Question

Consider a toroidal coil with a rectangular cross section (inner radius a, outer radius b, height h), which carries a total of N turns. using the expression for magnetic energy density u = B²/2mu₀

Answer 1

Let 'a' be the inner radius of toroidal coil

'b' be the outer radius of toroidal coil

and 'h' be the height of the toroidal.

Applying an ampere's law, To find the magnetic field B :

. = ₀ I_enc.                                                                              { eq.1 }

simplify an above eq.

B (2r) = ₀ N I

OR      B = ₀ N I / 2r                                                                                       { eq.2 }

Using an equation of magnetic flux through one turn of the toroid by integrating over the rectangular cross-section with dA :

_B = .                                                                                                                   { eq.3 }

integrate with limit b to a,

_B = (₀ N I / 2r) . (h dr)

_B = (₀ N I h / 2) ln (b / a)                                                                                                         { eq.4 }

The self-inductance may be defined as :

L = N _B / I                                                                                                                                   { eq.5 }

inserting the value of '_B' in eq.5,

L = N (₀ N I h / 2) ln (b / a) / I

L = (₀ N² h / 2) ln (b / a)                                                                                                             

using the expression for magnetic energy density, u_B = B² / 2₀    { eq.6 }

inserting the value of 'B' in eq.6,

Or     u_B = ₀ N² I² / 8² r²