## Gouy S Method

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# Gouy’s Method for Measurement of Susceptibility

Apparatus. It consists of an electromagnet having parallel and flat pole pieces with a gap between them. The electromagnet provides a strong horizontal magnetic field of nearly 1 Wb m-2.

The sample under test is taken in the form of a thin rod. It is suspended vertically from one arm of a sensitive micro balance in the magnetic field between the wedge shaped pole pieces of the electromagnet. The specimen is suspended in such a way that its lower end is near the midpoint of the magnetic field where the magnetic intensity H is large while its upper end is outside the field in the region of low intensity H0. The cylindrical rod is then weighed by suspended it from a sensitive microbalance in the two cases:

(i) When field is off. (ii) when field is on.

Theory. Let μ

_{1}and μ

_{2}be the permeability of air and specimen respectively. Consider a magnetic field of intensity H in a region. Let V be the volume of the specimen. Hence the energy of magnetic flux in this region is 1/2 μ

_{1}H

_{2}V before inserting the specimen and 1/2 μ

_{2}H

_{2}V, after inserting the specimen. The difference of these energies is equal to the work done is inserting the specimen,

. : Work done = 1/2 (μ

_{2}– μ

_{1}) H

_{2}V

This is stored as potential energy of field.

Change in potential energy of field,

U = 1/2 (μ

_{2}– μ

_{1}) H

_{2}V … (1)

We have, μr = 1 + χm or μ/μ

_{0}= 1+ χm

. : μ = μ

_{0}+ μ

_{0}χm

Let χ1 and χ2 be the susceptibilities of air and specimen. Then,

μ

_{1}= μ

_{0}+ μ

_{0}χ1 and μ

_{2}= μ

_{0}+ μ

_{0}χ

_{2}

. : μ

_{2}– μ

_{1}= μ

_{0}(χ

_{2}– χ

_{1}) … (2)

Substituting this in Eq. (1), we get

U = 1/2 μ

_{0}(χ

_{2}– χ

_{1}) H

_{2}V … (3)

Let Fx be the force acting on the specimen along x-axis. Then,

F = - dU/dx – d{1/2μ

_{0}(χ

_{2}– χ

_{1})H

_{2}V}/dx

= -1/2 μ

_{0}(χ

_{2}– χ

_{1}) d(H

_{2})V/dx

If H

_{x}, H

_{y}and H

_{z}are components H, then

H

_{2}= Hx

_{2}+ Hy

_{2}+ Hz

_{2}

. : F = - 1/2 μ

_{0}(χ

_{2}– χ

_{1}) {d(Hx

_{2}+ Hy

_{2}+ Hz

_{2})/dx}V … (4)

Now, considering X-axis to be vertical, the vertical force on a small element dx of rod of volume dV, at a distance x from origin is

dF

_{x}= - μ

_{0}(χ

_{2}– χ

_{1}) [H

_{x}dH

_{x}/d

_{x}+ H

_{y}dH

_{y}/d

_{x}+ H

_{z}dH

_{z}/d

_{x}] dV

Let α be cross-sectional area of rod. Then,

d

_{v}= α d

_{x}

. : dF

_{x}= - μ

_{0}(χ

_{2}– χ

_{1})[H

_{x}dH

_{x}/d

_{x}+ H

_{y}dH

_{y}/x + H

_{z}dH

_{z}/d

_{x}]adx … (5)

In the narrow gap between pole pieces, the magnetic flux will consists of straight lines direct from one pole face to another, in y direction say. Thus only the component Hy will be of significant magnitude while Hx and Hz will be negligible. Hence Eq. (5) is expressible as

dF

_{x}= - μ

_{0}(χ

_{2}– χ

_{1}) H

_{y}dH

_{y}/dz adx … (6)

The total vertical force on the rod due to whole field variation along the length under the limit H

_{y}= H to H

_{y}= H

_{0}is

Fx = - μ

_{0}(χ

_{2}– χ

_{1}) a ∫H

_{0}H

_{y}dH

_{y}dx/dx

= - μ

_{0}(χ

_{2}– χ

_{1}) a ∫H

_{0}Hy dH

_{y}

= - μ

_{0}(χ

_{2}– χ

_{1}) a [H

_{y}

^{2}/2]H

^{Ho}

= - 1/2μ

_{0}(χ

_{2}– χ

_{1}) a (H

_{02}– H

_{2}) … (7)

Let m

_{1}and m

_{2}be the weights to counterpoise the force on rod with magnetic field off and on respectively. Then,

F

_{x}= (m

_{2}– m

_{1})g. … (8)

Equation (7) and (8), we get

(m

_{2}–m

_{1})g = 1/2μ

_{0}(χ

_{2}– χ

_{1}) a (H

_{2}– H

_{0}

^{2})

(χ

_{2}– χ

_{1}) = 2(m

_{2}– m

_{1})g / μ

_{0}a(H

_{2}– H

_{0}

^{2}) … (9)

H

_{0}and H can be measured by the flux meter. Thus by putting the susceptibility of air χ

_{1}and other known quantities in the above equations, χ

_{2}is calculated.

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