Toroid Air Gap: MMF, B-H Curve, Flux Density & More

[sarahgarden]

Homework Statement

(I have done parts i) through v) correctly. It is the remaining two I'm stuck on.

A toroid of rectangular cross-section has a height of 10mm, an inner diameter of 68mm and an outer diameter of 82mm. It is made of iron and is wound with 118 closely-spaced turns. When a current of 2A flows, calculate;

i)the mmf around the toroid
ii)the magnetic field-strength within the toroid
iii)Using B-H curve determine the flux density and flux in the ring,
iv)the relative permeability at this point on the curve

A 1mm air gap is then cut in the toroid. The current is still 2A. Calculate;
v)the new flux density in the ring,
vi)the mmf around the ring, and across the air gap.
vii) It is then required to establish a flux density of 0.75T in the air gap. Calculate the current needed to do this.

Answers to parts vi) and vii)
vi) 53, 183 A-t
vii) 8.0 A

Homework Equations

$$m.m.f.=NI=BS(R_{m}iron+R_{m}air)$$

$$R_{m}=\frac{l}{\mu_{0}\mu_{r}S}$$

where

$$\mu_{r}=557$$,

I have $$B=0.21T$$ (although the answer gives it as 0.23)

$$R_{m}(iron)=4.787*10^6$$ and $$R_{m}(air)=11.365*10^6$$

$$S$$= cross-sectional area = $$7*10^{-5}$$

The Attempt at a Solution

I get the m.m.f. across the air gap correct if I use the value of B = 0.23T in answers. My answer was 0.21T and I found this was significant:

$$R_{m}(air)=\frac{l}{\mu_{0}S}$$

$$m.m.f.=BS(R_{m}(air))$$

$$=0.23*7*10^-5*11.365*10^6=183.5At$$

I did as before with the iron, subtracting the 1mm from \piD for l (in metres)

$$m.m.f.=BS(R_{m}(iron)$$ where

$$R_{m}(iron)=\frac{l}{\mu_{0}\mu_{r}S}$$

but I get 70, which is far from the answer of 53.

For the last part (vii) I don't know why it isn't simply

$$I=\frac{BS}{N}R_{m}(air)$$

with this I get 5A rather than 8Your help is much appreciated.

 

[AvroArrow]

For part vi), you have to take into account the fact that B (the flux density) has changed as a result of introducing the air gap. So, the value of B in your equation should be the new flux density, which is 0.75T.For part vii), you are correct that the equation you provided should work. However, you have not taken into account the fact that the relative permeability has changed due to the addition of the air gap. The relative permeability of air is much lower than the relative permeability of iron, so the reluctance of the air gap is much higher than that of the iron. This means that the current needed to produce a given flux density will be higher (because the reluctance is higher).