SSC JE Electrical Engineering 2015 Paper

Question : 171 of 199

Marks: +1, -0

A current of 2A passes through a coil of 350 kms wound on a ring of mean diameter 12 cm. The flux density established in the ring is 1-4 wb/m

^2

.Find the value of relative permeability of Iron.

Solution:

B = 1.4 web/m

^2

D = 1.2 m
N = 350
MMF (F) = NI = 350 × 2 = 700 AT

I= πD = π × 0.12

µ_0=4π×10^{-7}

At= B/{µ_ 0 µ_r} I

µ_r=B/{ATµ_ 0 } I={1.4 ×.12}/{4 ×700 ×10^{-7}}

{6 ×10^{-5}}/10^7=600

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