REAL TAKE HOME TEST 4 , 4B SP '12

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1.  (56 POINTS) CHAPTER 25 and 28 THEORY.  Magnetic field of a co-axial cable consisting of long wire surrounded by hollow tube.  REFERENCES: QUIZ 5, TEST #2 AND CH. 28 EXERCISES 45, 46 AND  47.

A very long, straight cylindrical conducting wire of radius a and length L is surrounded by a concentric hollow conducting tube of inner radius b,  outer radius c and same length.  See fig. 1 of schematic below.

The conducting material of  both wire and tube has uniform resistivity  ρ = 1.6800×10−8    Ω-m at 20 oC.  GEOMETRY:  L = 1000.000 mm.  Assume a = 2.000 mm, b = 4.000 mm and c = 6.000 mm and, as  you see, L is MUCH, MUCH  larger than either radii a, b or c.

Applied between the ends of the coaxial cable is a voltage difference V = 1.000x10 -3 ( V) = 1.000 mV  a shown in fig. 2.    

 

It may be helpful to use supplied symbols  ρ ,  V ,  a,  b  c or L or other constants until the very last step when you plug in numbers to compute  intermediate or final quantities.

 

(a) (14 points)  What is the magnitude B of the magnetic field  inside the wire  a  distance a/2 from the center? 


(b) (14 points) What is the magnitude B of the magnetic field  within the space between wire and tube  a distance (a + b)/2 from the center?


(c) (14 points) What is the magnitude B of the magnetic field on the surface of the outer tube  i.e., a distance c from the center?


(d) (14 points) Suppose  the temperature of the co-axial cable rises to 160.000 oC.   What is the new field magnitude B  inside the wire  a  distance a/2 from the center?  What is the difference between this answer and that of part (a)? In evaluating the difference, be mindful of  significant digits. The temperature coefficient of resistivity 0.00393 (oC)-1 . Thank you.

(e) EXTRA CREDIT.  (5 points) What is the magnitude B of the magnetic field  a distance (b + c)/2 from the center, i.e.,  within the walls of tube ?  

SOLUTIONS OUTLINE: 

(a) Find the total current on the inner wire using  I = V/(ρL/A), where A = pi*a2. Then find B

using B(2*pi*r) = uo*I(r2/a2), where  = =4*pi*10 -7,  numerically. Set r = a/2. Note: I = 0.75 (A) from computation. 

(b) Find the total current on the inner wire using  I = V/(ρL/A), where A = pi*a2. Then find B

using B(2*pi*r) = uo*I, where  = =4*pi*10 -7,  numerically. Set r = (a+ b)/2 = 3.000 mm.

(c) Find the total current I' of wire and tube using V/(ρL/A'), where A' = pi*(c2 - b2 + a2) . Note cross-sectional A' = area of tube [pi*(c2 - b2 ]  + area of inner wire [pi* a2]. Then find B using B(2*pi*r) = uo*I', where  = =4*pi*10 -7,  numerically. Note: I' = 4.49 (A) from computation. Set r = c.

(d) Find the new higher  resistance using info from Ch. 25. See SEC. 25.2  on temperature dependences; we assume area A and length L do not change appreciably during the heating process . Since the resistance increases, I decreases as does  magnitude B.

(e) Extra Credit. This is really easy when you think about it: Use B(2*pi*r) = uo*(I + I"), where I = total current in the inner wire which you already know from part  (a) or (b); the current I'' is in the section of tube between b  and (b + c)/2 = 5.000 mm.   NOTE: r  =  (b + c)/2 = 5.000 mm.   

The trick is getting the current I'' in the section of tube between b  = 4.000 mm and (b + c)/2 = 5.000 mm.  Here is how you get I" :  The total current in tube is Itot = V/(ρL/A''), where A" = pi*(c2 - b2 ).  Itot = 3.74 (A) from computation. NOTE: You can also  get Itot by subtracting  4.49 (A) - 0.75 (A) = current in part (c) - current in  part (a) = 3.74 (A).

ALSO NOTE: I'' is only a fraction of   Itot .    That is to say, I" =    Itot *pi*(r2 - b2)/(c2 - b2 ), where r = (b + c)/2 = 5.000 mm, b = 4.000 mm and c = 6.000 mm.  


2. (55 POINTS)  CH 25, 26 THEORY. Below in Figure 1 is an  RC circuit with  capacitor C in series with a  resistive “black box,” a network of identical resistors, each with value R = 10.000 Ω.  For network, see Figure 2 showing  the identical  individual 10.000 Ω- resistors.  Connected in series with equivalent resistor Rbox and capacitor C is a  battery of voltage ε =  50.00 (V). We see the main RC circuit is connected to terminals a and b of  “black box.”

As in you see in Figure  1, the switch is closed at t = 0.  Assume the capacitor is initially uncharged.  It takes 5.00 x10 -3 seconds for the voltage across the capacitor to reach a value of 10.00 (V). 

Answer the following problems,  keeping in mind it’s often helpful to use supplied or made up symbols until the very last step when you substitute into a formula on which they are based.

