- #1
Leonid92
- 45
- 2
- Homework Statement
- In the core made of electrical steel (Fig. 1), it is required to create a magnetic flux Φ = 4.2*10^(-3) Wb. Determine the number of turns of the winding, if the current is I = 5 A, and the dimensions of the core are specified in millimeters.
- Relevant Equations
- Φ = B*S
I*w = H*L
1) B = Φ/S = (4.2*10^(-3) Wb)/(2.5*10^(-3) m^2) = 1.68 T
2) Using electrical steel magnetization curve given in the textbook: magnetic field strength H corresponding to magnetic flux density 1.68 T is equal to 6000 A/m.
3) L is a length of the middle magnetic line of the core (Fig. 2).
L = 2*(200 - 50 + 200 - 50) = 600 mm = 0.6 m
4) According to Kirchhoff's second law for a magnetic circuit:
I*w = H*L,
where w is a number of turns of the winding.
Then:
w = (H*L)/I = 720.
Is it right solution?
The problem is that true answer given in the textbook is w = 240.
Could you please advise good reference book where one can find electrical steel magnetization curve? I'd like to check H value in another book.
2) Using electrical steel magnetization curve given in the textbook: magnetic field strength H corresponding to magnetic flux density 1.68 T is equal to 6000 A/m.
3) L is a length of the middle magnetic line of the core (Fig. 2).
L = 2*(200 - 50 + 200 - 50) = 600 mm = 0.6 m
4) According to Kirchhoff's second law for a magnetic circuit:
I*w = H*L,
where w is a number of turns of the winding.
Then:
w = (H*L)/I = 720.
Is it right solution?
The problem is that true answer given in the textbook is w = 240.
Could you please advise good reference book where one can find electrical steel magnetization curve? I'd like to check H value in another book.
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