Mid Term Exam 3. Apr 24, /40 2 /40 3 /30. Total /110

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1 Name: ID: Mid Term Exam 3 Phys 248 Apr 24, 2009 Print your name and ID number clearly above. To receive full credit you must show all your work. If you do not show your work you will only receive partial credit, even for a correct final answer. Problem Score 1 /40 2 /40 3 /30 Total /110 Useful constants: Planck constant h = x J s Reduced Planck constant =! =1.054 x J s Permittivity of free space:! 0 = "10 #12 C 2 N #1 m #2 hc = 1240 ev nm! c = ev nm c = 3 x 10 8 m/s 1 ev = 1.6 x J k e 2 = 1.44 ev nm Permeability of free space: µ 0 = 4! " 10 #7 T $ A #1 m m e = electron mass = 9.11 x kg = MeV/c 2 e = electron charge = -1.6 x C u=atomic mass unit= x kg= MeV/c 2 m p = proton mass = MeV/c 2 = u m n = neutron mass = MeV/c 2 = u

2 Problem 1: Magnetic Forces and Fields. (a) (10 points) A singly charged ion of mass m is accelerated from rest by a potential difference ΔV and deflected by a uniform magnetic field B (perpendicular to the ion s velocity) into a circular path of radius R. A doubly charged ion of mass m is accelerated through the same ΔV and is deflected by the same magnetic field into a circular path of radius R =2R. What is the ratio of the masses of the ions? For the singly charged ion, from q!v = 1 2 mv2, we have v = The circular path is given by R = mv qb = 2!V B charged ion, we have v! = 2 q!"v m! Therefore, and R! = R! R = 2 = m! q! m q = m! 2m, and m! m = 8. m q 2q!V m.. For the doubly m! v! q! B = 2"V B m! q!.

3 (b) (15 points) An infinitely long wire with current I=2 A is curved as shown below. Suppose an electron is located at point P and is moving with velocity! v =!v 0 ŷ at time t 0., where v 0 = 3! 10 4 m s. Determine the magnetic force (magnitude and direction) on the electron at time t=t 0. I d=10 cm P v y x I The magnetic field at point P can be obtained using the Biot- Savart law: the result is! B = µ 0I 4!d ẑ = 2 " 10#6 T. The magnetic force on the electron is F! = q v!! B! = e v 0 ŷ! µ I 0 4"d ẑ = e v 0µ 0 I 4"d ˆx = 9.6! 10 #21 Nˆx.

4 (c) (15 points) A long cylindrical shell with inner radius a and outer radius b carries total current I 1 with a uniform current density, as shown below. A long wire with current I 2 ( I 2 =2 I 1 ) in the opposite direction is located at the center of the cylindrical shell. Compute the magnetic field (magnitude and direction) everywhere, and plot it as a function of the radius r. I 1 a b I 2 The magnetic field is most easily obtained using Ampere s Law:! " B! d l! = µ 0 I enc. For r<a: I enc = I 2, and hence! B(r) = µ 0I 2 2!r ˆ" = 2µ I 0 1 2!r! B! d! l " = B2#r = µ 0 I 2, such that ˆ". (Here ˆ! is defined to be a unit vector that goes out of the page above the wire and into the page below). For a<r<b, I 1 = J!(b 2 " a 2 ). The current density J is then I J = 1. Applying Ampere s law inside the cylindrical shell,!(b 2 " a 2 ) # I enc = I 2 + J!(r 2 " a 2 ) = I 1 2 " r 2 " a 2 & $ % b 2 " a 2 (, and so B(r)! = µ 0I 1 # 2 " r 2 " a 2 & 2!r $ % b 2 " a 2 ( ˆ). For r>b, the total current enclosed is just I 1, so! B(r) = µ 0 I 1 2!r ˆ". See the plot on the next page (numerical values chosen: a=1, b=2, µ 0 I 1 /(2π)=1).

