Question

9. A toroid has center line radius 0.1 m. cross section area of 0.2 cm., determine the number of turns so that its inductance is 0.2 uF if it is filled with air. For the same parameters, what will be the inductance if it is filleod with a material of , -500? In the material-filling case, what is the magnetic flux density in the core if 2A current is applied? 10. For the toroid mentioned in problem 9, if there is an air gap of 1mm, how much will be the inductance (for the same parameters).

9 and 10 please

Answer 1

a) L = mu N_0 N^2 A/2 pi r A = cross sectional area r = torpid radius to center line L = 1 times 4 pi times 10^-7 times N^2 times 0.2 times 10^-4/2 pi times 0.1 = 0.2 times 10^-6 N^2 = 2500 N = 500 b) mu r = 500 L = mu_r mu_0 N^2 A/2 pi r = 500 times 4 pi times 10^-7 times n^2 times 0.2 times 10^-4/2 pi times 0.1 N^2 = 500 N = 22.36 TildeEqual 23 B = mu_r mu_0 N I/2 pi r = 500 times 4 pi times 10^-7 times 23 times 2/2 pi times 0.1 = 46 times 10^-3 Weber \r\n if air gap present then formula is L = n^2 times mu_r times mu_0 times H/l for air L = N^2 times 1 times 4 pi times 106-7 times 0.2 times 10^-4/1 times 10^-3 N^2 = 7.96 N = 2.83 TildeEqual 3 b) 0.2 times 10^-7 = L - N^2 times 500 times 4 pi times 10^-7 times 0.2 times