- #1

Leonid92

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- 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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