m, at r=3 m. If the maximum tension that the string can achieve before breaking is T=76 N, determine the maximum angular velocity w at which both blocks can turn around the center of the table. yan e E 4 Up ed a A 1 E se É aro 4=É y? E AL po a a 2 e 2 > A, da ¿aa > 38 27 E e = mw E Es Ine 4 frsC abel. 47 le ze mE > E 7 Mw ay 2444 = Tux = 0) WE q ul 17 PTOS ERA 2. Let us consider the set of charges of the figure. All of them are at the same distance from the point (P). What is the relationship between the charges values so that the net electric field E at P is zero? 6) LEA 27 =-%=- 43 210 Q E t=07 O) e Q A E | 40% E A ad E 3 o 550 | e Ed q sE¿=0 3. A Wire is curved forming a semicircle of radius R and it is centered at the origin. The linear charge density of the wire is 2=A8”. The electric potential at the origin is: dí: UE. A MEDER polFPI * PERES fa ade ARAS) , Y mR= Ur a La E => yz 4 pre 2 pers Ss A . E A AE | AE ere, HE)”. MIE da Ml 31748, 4. Two equal point charges Q=2 C are located at the pointsí0,a/2,0) and (0,-a/2,0). The electric flux through E triangle formed by the vertices (a,0,0), (0,a,0) and (0,0,a) is (Note: an octahedral is formed by eight Edo = S ZA. ul al, As He P.M E E pira x ión a 17 e a cd O = 0oseyresxfo? > 506x074 m> —_———. 5. Let us consider a parallel plate capacitor with a capacitor capacitance Co. The area each plate is A. 1f 2/3 of the space between the plates is filled with a dielectric with a relative permittivity e,=5, the new capacitor capacitance is: 7% canautor ou Le conndetd y" O commokd ml Jenes, 3 He EA e 4 % E 2 7 1 Je e, Je E Dl “eS 4: Es A 2 E > Ur e ela? O 2d/3 d at e Al Ir Va - 2 (24) a (21 $)- — ds (- É6A A = 4 3106. A Dl: 2 (es 2 E > da TONE] NO 6. Let us consider a cylindrical copper wire of radius r=0.654 mm, length 1=2m carryipe an electric current 120.5 A. The energy dissipated by Joule's effect after 3 minutes is (Datum: pcu=1,7:10 Qm): ESPA Tie distal /tmo.- Buses RI E, A= 12 Uliejaleg = RICC 7 AL R= t£ > UL= Plz -. = LE ESE ara xfa ÍabO, A car (0.6540 3)? = MISCIUIANOS Ads geo» ly AMY 9. Let us consider a conducting wire with a length L. Itis made a clos 1 1 i . . sed loop with the wire. If th dipolar moment of the closed loop is maximum, how should the loop be? s 4 ¿spain moZA > AL pe Dago try - 4 E ud as 0 de E > A,= [E] 2 M,= Més 7% 7 E 5% Lula Loops A Hs E ES mM 242 Losps : E ae 1d y turn carcubo A 'he fundamental frequency of the string is 10. A string of length L= 2 m is attached to the ends of a wall. T length A=1 m. The order of the harmonic v=100 Hz. A standing wave is observed in the string with a wave propagating in the string and the wave speed are A pa, 2 A ÍA a Ss 2x2 -4 pe domo ció 4 O E E dy Y ME eS E AS Pl > V= e Ax Yx d00= Yoo m5 pta A Thorifos n= Y, y= wo m/6 Problem 1 (20 points) A. parallel plate capacitor has square plates of side L=20 cm pda a distance d=4 mm. The capacitor is connected to a battery with a potential difference AV=100 V, as itis shown in the figure. a) Determine the surface charge density o acquired by the plates of the capacitor. - b)Find the potential at a point P located at a distance x=3 mm from the upper plate. Y =100V => V= loo — un |, Amen ' 2 Le |. Ama, A E A dielectric slab of Plexiglas (e;=3.4) with a ihickness w=2 mm is introduced in the middle of the capacitor. c) Determine the capacitor capacitance of the capacitor and the surface charge density o”. d) Determine the electric field E in the regions 1, 2 and 3.. €) Find the electrostatic energy stored in the capacitor. a) e poralel - flile- ayudo” > Ls Y A Pes Cade: > Es. 0 TA E > 2-54 2 %- Sa AV Tata dor bulo 530 E EL. as! sd T= sore ts ¿coge he e 02.24 0 Up? 0» 2214) = 22138 N 4/m? 3/ ue f He ut y El PS quee, E= Ca Es — la a 05 > po E- ar | pl : Vila a a e 7) ; - Er e ito Pe E E > E E MO, > = dE EE És a E q.” A S LE q Maciel Et Ls ES - 386363 E VI 8-£54]4x lo Me ¿q e he. $ d0 T 2 (1368. 