Exam 3 Solutions. ! r, the ratio is ( N ) ( ) ( )( ) 2. PHY2054 Spring Prof. Pradeep Kumar Prof. Paul Avery Prof. Yoonseok Lee Mar.

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1 PHY054 Spring 009 Prof. Pradeep Kumar Prof. Paul Avery Prof. Yoonseok Lee Mar. 7, 009 Exam 3 Solutions 1. Two coils (A and B) made out of the same wire are in a uniform magnetic field with the coil axes aligned in the field direction. Coil A has turns of loop with a radius of cm and coil B has 5 turns with a radius of cm. Now the field strength is doubled in s. What is the ratio of the emf induced in each coil (emf_1:emf_)? Answer: 1: 8:1 1:1 Solution: The induced emf in coil i is N A!" /! t. The ratio of emfs is thus independent of the field change. Since the area is i i B! r, the ratio is ( N ) ( ) ( )( ) 1A1 / NA = N1 / N r1 / r. An ideal transformer has a 100-turn primary coil and 10,000-turn secondary coil. 1 ma DC current is flowing through the primary coil and the current suddenly drops to zero in 1 µs. What is the induced voltage appearing on the secondary coil if the self inductance of the primary coil is 1. H 0.8 H 1.7 H? Answer: 10,000 V 80,000 V 170,000 V Solution: The emf in the primary coil is L! i /! t. The emf in the secondary coil is calculated from the ratio of turns so is equal to 100 L! i /! t. 3. A 100 Ω 160 Ω 90 Ω resistor, a 100 µf capacitor, and a 0.45 H inductor are connected in series to a 10 V rms 60 Hz power source. What is the average power consumption in this circuit? Answer: 47. W 50.0 W 45.4 W Solution: The average power consumption is from ave rms Irms = Vrms / Z = Vrms / R + X L, with X =! L = 10" L. P = I R, where the rms current is calculated L 4. A kg object is floating near the international space station. The crew members in the space station direct a 1. kw laser perpendicular to a flat surface of the object for an hour. What is the change of speed of the object? Assume that the object surface absorbs laser light completely. Answer: 0.01 m/s m/s m/s Solution: The force from the laser is IA / c = P / c, where P is the power of the laser. The change! v = at = F / m t = Pt / mc. in velocity Δv is obtained from acceleration time or ( ) 1

2 PHY054 Spring As a beam goes through layers of different materials (air, diamond, and ice), it bends as shown in the figure. The figure indicates the angles formed by each beam and vertical line. When the angle A (angle of incidence) is , what is the angle B (angle of refraction in ice) in degrees? Note that n air = 1.00, n diamond =.4 and n ice = Answer: Solution: We use Snell s law: nair sin! air = nice sin! ice. The angle in diamond can be ignored. 6. Yellow light has a wavelength of 589 nm. What is the frequency and speed of the light in a medium with an index of refraction (frequency, speed)? (in Hz and m/s) Answer: , , , Solution: The frequency is unchanged and given by Hz. The speed is c = c / n, which depends on n. n 8 " 9 14 f = c /! = 3# 10 / 589# 10 = 5.09# A solenoid consists of a tube of diameter 4 cm and length 60 cm that has,500 turns of wire wrapped around it. A current of 45 A flows in the wire. What is the inductance of the solenoid? Answer: 0.59 H 1.33 H.61 H Solution: The inductance is µ 0 AN / L, where N is the number of turns, L is the coil and A is the coil area. 8. Refer to the previous problem. What is the energy contained in the magnetic field of the solenoid? (You can obtain this without solving the previous problem.) Answer: 600 J 1350 J 640 J Solution: The energy U can be found from the B field. B = µ 0ni and U = B AL / µ 0, where n = N / L. 9. A 60 Ω 75 Ω 90 Ω resistor and an inductor L are hooked in series to a 4 V battery and a switch that is initially open, preventing current from flowing. A voltage of 17.0 V is measured across the inductor msec after the switch is closed. What is L? Answer:.1H 3.0 H 4. H Solution: The voltage across L is VL V ir and we can solve for L. tr / L =!. Thus V = Ve!, / =!, where i ( V / R)( 1 e!tr L ) L

