Deflection Magnetometer and its Application

Deflection Magnetometer

Deflection magnetometer is an instrument used to measure the magnetic moment and polar strength of a bar magnet. It is also used to compare the pole strengths of two bar magnet.

Construction

A deflection magnetometer consists of a small magnetic needle pivoted at the center of a graduated circular scale. Two light aluminum pointers are attached perpendicularly to its two sides to note its deflection. The circular scale id divided into four quadrants, each calibrated in degrees from 0^o to 90^o C.T arrangement is provided in a plastic case with a glass cover at the top and a plane mirror attached at the base. The plane mirror is used to remove the parallax during the measurement.

Tan-A Position

In this position, the magnetometer is placed with its arm along the east-west direction and bar magnet is placed with its arm along the east-west direction and bar magnet is placed on one of the arms symmetrically along it pointing the north pole to the east or west direction.

Suppose that ‘d’ be the distance from the center of the bar magnet to the centre of the needle. Let M be its magnetic moment and 2l its pole strength. Since the needle is in end-on position of the bar magnet, the magnetic field at the needle is

$$ B = \frac {\mu _o\: Md}{2\pi (d^2 – l^2)^2} $$

pointing towards the east or west. The horizontal component, H of earth’s magnetic field is perpendicular to B and the needle will be deflected through an angle \(\theta \). So the middle is under the action of two perpendicular fields and from tangent law,

$$\begin{align*} B &= H \tan \theta \\ \text {or,} \: B &= \frac {\mu _0 \: Md}{2\pi (d^2 – l^2)^2} \end{align*}$$

pointing towards the east or west. The horizontal component, H of the earth’s magnetic field is perpendicular to B and needle will be deflected through an angle \(\theta \). So the middle is under the action of two perpendicular fields and from tangent law,

$$\begin{align*} B &= H\: \tan \theta \\ \text {or,} \: \frac {\mu _0 \: Md}{2\pi (d^2 – l^2)^2} &= H \: \tan \theta \\ \text {or,} \: M/H &= \frac {2\pi}{\mu _o} \frac {(d^2 – l^2)^2}{d} H \tan \theta \\ \end{align*}$$

Putting the values of d, l and \(\theta \) noted in the experiment, we can calculate the value M/H.

Error in Deflation Magnetometer

1. While measuring M/H, there may come several errors that occur due to inaccuracy in l, q and d. These may be eliminated by special processes discussed below:
2. The pivot of the magnetic needle may not lie at the centre of the circular scale and the two pointers will show different angular readings. Such error is eliminated by taking the average of the two readings.
3. The magnetic centre and the geometric centre of a bar magnet may not coincide to each other. The error is removed by taking an average of deflection readings of the magnet by placing it pointing the north pole towards the east and then, by reversing it to point north pole towards the west through 180⁰ at the same distance.
4. The geometric axis of the magnet may not lie along its magnetic axis. The error is removed by taking the average of deflection reading of the magnet at one arm, and putting it upside down at the same distance on the same arm.
5. The zero of the linear scale may not coincide to the centre of the circular scale. This error is eliminated by repeating the above all processes by placing the magnet at the next arm at the same distance from the needle.
6. Tan-B Position of Gauss
7. In this position, the magnetometer arms are placed along north-south direction. The bar magnet is placed on one of its arms perpendicular to it pointing the north pole towards the east or west direction. The magnetic needle is under the action of two perpendicular fields and it will be deflected through an angleq. From tangent law,
   $$ B = \tan \: \theta $$
   Since the needle is in broadside-on position of the magnet, the field due to the magnet at the needle is
   $$\begin{align*} B &= \frac {\mu _0}{4\pi } \frac {M}{\sqrt {(d^2 + l^2)^3}} \\ \text {and so,} \\ \frac {\mu _0}{4\pi } \frac {M}{(d^2 + l^2)^{3/2}} &= H\: \tan\: \theta \\ \text {or,} \: \frac MH &= \frac {4\pi }{\mu _0} (d^2 + l^2)^{3/2} \tan \: \theta \end{align*}$$

Putting the known values in this equation we can calculate the value of M/H.

Application of Deflection Magnetometer

Comparison of magnetic moments of two bar magnets:
Two magnets are placed on two arms of the deflection magnetometer by adjusting the distance that magnetic needle shows zero deflection. In Tan-A position
$$\begin{align*} \frac {\mu_0 M_1d_1}{2\pi (d_1^2 – l_1^2)^2 } &= \frac {\mu_0 M_2d_2}{2\pi (d_1^2 – l_1^2)^2} \\ \text {or,} \: \frac {M_1}{M_2} &= \frac {d_2(d_1^2 – l_1^2)^2}{d_1(d_1^2 – L_1^2)^2} \\ \end{align*}$$
If Tan-B position is used, then
$$\begin{align*} \frac {\mu_0 M_1}{2\pi (d_1^2 + l_1^2)^{3/2} } &= \frac {\mu_0 M_2d_2}{2\pi (d_1^2 + l_1^2)^{3/2}} \\ \text {or,} \: \frac {M_1}{M_2} &= \frac {(d_1^2 + l_1^2)^{3/2}}{(d_1^2 – L_1^2)^{3/2}} \\ \end{align*}$$
In this way magnetic moments of two bar magnets are compared by using deflection meter.

Measurement of the magnetic moment and the pole strength of a bar magnet:
Measuring the ratio M/H at a place, the magnetic moment and the pole strength m is measured as
$$\begin{align*} M &= m \times 2l \\ \text {or,} \: m &= \frac {M}{2l} \\ \end{align*}$$

Comparison of horizontal components of the earth’s magnetic field at two places:
using the deflection magnetometer at a place, the ratio M/H₁ is measured where H₁ is the horizontal component of the earth’s magnetic field at that place. The same magnetometer and bar magnet are carried to another place and the ratio M/H₂ is measured where H₂ is the horizontal component at that place. The ratio of these measurements gives the ratio of two horizontal components H₁/H₂ at two places. For example, using tan-A position.
$$\begin{align*} \frac {M}{H_1} &= \frac {2\pi }{\mu _0} \frac {d^2 – l^2)^2}{d}\tan \: \theta _1 \\ \text {and at another place,} \\ \frac {M}{H_2} &= \frac {2\pi }{\mu _0} \frac {d^2 – l^2)^2}{d}\tan \: \theta _2 \\ \text {Then, we have} \\ \frac {H_1}{H_2} &= \frac {\tan \: \theta _2}{\tan\: \theta _1} \\ \end{align*}$$