# 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?

# 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?

## 1 Answers

### Tangent law

Consider a bar magnet with magnetic moment*m*, suspended horizontally in a region where there are two perpendicular horizontal magnetic fields, and external field*B*and the horizontal component of the earth’s field*B _{H}*. If no external magnetic field

*B*is present, the bar magnet will align with

*B*. Due to the field

_{H}*B*, the magnet experiences a torque

*t*, called the deflecting torque, which tends to deflect it from its original orientation parallel to

_{D}*B*. If

_{H}*?*is the angle between the bar magnet and

*B*, the magnitude of the deflecting torque will be,

_{H}The bar magnet experiences a torque*t _{R}*due to the field

*B*which tends to restore it to its original orientation parallel to

_{H}*B*. This torque is known as the restoring torque, and it has magnitude.

_{H}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)

where*I*= moment of inertia of the magnet about the axis of oscillation*m*= magnetic moment of the magnet*B _{H}*

_{}= 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) period*T*and calculating the moment of inertia*I*of the bar magnet,*mB _{H}*is determined using the vibration magnetometer. We will call this value

*x*.

### Working principle

### 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 poles*P*(though in reality no single magnetic pole can exist), one positive and one negative, separated by a distance*L = 2l*, with the positive pole labeled*N*and the negative pole labeled*S*. By analogy with Coulomb’s law, for each pole we would have a field.

,

and a magnetic dipole moment.

.

where*l = L/2*is the half-length of the magnet*m*= magnetic moment of the magnet*4p x 10 ^{-7}TmA^{-}*

^{1}- the magnetic permeability of free space, and

*?*= deflection of aluminium pointer.

Therefore, by the tangent law, at equilibrium

(6)

Solving for*m/B _{H}*

_{}we get:

(7)

We will call this value*y*.

### 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 distances*d*chosen 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,

which leads to,

(8)

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

Now using (5), (7) and (8) we can calculate*m*and*B _{H}*.

(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 placed*vertically*, 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 distance*d*from 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)

Where*P*is the pole strength in Amp-meters (A m) and*L*is the length of the bar magnet in meters. In equation (1), the horizontal component of the field from the bar magnet*B _{H}*bar corresponds to the external field

*B*, so we have . Substituting this in (13) and solving for the pole strength

*P*of the bar magnet,

(14)

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Rinkoo Gupta

AskIITians Faculty