# Why a deflection magnetometer should be placed in tan A and tan B position in order to calculate the magnetic moments of a bar magnet?

Rinkoo Gupta
8 years ago

### Tangent law

Consider a bar magnet with magnetic momentm, suspended horizontally in a region where there are two perpendicular horizontal magnetic fields, and external fieldBand the horizontal component of the earth’s fieldBH. If no external magnetic fieldBis present, the bar magnet will align withBH. Due to the fieldB, the magnet experiences a torquetD, called the deflecting torque, which tends to deflect it from its original orientation parallel toBH. If?is the angle between the bar magnet andBH, the magnitude of the deflecting torque will be,

The bar magnet experiences a torquetRdue to the fieldBHwhich tends to restore it to its original orientation parallel toBH. This torque is known as the restoring torque, and it has magnitude.

The suspended magnet is in equilibrium when,

(1)

TThe above relation, called the tangent law, gives the equilibrium orientation of a magnet suspended in a region with two mutually perpendicular fields.

### Vibration Magnetometer

The equation of motion of the bar magnet suspended horizontally in the earth’s magnetic field is

Thus its period of oscillation, for small?, is approximately.

(2)

whereI= moment of inertia of the magnet about the axis of oscillation
m= magnetic moment of the magnet
BH= horizontal intensity of the earth's magnetic field.
For a rectangular bar magnet,

(3)

Where
M= mass of the magnet
L= length of the magnet (longest horizontal dimension)
b= breadth of the magnet (shortest horizontal dimension)

Squaring equation (2)

(4)

which gives us,

(5)

Thus, by measuring vibration (oscillation) periodTand calculating the moment of inertiaIof the bar magnet,mBHis determined using the vibration magnetometer. We will call this valuex.

### Tan-A position

In Tan A position (Fig. 1), prior to placement of the magnet, the compass box is rotated so that the (0-0) line is parallel to the arm of the magnetometer. Then the magnetometer as a whole is rotated till pointer reads (0-0). Finally, the bar magnet (the same one that was previously suspended in the Vibration Magnetometer) is placed horizontally, parallel to the arm of the deflection magnetometer, at a distance d chosen so that the deflection of the aluminum pointer is between 30° and 60°.

The magnet is a dipole. Suppose that, analogous to an electric dipole, there are two magnetic polesP(though in reality no single magnetic pole can exist), one positive and one negative, separated by a distanceL = 2l, with the positive pole labeledNand the negative pole labeledS. By analogy with Coulomb’s law, for each pole we would have a field.

,

and a magnetic dipole moment.

.

wherel = L/2is the half-length of the magnet
m= magnetic moment of the magnet
4p x 10-7TmA-1- the magnetic permeability of free space, and
?= deflection of aluminium pointer.

Therefore, by the tangent law, at equilibrium

(6)

Solving form/BHwe get:

(7)

We will call this valuey.

### Tan-B position

In this position (Fig. 2), prior to placement of the magnet, the compass box alone is rotated so that the (90-90) line is parallel to the arm of the magnetometer. Then the magnetometer as a whole is rotated so that the pointer reads (0-0). Finally, the magnet is placed horizontally, perpendicular to the arm of the magnetometer, at distancesdchosen so that the deflection of the aluminium pointer is between 30°and 60°.

From Fig. 5, at point C,

Thus the field due to the bar magnet in the center of the compass is,

(8)

Equation (8) gives us a second value ofy, which we average with the first, from equation (7).

Now using (5), (7) and (8) we can calculatemandBH.

(9)

(10)

Hence, the magnetic moment of the bar magnet is,

(11)

And the horizontal component of earth’s magnetic field is,

(12)

### Tan C position

In this position (Fig. 4), the bar magnet is placedvertically, in contrast to the Tan A and Tan B positions, where it is placed horizontally. The bottom end of the bar magnet is placed a distancedfrom the center of the compass box, chosen so that the deflection of the aluminum pointer is between 30° and 60°. From Fig. 7, the horizontal component of the field from the bar magnet at the center of the compass is

which reduces to,

(13)

WherePis the pole strength in Amp-meters (A m) andLis the length of the bar magnet in meters. In equation (1), the horizontal component of the field from the bar magnetBHbar corresponds to the external fieldB, so we have . Substituting this in (13) and solving for the pole strengthPof the bar magnet,

(14)