Complete Question
The complete question is shown on the first uploaded image
Compare the percent error between your value and the accepted value of 1.26 times 10-6 T · m/A. (Use the accepted value given to three significant figures in your calculation.). 100% % error = %
Answer:
The permeability of free space is [tex]\mu_0 = 1.32*10^{-6} \ Tm/A[/tex]
The percentage error is % error = 5.25%
Explanation:
From the question we are told that
The slope is [tex]s= 3.9\ G/A[/tex] , and
Generally [tex]1 \ gauss = 10^{-4 } tesla[/tex]
So [tex]s = 3.9 *10^ {-4} T /A[/tex]
From the relation given in the question
[tex]\mu_0 = \frac{2R}{N} [\frac{B}{I} ][/tex]
Where R is the radius of the coil [tex]=\frac{Diameter \ of \ coil }{2} = 0.017m[/tex]
N is the number of loops of the coil = 10
Now from the question we are told that
[tex]s = \frac{B}{I}[/tex]
substituting into the equation above
[tex]\mu_0 = \frac{2R }{N} s[/tex]
Substituting values
[tex]\mu_0= \frac{2 * 0.017}{10 } * 3.9*10^{-4}[/tex]
[tex]= 1.326 *10^ {-6} \ Tm/A[/tex]
Generally the % error is mathematically represented as
%[tex]error = \frac{Measured \ value - accepted \ value}{accepted \ value } *100[/tex]
Given that the accepted value is [tex]\mu_o_ {acc} = 1.26 *10 ^{-6} \ T \cdot m /A[/tex]
Hence substituting values
%[tex]error = \frac{(1.326-1.26)*10^{-6}}{1.26 *10 ^ {-6}} *100[/tex]
[tex]= 5.24[/tex]
