Name: Resolving the Earth's Magnetic Field
Uploaded: 2019-07-19
Duration: 3 min 12 s
Description: A magnetic meridian contains the information of the Earth’s net magnetic field. Let's focus on the magnetic meridian and resolve the net magnetic field in th...

Question

Describe the experiment to compare the magnetic moments of two bar magnets, using deflection magnetometer by deflection method in tan B position under the following heads : Derivation of formula

Accepted Answer

Deflection magnetometer works on the principle of tangent law. According to tangent law when two uniform magnetic fields act at right angles to each other on a magnetic needle, it comes to rest in the way of the resultant field.

The deflection magnetometer is placed in Tan B position. A bar magnet of magnetic moment $M_{1}$ and length $2 l_{1}$ is placed at a distance $d_{1}$ from the centre of the magnetic needle, on one side of the compass box. Since, the sensitivity of the magnetometer is more at $45 °$ , the distance of the bar magnet should be chosen such that the deflection lies between $30 °$ and $60 °$ .

The readings corresponding to the ends of the aluminium pointer are noted as $θ_{1}$ and $θ_{2}$ . The magnet is reversed pole to pole and kept at the same distance. Two more readings $θ_{3}$ and $θ_{4}$ are noted. By placing the magnet on the other side of the compass box at the same distance, four more readings $θ_{5}$ , $θ_{6}, θ_{7}$ and $θ_{8}$ are noted as above. The mean of the eight readings gives a value $θ_{I}$ .

The experiment is repeated as above for the second bar magnet of magnetic moment $M_{2}$ and length $2 l_{2}$ by placing it at a distance $d_{2}$ . Now the mean of the eight readings gives a value of $θ_{I I}$ .

Applying tangent law, for the first magnet,

$\frac{μ_{0}}{4 π} . \frac{M_{1}}{{({d^{2}}_{1} - {I^{2}}_{1})}^{3 / 2}} = B_{h} \tan θ_{I}$ ----(1)

For the second magnet

$\frac{μ_{0}}{4 π} . \frac{M_{2}}{{({d^{2}}_{2} - {I^{2}}_{2})}^{3 / 2}} = B_{h} \tan θ_{I I}$ -----(2)

From the above equations (1) and (2), we get

$\frac{M_{1}}{M_{2}} = \frac{{({d^{2}}_{1} - {I^{2}}_{1})}^{3 / 2} \tan θ_{I}}{{({d^{2}}_{2} - {I^{2}}_{2})}^{3 / 2} \tan θ_{I I}}$

For, $d_{1} = d_{2} = d$

$\frac{M_{1}}{M_{2}} = \frac{\tan θ_{I}}{\tan θ_{I I}}$

Describe the experiment to compare the magnetic moments of two bar magnets, using deflection magnetometer by deflection method in tan B position under the following heads : Derivation of formula

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