 

(a) (11 points)  How long after  the switch is closed will  the voltage across the capacitor reach a value  of 35.00 (V)?

(b) (33 points)  What is the  value of the magnitude of the current  I at t = 5.00 x10 -3 seconds? 

(c)  (6 points) What is the total heat generated in the resistor from t = 0 to t = infinity?  To compute this , you must show the explicit integration of the RATE of heat generation from t = 0 to t = infinity. HINT: The formula you integrate includes the symbol Rbox.

(d) (5 points) What is the  value of the magnitude Q of the charge on the capacitor’s positive plate at t = 5.00 x10 -3 seconds?  

 

SOLUTIONS OUTLINE: 
REDUCE THE RESISTOR NETWORK TO ITS EQUIVALENT: USING CH. 26 THEORY, IT CAN BE SHOWN REQ =R' =  6.25 OHMS. THEN  SEE SOLUTION TO #2 OF  TEST 3 AS A GUIDE TO SOLVE THIS PROBLEM, ONLY A MINOR VARIATION OF THAT THEME.
(a) 10 = 50*(1 - e-t/R'C) leads to R'C = 0.0224 (s) when t = 0.005 (s) and R'  = 6.25 ohms.  I am frankly perplexed why a large proportion of students got this wrong and did not use this expression for the CHARGING capacitor.  Now, 35 = 50*(1 - e-t/R'C) leads to t = 0.027 (s)
(b) I = (50/6.25)2*R*e-t/RC leads to I = 6.4 (A) when t = 0.005 seconds and  R' = 6.25 ohms.. 
(c) INTEGRAL OF (50/6.25)2 *e-2t/R'C* dt from 0 to infinity leads to (1/2)*CVo2 = 4.48 J, where Vo = 50.00 (V).
(d) CVo = 0.0358 F.

3. (55  points) CHAPTER 26 and 27 THEORY.  The vertically aligned  RC circuit shown is designed to exert an upward   magnetic force  to accelerate objects  upward or downward  in a gravitational field. The mass to be accelerated is a horizontal bar that slides on a frictionless, vertical track made up of two conducting wires connecting it to the circuit. In other words, the bar is constrained  to slide up or down the track under the influence of  the gravitational and magnetic  force only; no friction.

The 0.500 kg- bar  is 0.500 m long, has a resistance of 1.00 Ω and is in a HORIZONTAL, uniform 1.000 T-magnetic  field  pointing OUT of  page. Capacitor value C = 1.0x10 -6 F and the  values of the other two resistors are shown.   The battery is shown  as a box  between terminals a and b. The magnitude of the battery’s potential difference is |ε| = 50.00 (V).  REFERENCES: QUIZ 6 (RC-CIRCUIT  PROBLEMS) AND  CH. 27 EXERCISES 42, 70

The capacitor is uncharged when the switch is closed at t = 0.

 (NOTE: We  ignore the magnetic force due to the  velocity of the bar, i.e., due to the change in flux discussed in Ch. 29. Below I am not asking you to find the bar's  velocity; I only want the instantaneous acceleration. Indeed, in this problem the time scales are so small and the bar mass so relatively large,  if the bar had zero velocity at t = 0  it would essentially remain at rest throughout the  problem.)

 

(a) (8  points)  Given that circuit causes an upward  magnetic force on the bar,  which point, a or b, should be the positive terminal of the battery?

(b) (15 points)  What is the magnitude and direction of the bar’s acceleration just after the switch is closed at t = 0?

(c) (15 points)  How long would it take (in seconds) for the acceleration of the bar to be ZERO?


(d) (14 points)  How long would it take (in seconds) for the acceleration to be directed downward with magnitude 4.900 m/s2.

(e) ( 3 points) What essentially is the magnitude and direction of the bar's acceleration at t = 2.500x10 -3 seconds  (2.500 ms)?


SOLUTIONS OUTLINE: 

(a) From the right hand rule, the  bar current must flow LEFT to exert an upward magnetic force on bar; thus b is the higher voltage (positive) terminal.  
(b) ma = I'LB - mg and total current I = (50.00/R)*e-t/RC  , where R is the equivalent formed from the 2.00 ohm in series with  0.50 ohm. NOTE:  0.50 ohm is the parallel equivalent of the 1.00 ohm resistor shown and the bar. Set t = 0 and find I and I'. NOTE: I' = I/2. ANSWER: a = 0.2 m/s2.  (One sig . fig.) 
(c) ma = I'LB - mg = 0,  so I' = mg/(LB) = 9.8 (A). Set  9.8 = I/2, where I = (50.00/R)*e-t/RC .  Find t. 
(d) FOR DOWNWARD MOTION, mg/2 = mg - I'LB, so I' = 4.9 (A). Set  4.9 = I/2, where I = (50.00/R)*e-t/RC .  Find t. 
(e) ma = mg - I'LB; when t = 2.5 ms, I = 0, essentially, and thus I' = 0. Thus,  ma = mg, so a = g.