There were two versions of the final exam. This is the solution to version A; the solution to the other version is similar, with some problems having different numbers.

 

1. A cylindrical capacitor has an inner conductor of radius 2.5 mm and an outer conductor of radius 4.0 mm. The entire capacitor is 3.5 m long. The potential of the outer conductor is 350 V higher than that of the inner conductor, and the potential is zero at distances far away from the capacitor. Find the magnitude and sign of the charge on the inner conductor.

Solution:

The electric field of a cylinder is 1/(2pe0) l/r, where l = Q/l is the linear charge density. The potential difference between the two cylinders can be calculated by integration of the 1/r function, and we get V=1/(2pe0) (Q/l) ln (r2/r1). Insert V=350V, r1=2.5mm, r2=4.0mm, l=3.5m and solve for Q=1.4x10-7C.

2. In the circuit shown in the Figure, find the value of the R resistor. Note that three currents are given.

Solution:

Due to Kirchoff's junction law, the current in the 3W resistor is 8.0A. The loop law for the left loop yields E1=12V+24V=36V, and for the right loop E2=30V+24V=54V. Finally, for the top loop:

54V-36V-2.00A R = 0, and we get R=9.0 W.

3. An L-R-C series circuit, R = 250W, L = 0.400H, and C = 20.0nF, is connected to an AC voltage source of 65V, operating at the resonance frequency of the circuit. What is the current in the circuit? What is the voltage on the capacitor?

Solution:

The resonance frequency is w0=(LC)-1/2 = 11200 rad/s. At this frequency the current is I= Vsource/R=0.26A, and the voltage on the capacitor is VC=I/(wC) = Vsource/(RwC)=1160V

4. A converging lens with a focal length of 7.00 cm forms an image of a 40.0-cm-tall real object that is to the left of the lens. The image is 5.00 cm tall and inverted. Where is image located in relation to the lens (left/right, distance)?

Solution:

The magnification is m= - 5/40 = - s'/s, so we get s=8s' (the negative sign is for inverted image). But 1/s+1/s'=1/f, so 1/(8s')+1/s'=1/7, resulting in s'=7.88cm. The positive sign indicates that the image is on the right side (opposite to the object)

5. The capacitors in the Figure are initially uncharged and are connected as in the diagram with switch S open. The applied potential difference is Vab = +360 V. What is the potential difference Vcd?

A 0V

B 120V

C 240V

D 360V

E 60V

F none of the above

Solution:

The charge on the capacitors in series is always the same. The charge in the upper and lower branch is also the same, determined by Vab = Q/3 mF +Q/6mF. We get Q= 0.72x10-3C. Vcd = Vad - Vac= Q/3 mF - Q/6mF = 120V.

7. A length of wire is cut in half and the two lengths are glued together side by side to make a thicker wire. How

does the resistance of this new combination compare to the resistance of the original wire?

A four times larger

B two times larger

C same

D half

E one quarter

F none of the above

Solution:

The new resistance is 1/4 (one quarter) of the original resistance

7. The area of an elastic circular loop decreases at a constant rate, dA/dt = -3.50 x 10-2m2/s. The loop is in a magnetic field B = 0.48T whose direction is perpendicular to the plane of the loop. At t = 0, the loop has area A = 0.285 m2 . Determine the magnitude of the induced emf at t = 2.00 s.

A 16.8mV

B 6.84mV

C 103mV

D 73mV

E 25mV

F none of the above

Solution:

The magnitude of the induced emf is V=dF/dt, and the flux is F=BA. Therefore the emf is

V=B dA/dt= 0.035x.48 V = 0.0168V=16.8mV

8. The critical angle of a certain piece of plastic in air is Qc = 37.3o. What is the critical angle of the same plastic if it is immersed in water?

A 38o

B 49o

C 28o

D 64o

E 54o

F none of the above

Solution:

The index of refraction of the plastic is n=1/ sin Qc = 1.65. The index of refraction of the water is n'=1.33 (from the Table). The new critical angle is determined by sin Qc'= 1.33/1.65 resulting in Qc =53.7o.

9. As viewed from the earth, the moon subtends an angle of approximately 0.50o. What is the diameter of the image of the moon that is produced by the objective of the Lick Observatory refracting telescope of focal length 18 m?

A 9.0m

B 16cm

C 9.0cm

D 1.6cm

E 9.0mm

F none of the above

Solution:

The size is 0.5ox18m=0.5x6.28/360x18m=16cm

10. Coherent light from a mercury-arc lamp is passed through a filter that blocks everything except for one spectrum line. It then falls on two slits that are separated by 0.380 mm. In the resulting interference pattern on a screen 2.50m away, adjacent bright fringes are separated by 2.27mm. What is the wavelength?

A 690nm

B 518nm

C 345nm

D 173nm

E 189nm

F none of the above

Solution:

For bright fringes: d~ Rl/Dy, resulting in l=dDy/R=3.45x10-7m=345nm

11. If you can read the bottom row of your doctor's eye chart, your eye has a resolving power of one arc-minute, equal to 1/60 degree. If this resolving power is diffraction-limited, to what effective diameter of your eye's optical system does this correspond? Use Rayleigh's criterion and assume that the wavelength of the light is 550 nm.

A 1.9mm

B 2.3mm

C 3.8mm

D 4.6mm

E 5.2mm

F none of the above

Solution:

D=1.22l/sinQ1= 1.22x5.5x10-7m/sin(1/60o)=2.3mm