# Air gap in toroid

#### sarahgarden

1. 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

2. 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 = $$7*10^{-5}$$

3. 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 8

Last edited:
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