Question 15.2: Flux Density in a Toroidal Core Consider the toroidal coil s...
Flux Density in a Toroidal Core
Consider the toroidal coil shown in Figure 15.8. Find an expression for the magnetic flux density B on the center line of the core in terms of the number of coil turns N, the current I, the permeability μ of the core, and the physical dimensions. Then, assuming that the flux density is constant throughout the core (this is approximately true if R >> r), find expressions for the total flux and the flux linkages.
By symmetry, the field intensity is constant in magnitude along the dashed circular center line shown in the figure. (We assume that the coil is wound in a symmetrical manner all the way around the toroidal core. For clarity, only part of the coil is shown in the figure.) Applying Ampère’s law to the dashed path, we obtain
Hl=H2\pi R=NI
Solving for H and using Equation 15.10 to determine B, we have
\mathrm{B}=\mu \mathrm{H} (15.10)
H=\frac{NI}{2\pi R} (15.16)
and
B=\frac{\mu NI}{2\pi R} (15.17)
Assuming that R is much greater than r, the flux density is nearly constant over the cross section of the core. Then, according to Equation 15.6, the flux is equal to the product of the flux density and the area of the cross section:
\phi=BA=\frac{\mu NI}{2 \pi R}\pi r^2=\frac{\mu Nir^2}{2R} (15.18)
Finally, we note that all of the flux links all of the turns, and we have
\lambda = N\phi=\frac{\mu N^2Ir^2}{2R} (15.19)