Question

3. The field coil has N, turns and a finite thickness to it, with turns of radius varying from inner to outer radii, Ri and Ro. Show that when
Nf = 200 turns, I0 = 3.342E-3A, R = 0.105m

Answer 1

Consider a ring of radius r thickness Lr & R_i lessthanorequalto r lessthanorequalto R_0 Total no of turns = M_f No of turns per unit length, n = N_f/(R_0 - R_1) (A turns are distributed from R_1 to R_0) No of turns in ring of thickness dr n_r = n de we know, magnetic field at the center of circular loop, B(z) = mu_0 2 pi r^2 N_f I_0/4 pi (R^2 + 2^2)3/2 if z = 0 (then at center) rightdoublearrow B_center = mu_0 2 pi R^2 N_fI_0/4pi R^3 = mu_0N_fI_0/2R unity this relation, magnetic field of ring dB = mu_0 Inr/2 r = mu_0 I N_f/2 (R_0 - R_i) - dr/r for loop, r form R_i to R_o All dB^rightarrow will be out of page for countor clock current so B = integrtal^R_0_R_1 dB = integral^R_0_R_1 mu_0 I N_f dr/2(R_0 - R_1) r rightdoublearrow B = mu_0 I N_f /2(R_0 - R_1) integral^R_0_R_1 dr/r rightdoublearrow B = mu_0 I N_f /2(R_0 - R_1) ln R_0/R_i