(i) Mention two properties of soft iron due to which it is preferred for making an electromagnet.
(ii) State Gausss law in magnetism. How is it different from Gausss law in electrostatics and why?
OR
Derive an expression for the axial magnetic field of a finite solenoid of length 2l and radius r carrying current I. Under what condition does the field become equivalent to that produced by a bar magnet?
To make an electromagnet, the choice of the core material is very important as it enhances the magnetic capabilities. Thus, soft iron is used as:
(i) It has high permeability. This means that it provides considerable magnetization for small magnetizing fields.
(ii) It has low retentivity. This property allows the electromagnet to lose the magnetism immediately as the magnetizing field is removed.
Gauss Law in magnetism states that the net magnetic flux through any closed surface is zero. This is because the magnetic field lines are continuous loops, hence for each magnetic field line going in we have one that’s coming out giving the net flux zero.
Its stated as, ![]()
Gauss law in electrostatics states that the net electric flux through a closed surface is proportional to the charge enclosed by the close surface. Hence, it’s zero only if the charge enclosed is zero. It’s given by,
.
In magnetism, monopoles do not exist and only dipoles exist. Hence the total flux is always zero. Whereas, charges can exists separately, thus giving a definite flux.
OR
Let there be a solenoid of radius an and of length 2l. Let the number of turns be n and the current through the solenoid be I.
Here, OP=r.
Let there be a small element dx at a distance x from O.
The number of turns in the element = ![]()

Then the magnitude of magnetic field at P is
![]()
If
and ![]()
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If M is the magnetic moment of the solenoid,
![]()
This will give the magnetic field as
![]()
Thus, the magnetic field due to a solenoid resembles that of a bar magnet.
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