For measuring the magnetic moment of a bar magnet, one could use

The time period of oscillation of a bar magnet in an oscillation magnetometer is given by $T = 2\pi \sqrt{\dfrac{I}{MB_H}}$. This gives the product $MB_H = \dfrac{4\pi^2 I}{T^2}$. If the earth's horizontal magnetic field $B_H$ and the moment of inertia $I$ are known, the magnetic moment $M$ can be determined.

In a deflection magnetometer, the deflection $\theta$ is related to $M$ and $B_H$ by $\dfrac{M}{B_H} = k \tan\theta$, where $k$ is a constant depending on the position (Tan A or Tan B). If $B_H$ is known, $M$ can be determined.

If $B_H$ is not known, we can use both instruments. The oscillation magnetometer gives the product $MB_H$, and the deflection magnetometer gives the ratio $\dfrac{M}{B_H}$. Multiplying these two quantities yields $M^2$, allowing us to find $M$ without knowing $B_H$.

A tangent galvanometer is primarily used to measure electric current, not the magnetic moment of a bar magnet.