Question

Two coils having 100 and 150 turns respectively are wound side by side on a closed iron circuit of cross-sectional are of 125cmsquare and mean length 200m. If the permeability of iron is 2000. calculate, (i)Mutual inductance between two coils and (ii)The e. m. f. induced in the second coil, if current in first coil changes from 0 to 5A in 0. 02S​

Answer 1

Answer:

Formula used : M = N1N2/(l/μ0μrA)H, N1 = 30, N2 = 600; A = 100 x 10-4 = 10-2m2, l = 2m ∴ M = μ0μr A N1N2/l = 4π × 10−7 × 2000 × 10−2 × 30 × 600/2 = 0.226 H dI1 = 20 − 0 = 20 A; dt = 0.02 s; eM = MdI1/dt = 0.226 × 20/0.2 = 226VRead more on Sarthaks.com - https://www.sarthaks.com/492125/coils-having-turns-respectively-are-wound-side-side-closed-iron-circuit-area-cross-section

Explanation:

Two coils having 30 and 600 turns respectively are wound side-by-side on a closed iron circuit of area of cross-section 100 sq.cm. and mean length 200 cm. Estimate the mutual inductance between the coils if the relative permeability of the iron is 2000. If a current of zero ampere grows to 20 A in a time of 0.02 second in the first coil, find the e.m.f. induced in the second coil.Read more on Sarthaks.com - https://www.sarthaks.com/492125/coils-having-turns-respectively-are-wound-side-side-closed-iron-circuit-area-cross-section