PHY 2049 SPRING 2001 FINAL EXAM

Transcription

1 PHY 049 SPRING 0 FINA EXAM 1 Three charges of the same sign and value q are placed in the corners of an equilateral triangle and free to move One more charge Q is placed in the center of the triangle so that the entire system of the four charges has became statically stable, ie the net force on each of the four charges is equal to zero? Find the value of charge Q in terms of q Answer: 058q q 60 o Q? q r 30 o F q 3 o r cos 30 1/ r 3 r 3 1/ F F 1 By symmetry, the charge Q will be in equilibrium regardless of it s magnitude or sign Therefore, we have to consider one of the corner charges et s consder the lower right charge Since the other corner charges will repel this charge, Q must attract the corner charge Since Q attracts the corner charge, it must have the opposite sign as q The three forces acting on the lower right corner charge are indicated in the figure The angles are found from recalling that the angles in an equilaterial triangle are 60 ) F F 1 + F + F 3 F 1 î +F cos 60 î F sin 60 ĵ)+ F 3 cos 30 î + F 3 sin 30 ĵ) F 1 + F cos 60 F 3 cos 30 )î + F sin 60 + F 3 sin 30 )ĵ Now we must find expressions for F 1, F,and F 3 F 1 k q q F k q q kq kq F 3 k q Q / 3) 3k q Q

2 It is given that F 0 and so either component must vanish component, Choosing the simpler y 0 F y F sin 60 + F 3 sin 30 kq 3 + 3k q Q 1 3 q +3 Q Q q 3 Recalling that Q must have the opposite sign of q, theanswerisq q/ 3 058q Two light bulbs one labeled 100 W / 10 V and the other 0 W / 10 V) are connected by a curious student in series and plugged into a 10 V outlet How much total power does the two light bulbs dissipate? Answer:17 W The power is related to the potential and resistance by P V /R Solving for R, R V /P For the two light bulbs, R 1 V /P 1 10 V) /100 W) 144 Ω and R V /P 10 V) /0 W) 70 W When the resistors are connected in series R R 1 + R 864 Ω The total power is now P V /R 10 V) /864 Ω) 167 W 3 An infinitely long wire carrying a current of 15 A is bent 70 with a radius of curvature of the bend equal to 1 cm see figure) What is the magnitude of the magnetic field at the center of point of the bend O? Answer:1 µt R O This figure can be broken into a semi-infinite wire, 3/4 of a circular loop, and another semi-infinite wire Using the right hand rule for each section, it is seen that the magnetic field due to each is in the same direction Therefore, B B 1 + B + B 3 B B 1 + B + B 3

3 A semi-infinite wire is 1/ of an infinite wire B 1 B 3 1 ) µ0 i πr and for 3/4 of a loop B 3 ) µ0 i 3µ 0i 4 R 8R Therefore, B B 1 + B + B 3 µ 0i 4πR + 3µ 0i 8R + µ 0i 4πR µ 0i 1 R π + 3 ) 8 4π 10 7 Tm/A)15 A) m) µ 0i 4πR 1 π + 3 ) T 4 A parallel plate capacitor has circular plates of radius R and plate separation d The potential difference between the plates is given by the time dependent function V t) E sinπft) What is the maximum value of the magnetic field induced between the plates at a radial distance R from the axis of symmetry? Answer: The Ampere-Maxwell law states B dφ E d s µ0 i enc + µ 0 ɛ 0 dt In the gap of the capacitor, there is no moving charge and i enc 0 The electric flux is Φ E EA EπR The flux extends only to the edge of the capacitor and that is why the area is πr The product of the electric and magnetic constants can be rewritten using µ 0 ɛ 0 1/c,wherec is the speed of light Substituting into the Ampere-Maxwell law, B dφ E d s µ0 i enc + µ 0 ɛ 0 dt 0+ πr de c dt The left hand side is found by integrating the magnetic field around a circle of radius r R concentric with the capacitor, B d s B ds Bπr BπR 4πBR Plugging into the left side of the Ampere-Maxwell law B πr de d s c dt 4πBR πr de c dt B R de 4c dt πrfe c d

