A coil of 500 turns is wound over a circular iron ring of diameter 20 cm and area of cross section 3 cm². It produces a flux of 0.5 mWb when carrying a current of 2.09 A. Calculate the permeability of the material.

Question

Question

This question was previously asked in

MPPGCL JE Electrical 19 March 2019 Shift 2 Official Paper

Answer 1

Concept

The permeability of the material is given by:

\(μ={B\over H}\)

where, B = Magnetic flux density

H = Magnetic field strength

Calculation

Given, d = 20 cm

N = 500

A = 3 cm²

ϕ = 0.5 mWb

I = 2.09 A

\(H={Ni\over l}={500× 2.09\over \pi × 0.2}=1663.2\space A/m\)

\(B={\phi \over A}={0.3× 10^{-3}\over 3× 10^{-4}}=1.667 \space T\)

\(μ={B\over H}={1.667\over 1663.2}\)

μ = 83.34 × 10^-6 H/m