39484 10 Pxp= 631 a : U,= cuero + Problema 2 (20 puntos) Un cilindro muy largo de radio 3 cm conduce una intensidad de corriente de 4 A distribuida uniformemente. Un hilo rectilineo infinito paralelo al cable y de radio despreciable está situado a 12 cm y conduce la misma corriente pero en sentido contrario. Vista desde arriba a A y T 3 cm a] => 154 A E l hacia adentro > y EP Xx 12 cm , I hacia afuera a) Obtenga el campo magnético B en el punto P¡(-1.5 cm, 0) b) Obtenga el campo magnético B en el punto Pa(6 cm, 4cm) c) Obtenga la fuerza F que ejerce el cilindro sobre el hilo por unidad de longitud Fuera del cilindro: a , o Ls B. do ss ; BPE digo dl Exg nr a. ds Mu rrano a q * Milo. Si susppre tamos fruua r> radio eel lulo Ga ta Aespraaste ET hada fura olas papel : > sn ) Si x IB]. A - pto, gol” JE 2.n-0,03 2p-003* A =4,B-4%(2) T Da Lula douda E ¿clar umslaito dbicec. y auch + B uh, E y Fecia Aute, del pod HS. = 43, E a ke YT E lo7. Ansa, Y 201 $925. Jo? A) ¿nr 20-13, lo 13,510? Ep = $, 785 JD ELIT Priucipoio ae Aupa pr cad : R -S E E=5, -. 8, “A o): ” e Lab Pr $ A A E DO e poda TP (Ca, da) Bai Bul a y 3 6 3 A E y pos ue. TI bane aduuls o e Bono 2 práouga ce de le le qu une / | lo pia - Problem 3 (20 points) Two wires are connected by two conducting bars of length 10 cm. The bars move with a constant speed vi=0.1 m/s and v,=0.2 m/s (see figure). In the region, there is a homogeneous and constant magnetic field B=10 mT perpendicular to the plane of motion, as itis indicated in the figure. < do ip xxx xx xxxx Y, A Xx xx xx xx | | a) If at t=0 the bars are d¿=10 cm apart, find the magnetic flux through the closed loop formed by the wires and the bars for any time. b) Calculate the electromotrive force induced at any time. c) If the resistivity of the bars is p=1.7:10* Qm (the resistivity of the wires is negligible) and its cross-section area is S=10% m?, find the electric current flowing through the loop indicating the flowing direction. d) Determine the force that is necessary to exert on the bars so that they can move at a constant velocity. a) The magnetic flux is: 9=[[ Bas As the magnetic field is homogeneous, perpendicular to the surface and constant, the magnetic flux expression ca be written as: 9 =[[[F-4S = BS As the magnetic field is known, it is only necessary to calculate the area defined by the wires and the bar. The area ¡is (rectangle area) just height times (I=10 cm) the width, which is time-varying. At t=0, the width ¡is dy. The width increases on increasing time as the left and right bars move with v1= 0.1 and v2=0.2 m/s, respectively. The calculation can also be done if the relative velocity of one of the bars is considered with respect to the other one, v=v,-v,=0.1 m/s, (left bar is at rest and the right one moves with v). Acoording to this the magnetic flux is: $ = BS = Bl(d, +vt) Ifthe values of the different parameters are taken into account: $=BS= BL(d, +vt)=10107?x0.1x(0.1+0.11) =10*+10*7 Time is in second and the flux is in weber (Wb). b) The electromotive force is determined from the Faraday's law. d d =L4=-(10*+10*1)=-10*V E En e ) c) Now we determine the electric current intensity flowing through the closed loop. For it, we need to find determine the resistance of the bars. For a bar wit a resistivity p, cross-section S and length d, the resistance is: E 0.1 R=pÍ=1.710*= E 10% 1.7100 The close loop consists of two bars connected in series, so the net resistance is two time the resistance of one bar. Therefore R;yr=3.4-10%0. Once the net resistance is known, the electric current can be obtained from the Ohm's las as: El 10% LS El HE pa R 3.4107 TOY =0.02944 = 29.4mA The electric current flows in the direction such as the magnetic field due to it try to compensate the changes in the magnetic flux associated to the bars motion. On increasing the time the magnetic flux increases, so the induced electric current must originate a magnetic field opposite to the magnetic field that already exists. The 10 mT magnetic field is directed to-k, so the induced electric current originates a magnetic field directed to +k. It means that the induced electric current is flowing anticlockwise. d) The bars move at constant speed if the net forcé on them is zero (the acceleration must be zero).The only force present on the bars is the magnetic force. The magnetic force is: Frog = Xx B mi 
