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# Aim

To determine the magnetic dipole moment (*m*) of a bar magnet and horizontal intensity (*B*_{H}) of earth’s magnetic field using a deflection magnetometer.

## Apparatus

Deflection magnetometer, box-type vibration magnetometer, timer, bar magnet.The deflection magnetometer consists of a large compass box with a small magnetic needle pivoted at the center of a circular scale so that the needle is free to rotate in a horizontal plane .

A large aluminium pointer is rigidly fixed perpendicular to the magnetic needle. The circular scale is graduated in degrees. (0-0) and (90-90) readings are marked at the ends of two perpendicular diameters. The compass box is placed at the center of a wooden board one meter long. The wooden board has a millimeter scale along its axis. The zero of this scale is at the center of the compass box.

The box-type vibration magnetometer consists of a rectangular box with glass sides. A vertical tube is fitted at the top and a torsion-free fiber is suspended from the top of this tube carries a light aluminum stirrup in the box below. The bar magnet can be placed horizontally in this stirrup. At the bottom of the box two leveling screws are provided .

## Theory

The horizontal component of earth's magnetic field,* B*_{H}, is the component of the magnetic field of the earth along a horizontal plane whose normal vector passes through the center of the earth. *B*_{H}_{ }is measured in Tesla, *T*.

The magnetic dipole moment *m* of a magnetic dipole is the property of the dipole which tends to align the dipole parallel to an external magnetic field.* m* is measured in Ampere-square meters (A m^{2}) or, equivalently, in Joules per Tesla (J/T).

### 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*_{H}. Due to the field *B*, the magnet experiences a torque *τ*_{D} , called the deflecting torque, which tends to deflect it from its original orientation parallel to *B*_{H}. If *θ* is the angle between the bar magnet and* B*_{H}, the magnitude of the deflecting torque will be,

The bar magnet experiences a torque* τ*_{R} due to the field *B*_{H} which tends to restore it to its original orientation parallel to *B*_{H}. 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 *θ* , 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

*4π x 10*^{-7} TmA^{-}^{1} - the magnetic permeability of free space, and

*θ* = 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)