Login

a magnetising field of 1600 A/m produces a magnetic flux of 2 .4 x 10-5 Wb in a bar of iron of cross section 0 .2 cm2 . Calculate the Permeability and Susceptibility of the bar ?

u = 0.75 × 10^-3

& chi = 624 ,

is it correct i dont thik so
  • -19
View Full Answer
Hi let us check friend Mukul's answer by carrying out the steps 
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) 
Here Singh goes wrong. I don't know why
  • 42
Hii
  • -35
Question

  • 2
Hope this helpful

  • 2
Hi let us check friend Mukul's answer by carrying out the steps 
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) 
Here Singh goes wrong. I don't know why
  • 1
What are you looking for?