The Second Midterm: Problem #2


Statement of the Problem

Solution to the Problem

I began by running my fellow colleague's, Paul Shestople, solution code from assignment NP 5.02 to get a feel for the program. I tested it for various a's between 0 and 1. The results seem reasonable to me. This code will be helpful to begin building my field strength analyzer. So, onwards to my mission of shielding factor determination!

Part A

Using the program that plots the magnetic field lines of a cylindrical object of radius b made of permeable material with a hole of radius a in the center, I calculated the number of field lines in the hole for a diameter of one centimeter (n1) and the number of field lines in the exterior in the range of one centimeter (n3). The samples where drawn at x=-0 and x=-2 respectively. In the program, I set b=1, and a=b/2. The field strength (alpha) is determined to be: alpha=n1/n3 Let's take a look at various choices of permeability constants (mu).

  • For mu = 1: alpha=100.000/100.000=1.00000

  • For mu = 100: alpha=3.00000/111.000=0.0270270

  • For mu = 600: alpha=3.00000/1113.00=0.002695

  • For mu = 1000: alpha=3.00000/1115.00=0.002691

    Reviewing the above results: If one keeps both the interior and the exterior of the cylinder constant, as mu increases alpha decreases as is expected. This is good since the smaller alpha gets, the better the shield. Note: Graph accuracy and determination of alpha become increasingly more difficult as mu is increased due to time constraints resulting from an increase in the array size and the number of computations that the computer needs to compute. These factors increase the time it takes the computer to complete the task.

    Part B

    The object is to determine which metal iron or mu metal will produce a better, cost efficient shield with a shielding factor greater than or equal to 10 and an interior radius of 5 cm. The factors involved are listed in Table 1.

    	Price		$2/kg		$20/kg
    	Density		7900 kg/m^3	7900 kg/m^3
    	mu		100		1000
    Table 1:  Given information for iron and mu metal

    After much time and many computations I found the required information listed in Table 2.

    			 	IRON            MU METAL
    	Outter radius,b		
    	Inner radius, a		0.05m		0.05m 
    Table 2:  Computer determined information for iron and mu metal 

    Comparing the information listed in table 2, the most cost effective material appears to be ?

    Well, I made corrections to the program but it runs very, very slowly. I do not have anymore time to spend on this problem. My guess is that mu=1000 for an extremely small b is probably the best. I did run the program twice for mu=100. The results are:

    	mu=100:     b=0.5          b=1.0
    	test n1     0.8            1.5
    	test n3	    0              0
    	n1=	    9.00000	   7.00000
    	n3=	    23791.0        47315.0  
    	alpha=	    3.78294E4      1.47945E4
    	$/m	    100            100