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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 μ1 and μ2 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 μ1H2V before inserting the specimen and 1/2 μ2H2V, after inserting the specimen. The difference of these energies is equal to the work done is inserting the specimen,
        . : Work done = 1/2 (μ2 – μ1) H2V
This is stored as potential energy of field.
Change in potential energy of field,
        U = 1/2 (μ2 – μ1) H2V                                … (1)

We have, μr = 1 + χm     or    μ/μ0 = 1+ χm

. :    μ = μ0 + μ0 χm
Let χ1 and χ2 be the susceptibilities of air and specimen. Then,
    μ1 = μ0 + μ0χ1 and μ2 = μ0 + μ0 χ2
. :    μ2 – μ1 = μ02 – χ1)                                    … (2)

Substituting this in Eq. (1), we get
    U = 1/2 μ02 – χ1) H2V                                    … (3)

Let Fx be the force acting on the specimen along x-axis. Then,
    F = - dU/dx – d{1/2μ02 – χ1)H2V}/dx
        = -1/2 μ02 – χ1) d(H2)V/dx

If Hx, Hy and Hz are components H, then
    H2 = Hx2 + Hy2 + Hz2
. :    F = - 1/2 μ02 – χ1) {d(Hx2 + Hy2 + Hz2)/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
    dFx = - μ02 – χ1) [Hx dHx/dx + Hy dHy/dx + Hz dHz/dx] dV

Let α be cross-sectional area of rod. Then,
    dv = α dx
. :    dFx = - μ02 – χ1)[Hx dHx/dx + Hy dHy/x + Hz dHz/dx]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
    dFx = - μ02 – χ1) Hy dHy/dz adx                                … (6)

The total vertical force on the rod due to whole field variation along the length under the limit Hy = H to Hy = H0 is
    Fx = - μ02 – χ1) a ∫H0 Hy dHy dx/dx
        = - μ02 – χ1) a ∫H0 Hy dHy
        = - μ02 – χ1) a [Hy2/2]HHo
        = - 1/2μ02 – χ1) a (H02 – H2)                             … (7)

Let m1 and m2 be the weights to counterpoise the force on rod with magnetic field off and on respectively. Then,   
        Fx = (m2 – m1)g.                                    … (8)
Equation (7) and (8), we get
        (m2 –m1)g = 1/2μ02 – χ1) a (H2 – H02)
    (χ2 – χ1) = 2(m2 – m1)g / μ0 a(H2 – H02)                            … (9)

H0 and H can be measured by the flux meter. Thus by putting the susceptibility of air χ1 and other known quantities in the above equations, χ2 is calculated.

Gouy Method

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