Question

An iron core shaped as a doughnut with round cross-section of radius `a = 3.0 cm` carreis a winding of `N = 1000` turns through which a current `I = 1.0 A` flows. The mean radius of the doughnut is `b = 32 cm`. Using the polt in Fig. Find the magnetic energy strored up in the core. A field strength `H` is suposed to be the same throughout the cross-section and equal to its magnitude in the centre of the cross-section.

Answer 1

We apply circulation theorem,
`H.2 pi b = N I`, or, `H = N I//2 pi b`.
Thus the total energy,
`W = (1)/(2) BH. 2 pi b. pi a^(2) = pi^(2) a^(2) b BH`.
Given `N,I, b` we know `H`, and can find out `B` from the `B -H` curve. Then `W` can be calculated.