Question

a magnetising field of 1500A/m produces flux of 2.4*10^-5 weber in a iron bar of cross section area of .5 cm^2 . the permeability of iron bar is

Answer 1

First magnetic induction B = flux / area is to be found out. B = 2.4 x 10-5 / 0.2 x 10-4 = 1.2 T
Now permeability u = B / H
Given H = 1600 A/m, and found B = 1.2 T
So u = 7.5 x 10-4 H/m
CONGRATS! Friend Mukul is right
Now with susceptibility chi we have to follow u / uo = 1 + chi
uo = 4 pi x 10-7
By the by chi = 596 (nearly) 

Answer 2

Answer:

H=1500A/m. ;. Flux (Phi) =2.4*10^-5

Area of cross section A=0.2 cm^2

Magnetic induction B=flux/area

B=2.4*10^-5 / 0.2*10^-4

B=1.2 T

Permeability u=B/H

u=1.2/1500

u=0.8*10^-3