5 B r Problem 2: RC circuit. In the circuit below, the switch S has been closed for a long time and the circuit carries a constant current. Take C 1 = 6.00 µf, C 2 = 2.00 µf, R 1 = 2.00 kω, and R 2 = 5.00 kω. The power delivered to R 2 is 3.20 W. (a) (10 points) Find the charge on C 1 and C 2. (b) (10 points) Now the switch is opened. After equilibrium is reached in the circuit, what is the charge on C 1 and C 2? Suppose now the battery is disconnected from the circuit while S is kept open, such that the two capacitors fully discharge. (c) (10 points) Determine the total charge stored on the capacitors as a function of time. (d) (10 points) Compute the total energy dissipated during the discharge process. [Hint in c-d: reduce the circuit to a circuit with an equivalent resistor in series together with an equivalent capacitor.]

6 Solutions: (a) When the switch is closed since a long time the capacitors are fully charged and they behave like open circuits, so the current flows in the two resistors that are in series. The power delivered P to R 2 is: P = R 2 I 2, hence I = = 25.30mA. The potential difference on R 2 R 1 is:!v 1 = R 1 I = 50.60V. The charge on C 1 is Q 1 = C 1!V 1 = µC and on C 2 it is Q 2 = C 2!V 2 = C 2 R 2 I = µC. The electromotive force or potential difference at the battery terminals is! = "V 1 + "V 2 = (R 1 + R 2 )I =177.10V. (b) After a long time the switch is opened the current drops to zero because the two capacitors are fully charged and they behave like open circuits so no current will flow in R 1 and R 2. The charge on the capacitors has changed and the potential difference across R 1 and R 2 is zero because the current is zero. Hence: Q 1 = C 1! =1.06mC and Q 2 = C 2! = 0.35mC since the potential difference of the battery is appears across each capacitor (they are in parallel with the battery). (c)-(d). Once the battery is disconnected, the two capacitors are still in parallel and still there is no current in the circuit. Hence, the charge on the capacitors stays constant and the sum has the value Q 1 +Q 2 computed in (b). Since there is no flow of charge there is no dissipation, so the dissipated energy is zero. The hint to consider an equivalent circuit with two capacitors and two resistors in series is useful to calculate the time constant! C! = R eq C eq, where R eq =7 kω, and C eq = 1 C 2 $ " # C 1 + C 2 % & = 12 µf = 1.5µF. The 8 value of the time constant is useful to demonstrate mathematically what is written above from the equation of the circuit, Q 1 C 1! IR 1! Q 2 C 2! IR 2 = 0.

7 Problem 3: Quantum and nuclear physics. (a) (10 points) A hydrogen atom in the 3s state has the following normalized wave function:! 300 (r) = # 1 & " $ % 3a 0 ( 1) 2 r # # r & & % 3 a 0 27 $ % a 0 ( ( e )r (3a 0 ). $ What is the probability of finding the electron in the range!r = 0.01a 0 of r = 2a 0, where a 0 is the Bohr radius? The probability of finding the electron in this range is given by Prob=P(r = 2a 0 )!r, where the radial probability density for this state is P(r) = 4!r 2 " 300 (r) 2. Evaluating it, one obtains Prob=2.14! 10 "6.

8 (b) (10 points) You are given two nuclides, X and Y. Initially, there are 2.5 times as many nuclei of type X as there are of type Y. Three days later, there are 4.20 times as many type X nuclei as type Y nuclei. If the half-life of the type Y nuclei is 1.60 days, what is the half-life of the type X nuclei? The decay is governed by N i (t) = N i0 e!" it, where for each nuclide,! = ln2 T 1 2. Given that N X 0 = 2.5N Y 0, N X (t = 3d) = 4.2N Y (t = 3d), and! Y = ln2 T 1 2Y = ln2 1.6d, we obtain the condition e!" X 3d = e! 3ln Taking the ln of both sides, we have!" X (3d) = ln ! 3ln2 =! #!$, 1.6 such that T 1/2 X = 3ln2! = 2.66d. (c) (10 points) Write down the explicit reaction and determine the Q value (in MeV) associated with the spontaneous fission of 236 U (Z=92) into 90 Rb (Z= 37) and 143 Cs (Z=55). The masses of 236 U, 90 Rb, and 143 Cs are u, u and u. The reaction proceeds as 236 U! 90 Rb Cs + 3n. The Q value is Q = (m U!m Rb! m Cs! 3m n )c 2. Evaluating it yields Q = MeV.