Example problems of magnetic circuits: Difference between revisions

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Now with this, the length and cross sectional area of the core we can solve for reluctance <math> R_c </math> by: <br>
Now with this, the length and cross sectional area of the core we can solve for reluctance <math> R_c </math> by: <br>


<math> R_c = \frac{L}{\mu A} = \frac{1}{1.2566 \times 10^{-6}\times .1} = 7.96 \times 10^{6}
<math> R_c = \frac{L}{\mu A} = \frac{1}{1.2566 \times 10^{-6}\times .1} = 7.96 \times 10^{6} </math> <br> <br>
Similarly to get the reluctance of the gap <br>
 
<math> R_g = \frac {g}{\mu_0 (\sqrt{A} + g)^2} </math>

Revision as of 18:56, 10 January 2010

Given:

A copper core with susceptibility χm=9.7×106

length of core L = 1 m

Gap length g = .01 m

cross sectional area A = .1 m

current I = 10A

N = 5 turns


Find: B

Solution: First we need to find the permeability of copper μ given by the equation
μ=μ0(1+χm)

Which yeilds μ=4×π×107(1+9.7×106)=1.2566×106

Now with this, the length and cross sectional area of the core we can solve for reluctance Rc by:

Rc=LμA=11.2566×106×.1=7.96×106

Similarly to get the reluctance of the gap

Rg=gμ0(A+g)2