Question

Discuss magnetization of a ferromagnetic material with help of Rowland ring.

Answer 1

Hint: Use the formula for the magnetic field of a toroid without the iron core and the formula for the additional magnetic field due to the presence of the iron core. Derive the equation for the magnetic field of the Rowland ring with the iron core.

Complete step by step answer:

The magnetic permeability is given by
\[{\mu _0} = \dfrac{B}{H}\] …… (1)

Here, \[{\mu _0}\] is the magnetic permeability, \[B\] is the magnetic field and \[H\] is the intensity of the magnetizing field.

The additional magnetic field is given by
\[{B_M} = {\mu _0}M\] …… (2)

Here, \[{B_M}\] is the additional magnetic field, \[{\mu _0}\] is the magnetic permeability and \[M\] is the magnetization.

Complete step by step answer:

The ferromagnetic material such as iron undergoes magnetization which can be studied with the help of a Rowland ring.

The Rowland ring has the same shape as that of the toroid. The diagram of the Rowland ring is as follows:

In the above figure, a Rowland’s ring is shown in the centre of which there is an iron core and the outer toroid ring is wrapped with \[n\] turns of the coil carrying a current \[I\].

For the toroid with \[n\] number of turns (without the iron core), the magnetic field \[{B_0}\] should be
\[{B_0} = {\mu _0}nI\]

Here, is the permeability of the vacuum.

For the Rowland’s ring with the iron core, the magnetic field \[B\] is the sum of the magnetic field \[{B_0}\] of the toroid and the additional magnetic field \[{B_M}\] due to the iron core.
\[B = {B_0} + {B_M}\] …… (3)

Rewrite equation (1) for the magnetic field \[{B_0}\] of the toroid without the iron core.
\[{B_0} = {\mu _0}H\]

Rewrite equation (2) for the additional magnetic field \[{B_M}\].
\[{B_M} = {\mu _0}M\]

From the above equation, it can be concluded that the additional magnetic field \[{B_M}\] is directly proportional to the magnetization \[M\].

Substitute \[{\mu _0}M\] for \[{B_M}\] and \[{\mu _0}H\] for \[{B_0}\] in equation (3).
\[B = {\mu _0}H + {\mu _0}M\]
\[ \Rightarrow B = {\mu _0}\left( {H + M} \right)\]

Note:The magnetic field of a toroid due to current carrying coils is less than the magnetic field of the Rowland ring due to the current carrying coils and the iron core.