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Aisling1993
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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