3 PHY054 Spring A 50 Ω resistor, a 150 mh inductor and a capacitor C are connected in series to an AC emf source with V rms = 36 V and variable frequency. It is found that when the source frequency is set to 180 Hz 10 Hz 40 Hz the power delivered to the circuit is maximized. What is C? Answer: 5.1 µf 3.83 µf.93 µf Solution: The power is maximized at resonance! = 1/ LC, which allows us to solve for C. 11. Refer to the previous problem. At resonance, what is the rms voltage across the capacitor? Answer: 1 V 143 V 163 V Solution: The voltage across C is the same as the voltage across L at resonance. The voltage across C and L is given by VC = VL = ( Vrms / R) X L = ( Vrms / R)! 0L, where! 0 = " f0. 1. A type of glass has an index of refraction that varies from 1.6 at the violet end of the visible spectrum to at the red end of the spectrum. If white light of all visible wavelengths shines on a flat surface of the glass, what is the maximum angular dispersion of the refracted light? Answer: Solution: The maximum dispersion is the maximum difference in refracted angle, which occurs " when the incident angle is 90. Thus the dispersion is 1 "! sin ( 1/ n ) sin 1 ( 1/ n ) = ". disp violet red 13. Microwaves of frequency 5.50 GHz travel through a particular material of length m at a speed 9% slower than the speed of light in vacuum. Approximately how many complete wavelengths fit inside the material? Answer: Solution: The wavelength inside the material is! n = 0.91 c / f. The number of wavelengths that fit inside the material is N! = L /! n. The numbers shown are actually off by about 1% from the true answers. 14. A single loop of wire with a radius of cm rotates about a diameter in a uniform magnetic field of T. To produce a maximum emf of 1.0 V, it should rotate at: Answer: rad/s Solution: The maximum emf is given by! AB, where ω is the angular velocity, area and B is the magnetic field strength. A =! r is the 3

4 PHY054 Spring A 0.0-cm 0.10-cm 0.0-cm radius cylinder, cm long, is wrapped with wire to form an inductor. At the instant the magnetic field in the interior is mt the energy stored in the field is about: Answer: J J J Solution: The energy stored in the field is B AL / µ 0, where A is the area of the solenoid, B is the magnetic field strength and L is the length. 16. A solar cell has a light-gathering area of cm and produces A at V (dc) when illuminated with sunlight of intensity 1000 W/m. What is the efficiency of the solar cell? Answer: 16% 44% 8% Solution: The total power incident on the panel is IA, where I is the intensity and A is the area. The output power is Vi, where i is the current and V is the voltage. The efficiency is given by the ratio of these or! = Vi / IA. 17. A laser beam of wavelength nm has intensity W/m. The amplitudes of the associated electric and magnetic fields, in V/m and µt, respectively are: Answer: 100, , , 0.46 Solution: The intensity is given by max / µ 0 max / µ 0 I = cb = E c. Note Emax = cbmax. 18. A microwave oven heats food by using electromagnetic waves that form a standing wave in the oven. When.5 GHz waves are used this way, it is found, in a sample of food, that there are burn marks spaced cm apart. What is the index of refraction of the food sample for these electromagnetic waves? Answer: Solution: This was taken from a chapter problem. The wavelength in air is! = c / f = 1 cm. The power is maximum twice within each cycle, so the difference between the burn marks (L) is half the wavelength in the material. Thus L =! / =! / n and we can determine n. n 4

5 PHY054 Spring The front end or antenna input of radio and television receivers is a resonant LC circuit (similar to that discussed in Chapter 1. Tuning to different channels (frequencies) is accomplished by varying L or C until the circuit resonates at the station's frequency. If a particular circuit includes a 1.0 microhenry inductor, what capacitance (in picofarads) is required to tune it to an FM station that broadcasts on a wavelength of meters? Answer: Solution: The resonant angular frequency ω 0 is! 0 = 1/ LC and is also equal to! f, where f = c /!. Thus we can determine C. 0. An RL circuit with resistance 100 Ω 10 Ω 150 Ωis connected in series to a battery with EMF 1V. If the current builds up to 1/3 of its final steady-state value in 6.0 sec, what is the inductance L in Henrys? Answer: / Solution: The current in an RL circuit is i ( V / R)( 1 e!tr L ) =!. The condition that 1/3 of the tr / L maximum power is reached means that we have to solve e! = / 3 for L. 5