Physics 11b: Midterm Exam I with Solutions
March 13, 2008
Name: Please place a check next to your section leader: Ognjen Ilic Subhaneil Lahiri Josh Lapan Daniyar Nurgaliev Roberto Martinez Problem 1 2 3 4 5 6 Score Total Points 15 15 15 15 15 15
Total Sco
6. Due to the time-dilation effect, the time between initial and final ages for the daughter is longer than the four years experienced by her father: tf daughter ti daughter = (4.000 y) where is Lorentz factor (Eq. 37-8). Letting T denote the age of the f
13. (a) We choose a horizontal x axis with its origin at the left edge of the plastic. Between x = 0 and x = L2 the phase difference is that given by Eq. 35-11 (with L in that equation replaced with L2). Between x = L2 and x = L1 the phase difference is g
49. The saturation magnetization corresponds to complete alignment of all atomic dipoles and is given by Msat = n, where n is the number of atoms per unit volume and is the magnetic dipole moment of an atom. The number of nickel atoms per unit volume is n
10. Fig. 30-43(b) demonstrates that dB / dt (the slope of that line) is 0.003 T/s. Thus, in absolute value, Faraday's law becomes
=-
dB d ( BA) dB =- = -A dt dt dt
where A = 8 10-4 m2. We related the induced emf to resistance and current using Ohm's law.
8. Letting F = q E + v B = 0 , we get vB sin = E . We note that (for given values of the fields) this gives a minimum value for speed whenever the sin factor is at its maximum value (which is 1, corresponding to = 90). So
d
i
vmin =
E 1.50 103 V/m = = 3.7
6. The current in the circuit is i = (150 V 50 V)/(3.0 + 2.0 ) = 20 A. So from VQ + 150 V (2.0 )i = VP, we get VQ = 100 V + (2.0 )(20 A) 150 V = 10 V.
10. (a) The work done by the battery relates to the potential energy change:
qV = eV = e (12.0V ) = 12.0
16. We note that the voltage across C3 is V3 = (12 V 2 V 5 V ) = 5 V. Thus, its charge is q3 = C3 V3 = 4 C. (a) Therefore, since C1, C2 and C3 are in series (so they have the same charge), then 4 C C1 = 2 V = 2.0 F . (b) Similarly, C2 = 4/5 = 0.80 F.
24.
7. To exploit the symmetry of the situation, we imagine a closed Gaussian surface in the shape of a cube, of edge length d, with a proton of charge q = +1.6 10-19 C situated at the inside center of the cube. The cube has six faces, and we expect an equal
4. The fact that the spheres are identical allows us to conclude that when two spheres are in contact, they share equal charge. Therefore, when a charged sphere (q) touches an uncharged one, they will (fairly quickly) each attain half that charge (q/2). W
Physics 11b: Midterm Exam II with Solutions
April 17, 2008
Name: Please place a check mark ( ) next to your section leader: Ognjen Ilic Subhaneil Lahiri Josh Lapan Daniyar Nurgaliev Roberto Martinez Problem 1 2 3 4 5 6 Score Total Points 15 20 15 10 20 10
13. The rate at which photons are absorbed by the detector is related to the rate of photon emission by the light source via
Rabs = (0.80) Aabs Remit . 4 r 2
Given that Aabs = 2.00 10-6 m 2 and r = 3.00 m, with Rabs = 4.000 photons/s, we find the rate at