A) Calculate the inductance of the solenoid. (You can ignore thevariation of the

Question

A) Calculate the inductance of the solenoid. (You can ignore thevariation of the magnetic field across the cross section of thetoroid.)
B)If the material is removedfrom inside the solenoid and replaced by air, what is theinductance of the solenoid? A toroidal solenoid has a cross-sectional area of 0.460 cm^2, a mean radiusof 11.5 cm, and 1900 turns.The space inside the windings is filled with a ferromagneticmaterial that has a relative permeability of 510. A) Calculate the inductance of the solenoid. (You can ignore thevariation of the magnetic field across the cross section of thetoroid.) B)If the material is removedfrom inside the solenoid and replaced by air, what is theinductance of the solenoid?

Explanation / Answer

Given that a toroidal solonoid has a cross sectional area A=0.460 cm 2 or 0.460 *10^-4 m2 Also given mean radius as r = 11.5 cms or 11.5*10 ^-2mts No. of turns in a solonoid is given by N = 1900 Relative permeability of ferromagnetic material isgiven by k m =510 A) the inductance of the solonoid is given by L= (0N 2A)/2r where is the permeability of the ferromagneticmaterial = 0 km          0 is the permeability in vacuum  0= 4 *10^-7              now by substituting the above values we get inductance ofsolonoid as L= 0.014 H B)when the material is removed and the space is filled by theair then inductance is given by L = 0N2A/2r                            L=288 H          0 is the permeability in vacuum  0= 4 *10^-7              now by substituting the above values we get inductance ofsolonoid as L= 0.014 H B)when the material is removed and the space is filled by theair then inductance is given by L = 0N2A/2r                            L=288 H

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