JEE Main & Advanced Physics Magnetism Deflection Magnetometer

It's working is based on the principle of tangent law. It consists of a small compass needle, pivoted at the centre of a circular box. The box is kept in a wooden frame having two meter scale fitted on it's two arms. Reading of a scale at any point directly gives the distance of that point from the centre of compass needle.

(1) Tan A position : In this position the magnetometer is set perpendicular to magnetic meridian. So that, magnetic field due to magnet, is in axial position and perpendicular to earth's field. Hence \[{{B}_{H}}\tan \theta =\frac{{{\mu }_{0}}}{4\pi }.\frac{2Mr}{{{({{r}^{2}}-{{l}^{2}})}^{2}}}\] or \[{{B}_{H}}\tan \theta =\frac{{{\mu }_{0}}}{4\pi }.\frac{2M}{{{r}^{3}}}\]

(2) Tan B position : The arms of magnetometer are set in magnetic meridian, so that the magnetic field due to magnet is at it's equatorial position. Hence \[{{B}_{H}}\tan \theta =\frac{{{\mu }_{0}}}{4\pi }.\frac{M}{{{({{r}^{2}}+{{l}^{2}})}^{3/2}}}\] or \[{{B}_{H}}\tan \theta =\frac{{{\mu }_{0}}}{4\pi }.\frac{M}{{{r}^{3}}}\]

(3) Comparison of magnetic moments : According to deflection method \[\frac{{{M}_{1}}}{{{M}_{2}}}=\frac{\tan {{\theta }_{1}}}{\tan {{\theta }_{2}}}\]

According to null deflection method \[\frac{{{M}_{1}}}{{{M}_{2}}}={{\left( \frac{{{d}_{1}}}{{{d}_{2}}} \right)}^{3}}\]