4 Now it is necessary to consider the derivative of the electric field It is given that the potential across the capacitor depends on time according to V t) E sinπft) The electric field can be found from E V/d Therefore, de dt 1 dv d dt 1 πfe cosπft) d πfe cosπft) d The maximum value for the derivative will occur whenever cosπft) 1 Substituting into the expression for the magnetic field ) ) R πfe B max 4c d πrfe c d 5 When an object is observed through a lens, the upright image appears to be cm further from the lens than the object If the image appears to be magnified, find the focal length in cm) of the lens Answer:400 The upright image is virtual and will have a negative image distance If the image appears to be cm further away, we have i p + cm)) The magnification is given by m i/p This yields m i p p +cm) p p p +cm) p cm Accordingly, i p + cm)) 4 cm Substituting into the thin lens equation 1 f 1 p + 1 i 1 cm cm 4cm 1 4cm f 4 cm 6 Suppose a flat Frisbee is placed directly in front of a concave mirror, with the face orthogonal to the mirror s central axis If the mirror has a focus at position f, whereon the central axis would we have to place the Frisbee in order to create a virtual image that

5 has a surface area four times greater that the actual Frisbee? Answer:f/ A Frisbee is a disk and its area is πr It is given that the area of the image is 4 times the area of the object, A i 4A p This implies 4 A i A p πr i πr p R i R p This implies that the magnification is + The positive sign comes from the information that the image is virtual) The magnification equation states m i/p Sincem this implies i p Substituting into the mirror equation, 1 f 1 p + 1 i 1 p + 1 p p 1 p 1 p f p p f 7 One has two small identical converging lenses with a focal distance of 1 cm How far apart must the two lenses be separated, if one wants to make from them a microscope with magnification of 100? Answer:6 cm The magnification of a compound microscope is M s 5 cm f ob f ey Solving for s s Mf obf ey cm)1 cm) 100)1 4cm 5 cm 5 cm The magnification was taken to be negative since the compound microscope creates an inverted final image The distance from one lens to the other is d f ey + s + f ob 6cm 8 What is the minimum thickness of a transparent material n 140) coating a flat glass lens n 160) so that reflected light of wavelength 560 nm is eliminated by interference? colorred Answer:100 nm

6 The indices of refraction are n air), n 140 film), and n glass) The light reflected from the air/film interface will be shifted by half a wavelength since the index of refraction increases at the boundary For similar reasons, the light reflected from the film/glass interface will also be shifted by / For destructive interference, m + 1 ) n t + + n n t t m + 1 ) n m 1 ) n m 1 ) n The smallest film will be found when m has the smallest physically realistic value, m 1 The choice m 0 is not possible as it will yield a negative thickness) Solving for t t 1 1 ) ) nm n 140) 100 nm 9 Monochromatic light of unknown wavelength shines on two vertically aligned pinholes 4 cm apart and forms an interference pattern on a screen 1 m from the pinholes When a glass plate n 150) is inserted in front of the upper hole, one observes that the brightest fringe moves up by cm see figure) The thickness of the plate must be: Answer:16 mm d 4 cm d d 1 y cm 1 m The initial bright fringe is located at a position where the path difference from the screen to the two sources is zero Stated a different way, the total number of wavelengths needed to travel from one source to the screen is the same as the number of wavelengths needed to travel from the other source to the screen When the glass plate is inserted, the bright fringe still results from the equivalence of the number of wavelengths The bright fringe moves up because some wavelengths are shortened when the light passes through the glass From the figure d d 1 +y + d/) +4y

7 where the fact that d/ y cm was used The number of wavelengths from the top slit is N 1 t /n + t nt + t and the number of wavelengths from the bottom slit is Equating the number of wavelengths N +4y N 1 N nt + t +4y nt + t +4y n 1)t +4y 1 + 4y/) ) ) y 1/ 1+4 ) 1 4y 1+ ) y t y n 1) 00 m) 15 1)1 m) m The binomial theorem 1 + x) n 1+nx) was used to simplify the expression 10 A diffraction grating 5 cm wide produces the second order maximum at 30 with light of 500 nm wavelength What is the total number of lines on the grating? Answer: The diffraction grating obeys d sin θ m Solving for d, d m sin θ m) m sin 30 The spacing between adjacent lines is m For a side 5 cm wide the number of lines will be n w d m m 5 104