How Does a Toroidal Coil's Magnetic Field Change with an Air-Gap Insertion?

[Aisling1993]

Homework Statement

A toroidal coil,with N turns of wire carrying a current I, is uniformly wound around a
toroidal core that has a mean radius a, a cross-sectional area A= ∏(d/2)^2, and is made of a linear, isotropic homogeneous material with relative permeability μr. You may assume that the cross-sectional diameter, d, is very much smaller than a, d < a.

a, Use Ampère’s law to derive a formula for the magnetic field B inside the torus, and hence the total magnetic flux through the toroidal solenoid

b, A narrow air-gap of width w is now made, by removing a small sector of the toroidal core, so that the gap is a fiftieth of the circumference of the toroid. Show that the magnetic field in this gap is given by;

B = μrμ0NI/ 2∏a - w + wμr

Homework Equations

∫B.ds= B∫ds

flux= B*A

The Attempt at a Solution

For part a I got B=μ0μrNI/2∏a

and The flux ,phi = B*A = μ0μrNId^2/8

b, μ0μrNI = Bc.dsc + Bg.dsg

where Bc is the magnetic field in the toroidal core and Bg in the gap. Dsc is the area of the core and Dsg the area of the gap

Any tips on where to begin with part b would be much appreciated,

Thanks

 

[mfb]

B=μ0μrNI/2∏a

Okay.

flux ,phi = B*A = μ0μrNId^2/8

The right side is not B*A.

b, μ0μrNI = Bc.dsc + Bg.dsg

where Bc is the magnetic field in the toroidal core and Bg in the gap. Dsc is the area of the core and Dsg the area of the gap

Good. What is the relation between B- and H-field in the material / in the air gap? You can assume that the field is perpendicular to the surface of the material.

 

[Aisling1993]

Hello, thanks for helping, so now

Flux , phi = ∫B.dA = μ0μrNId^2/8

B=μ0(H + M) for the toroidal material but we have magnetic linearity so:

Bc= μ0μrHc and Bg=μ0Hg

where Bc and Hc represent the B-field and H-field in the toroid and Bg and Hg in the air gap.

but the total flux trough each area of the circuit must be the same

so Acμ0μrHc = Agμ0Hg

where Ac and Ag represent the cross sectional area of the toroid and air gap respectively.

So Hg = Acμ0μrHc / Agμ0

Is this the right direction? it gets quite messy after this point

 

[mfb]

Flux , phi = ∫B.dA = μ0μrNId^2/8

Again, the left side and the right side are different things.

At some point, it gets interesting to consider a line integral along the toroid core...