How to calculate earth's magnetic field by time period of suspended magnet? Assuming that torsion of the suspension wire small enough to be neglected, how can we derive earth's magnetic field by measuring time period of an oscillating magnet suspended on a string?
 A: If we put a magnetic dipole, $\mu$, in a uniform electric field of strength $B$ then the torque on the dipole is given by:
$$ T = \mu \times B \tag{1} $$
where the $\times$ symbol indicates the cross product. If the angle between the dipole and the field is $\theta$ then equation (1) can be written as:
$$ T = \mu B \sin\theta \tag{2} $$
For small angles $\sin\theta \approx \theta$, so if the oscillation is small, i.e. $\theta$ is small, equation (2) becomes:
$$ T \approx \mu B \theta \tag{3} $$
and this is the equation for a simple harmonic oscillator. The equivalent to Newton's second law for rotational motion is:
$$ T = I\ddot{\theta} $$
and substituting this into equation (3) gives us:
$$ \frac{d^2\theta}{dt^2} = -\frac{\mu B}{I} \theta \tag{4} $$
where I've put in the minus sign because if the magnet starts aligned with the field, $\theta = 0$, the torque will tend to restore the alignment. The period of this oscillator, $\tau$, is:
$$ \tau = \sqrt{\frac{I}{\mu B}} $$
and rearranging this we get:
$$ B = \frac{I}{\tau^2 \mu} \tag{5} $$
So, if you time the oscillations of your magnet to get the period, and you know the strength of the magnet and it's moment of inertia, you can use equation (5) to caculate the field strength of the Earth.
A: The method suggest by @JohnRennie is a good one but the problem is that you need to know the value of the moment of inertia of the magnet, which can be founds by measuring the dimensions of the magnet and its mass, and the value of $\mu$ the magnet moment of the magnet which can be found by measuring the period of the magnet's oscillations in a known magnetic field.
A Helmholtz coils produce a fairly uniform magnetic field in the central region.  The magnitude of the magnetic field that they produce $B_C$ can be found from the dimensions of the coils and the current passing through them.
Then one method of finding the horizontal component of the Earth's magnetic field $B_H$ would be to set up the Helmholtz coils so that their plane is at right angles to the horizontal component of the Earth's magnetic field and the magnetic field direction produced by the coils is the same as that of the Earth.
The period of the magnet's oscillations with a known current passing through the coils is measured $\tau_+$.
Keeping the magnitude of the current the same but reversing the current direction will produce a different period of oscillation $\tau_-$.
$\Rightarrow \tau_+ = \sqrt{\frac{I}{\mu (B_H+B_C)}}$ and $\tau_- = \sqrt{\frac{I}{\mu (B_H-B_C)}}$
$\Rightarrow \dfrac{\tau^2_- - \tau^2_+}{\tau^2_- + \tau^2_+}\cdot B_C$
Thus $B_H$ can be found for a range of different currents.
Next the angle of dip $\delta$ (the angle the Earth's magnetic field makes with the horizontal) has to be measured using a dip circle and from that a value of the Earth's magnetic field $B_E$ can be found.
$B_E = B_H \cos \delta$.
