Question

A cylindrical coil with a length L = 30 mm with n = 75 turns carries a current I = 0.1 A. What is the field in the electromagnet? A silicon-iron core with a relative permeability μr = 3000 is placed in the core of the coil. What is the magnetization of the silicon-iron?

Answer 1

The given values in the questions are as follows:-

The cylindrical coil has a length (L)=30mm=30×10^-3m

It carries a current (I)=0.1A.

It has turns (n)=75.

And the relative permeability of silicon-iron core is (_r)=3000.

So,we can calculate magnetic field produced by electro magnetic by the below formula,

H=ni/L

By substituting all the values we will get the answer

H=(75×0.1)/30×10^-3 A/m

=>H=(75×1/10)/(30×10^-3) A/m

=>H=(75×1/10)/(3×10^-2) A/m

=>H=(75)/(3×10^-1) A/m

=>H=250 A/m

Now,the relative permeability of silicon-core(ferrite) is given in the above question is _r=3000

So by this formula we can find induced magnetic field

B=_r0H

=>B=3000×4π×10^-7×250 Tesla

=>B=3×10³×4π×10^-7×25×10 Tesla

=>B=3×4π×25×10⁴×10^-7 Tesla

=>B=75×4π×10^-3 Tesla

=>B=0.942 Tesla

=>B=0.94 Tesla

And now we can find the magnetisation for it by the help of below formula

=>M=(B/₀)-H

=>M=(0.94/4π×10^-7)-250 A/m

=>M=748157.643312102 A/m

=>M=748158 A/m

=>M=7.48×10^-5 A/m

=>M=7.5×10^-5 A/m

Summary:-

1.So,the 250 A/m is the field in the electromagnet.

2.And the magnetisation of silicon-iron is 7.5×10^-5 .

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