A thin ring made of magnetic has a mean diameter `d = 30 cm` and supports a winding of `N = 800` turns. The cross-sectional area of the ring is equal to `S = 5.0 cm^(2)`. The ring has a cross-cut of width `b = 2.0 mm`. When the winding carries a certain current, the permeability of the magnectic equals `mu = 1400`. Neglectign the disipation of magnetic flux at the gap edges, find:
(a) the ratio of magnetic energies in the gap and in the magnetic,
the inductance of the system, do it two ways: using the flux and using the energy of the field.

Explanation

From `oint vec(H).d.vec(r) = N I`,
`H.pi d + (B)/(mu_(0)) .b = N I, (d gt gt b)`
Also, `B = mu mu_(0) H`. Thus, `H = (NI)/(pi d + mu b)`.
Since `B` si continous across the gap, `B` is given by,
`B = mu mu_(0) = (N I)/(pi d + mu b)`, both in the magnetic and the gap
(a) `(W_(gap))/(W_(magn etic)) = ((B^(2))/(2 mu_(0)) xx S xx b)/((B^(2))/(2 mu mu_(0)) xx S xx pi d) = (mu b)/(pi d)`
(b) The flux is `N int vec(B).d. vec(S) = N mu mu_(0) (N I)/(pi d + mu b). S = mu_(0) (S N^(2) I)/(b + (pi d)/(mu))`
Energy wise, total energy
`= (B^(2))/(2 mu_(0)) ((pi d)/(mu) + b)S = (1)/(2) (mu_(0) N^(2) S)/(b + (pi d)/(mu)) . I^(2) = (1)/(2)L I^(2)`
The `L`, found in the one way, agress with that, found inthe other way. Noe that, in caluculating the flux, we do not consider the field in the gap, since it is not linked to the winding. But the total energy includes that of the gap.