Here are some equations you may need. Q142. E = E =ke PE F =k2 V = APE = qAV 72 V2 V = IR; R=! PL P = IV = 12 R = Irms = A R V2 mu E F = IlB sin 0; F = quB sin0; r= V = B ' v = rw 9B ' BHOI B - HoIN VP Vs Ip Ns 2 T r -; F _ Molll2 (2; 2rd NP Ns Is NP APB DB = BIA = BA cos 0; ADB E = ( = - E = Blv At At 0; = 0,; n1 sin 01 = n2 sin 02; n = C Avacuum = Amedium m hi dix = c XT = C; k = - 1 ~ 9 x 109 Nm / C2 do ho do 4 TEO Elementary charge an electron: e = 1.602 x 10-19 C. Permittivity of free space : co = 8.85 x 10-12 F/m. Permeability of free space Mo = 47 x 10-7 Tm/A. The speed of light: c = 3.0 x 108 m/s Magnetism Chapter 20 25 points Electromagnetic Induction Chapter 21 30 points Electromagnetic Waves Chapter 22 15 points Light: Geometric Optics Chapter 23 30 points1 Chapter 20 5 pts. (1) Two magnets (1) and (2) with a magnetic eld strength 31 = 0.80 T and B; : 0120 T at a equal distance When the magiet (1) is placed at a certain distance from magnet (2), it pulls the magnet (2) with a force F}, of magnitude F1 = 2.50 N . What is the magnitude (F2) of the force F2 does the magnet (2) pull on the magnet (1)? 5 pts. (2) A wire carrying a 12.0 A current has a. length l = 2.5 m between the pole faces of a magnet at an angle 9 : 300, feels a force F' of magnitude F : 1.2 N 1 Suppose that the magnetic eld is approximately uniform between the pole faces of the magnet. What is the magnitude B of the magnetic eld E between the pole faces of this magnet Y 15 pts. (3) A horizontal wire carries a current I1 = 12.0 A DC. A second parallel wire at d : 2.5 cm below the rst wire, as shown in Figure 1, is set so that it does not fall due to gravity. I. > A |,7 x Figure 1: Force on parallel wires carrying a current a- Using the point A as reference point in Figure 1, draw in Figure 1 all the forces acting on the second wire b- What should be the direction of current in the second wire? 0 How much current 13 must the second wire carry so that it does not fall due to grav- ity? Given: The second wire has a mass of 0.12 g per meter of length. 2 Chapter 21 5 pts. (4) Explain what you know about Lenz's Law 10 pts. (5) A horizontal square wire loop of side d has in a plane perpendicular to a uniform magnetic eld I; pointing from above into the plan of the loop. The magnitude of magnetic eld E is: B : 0.80 T If in At : 4.0 ms the wire is reshaped from a square to a rectangle of dimensions: L = 10.0 cm; W = 6.0 cm, but remains in the same plane. a What is the side it of the square? b What is the magnitude of the average induced emf, e in the wire during this time? 5 pts. (6) Why do we sometimes need to use a transformer? 10 pts. (7) An ideal transformer has a primary with N1 turns and secondary with N2 turns. The load resistor is R and the source voltage is unm. Given: N1 = 40; N2 = 10; R2 : 10.0 0 and me = 120.0 V. Figure 2: Ideal step-down transformer If a voltmeter across the resistor R; reads a Vm, voltage of V2 = 610 V, what is the source resistance R1? 3 Chapter 22 5 pts, (8) Explain in your own words the importance of electromagnetic waves in our modern life, give at least three concrete examples. 5 pts (9) Consider the antenna in the gure below. Draw the radiation elds( electromagnetic waves) far far away from the antenna produced by a sinusoidal signal on the antenna .l 'I 5 pts. (10) Calculate the wavelength (in km) of a 1500 H z electromagnetic wave. 4 Chapter 23 15 pts. (11) An incident ray of monochromatic light traveling in a medium 1 of index of refraction n1 strikes a surface separating medium 1 from medium 2 of index of refraction n; as shown in Figure 4. The index of refraction n1 is greater than the index of reection n; (m > Hz) a.- Draw the reected ray, justify your answer. b- Drew the refracted ray, justify your answeri c If the index of refraction m = 150; 7L2 = 1.00, and the incident angle 9i = 91 = 41'8", 4 n2 I calculate the refracted angle 92. 5 pts, (12) An object 00' is placed at the point 0 in front of a concave mirror (Figure 4) . The distance object is 0A : 8.0 cm, the height of object is 00' : 9.0 cm. The radius of curvature is CA = 28.0 cm. Determine graphically the position of the image, Figure 4: 5 pts. (13) A concave spherical mirror forms a real image 0.25 times the size of the object. The distance d between the object and the image is 300 mi Find the magnitude of the radius of curvature of the mirror. Bonus question 1 for 6 points: A 10.00 cm high object is placed 70.0 cm from a double-convex lens (Figure 5:) with a focal length f = 40.0 cm. a Where the image of the object placed at the point 0 will be formed. Figure 5: b- How is the image? Nature of the image Real Virtual Position Up right Inverted or up side down Size Smaller Bigger Bonus question 2 for 4 points: Consider the divergent lens in the gure below (Figure 6:) Where the image I I ' of the object 00' will be formed? ____. ___________ l 3' _____________________________ . _____ F o F' Figure 6