Example problems of magnetic circuits: Difference between revisions

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Solution:
Solution:
First we need to find the permeability of copper <math> \mu </math> given by the equation <br> <math> \mu = \mu_0 (1 + \chi_m)</math> <br> <br>
First we need to find the permeability of copper <math> \mu </math> given by the equation <br> <math> \mu = \mu_0 (1 + \chi_m)</math> <br> <br>
Which yeilds <math> \mu = 4 \times \pi \times 10^{-7}(1+-9.7 \times 10^{-6}) = 1.2566 \times 10^{-6} </math> <br><br>
Which yeilds <math> \mu = 4 \times \pi \times 10^{-7}(1+-9.7 \times 10^{-6}) = 1.2566 \times 10^{-6} \frac{N}{A^2} </math> <br><br>
 
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>



Revision as of 19:11, 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: Bg

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)=1.2566×106NA2

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=.014×π×107(.1+.01)2=74.8×103

Now Using Bg=NI(RgRc)((A+g)2
Yields <math> B_g = \frac{5 \times 10}{74.8 \times 10^{3} \times 7.96 \times 10^{6} \times (\sqrt{.1} + .01)^2} = .789 \times 10^{-9}