Current to cancel Earth's magnetic field I'm making a lab report where I've used a Helmholtz coil and a magnet to find Earth's magnetic field intensity in the area by relating the frequency of oscillation of the magnet when a certain current is going through the coil, using the equation:
$$ f^2 = \frac{μ_o}{4π^2m_i} \left(\frac{8\mu_0N}{5^{3/2}R}I \pm B\right). $$
Everything went ok and I found the experimental value of the magnetic field to be $17\ \mu\mathrm{T}$; the teacher said around $2\times10^{-5}\ \mathrm{T}$ would be fine. 
One of the questions was, "What is the current needed to cancel Earth's magnetic field?"
I did some research and found that if the field is cancelled then the period of oscillation goes to infinity and so concluded that the frequency of oscillation ($1/T$) must go to zero.
By setting $f^2 = 0$ in the previous equation I found that
$$ I = \frac{5^{5/3}BR}{8\mu_0N}$$
Are these assumptions correct? I know you're not supposed to do my homework; if someone could only answer if $f = 0$ when the field is cancelled is good enough.
Notation:
$B$ is Earth's field;
$I$ is the current;
$R$ is the coil's radius;
$N$ is the number of loops/turns;
$\mu_0$ is $4\pi\times10^{-7}$;
$m_i$ (not needed in the second equation) is the moment of inertia of the magnet.
 A: This is super belated, but I'll try to give an answer to this question, should it be helpful for future homework doers. As is, the question is a bit under-specified because it may not be evident what frequency is being mentioned (resonance, precession, etc., though essentially the same number), which is probably why answers have not been forthcoming. 
The short answer is YES, your approach is correct, but I'll make a reasonable assumption of precessional motion being detected, and explain why, based on the assumptions. In this experiment, one uses the Helmholz coil to do the cancellation of Earth's field (plus any other sources), but the magnet is used to provide detection. Basically, a magnet at some small angle with a total net magnetic field $B_T$ will undergo what is known as 'precession' (wobble) about the total magnetic field at a frequency $\omega_p$ proportional to 
$\omega_p = 2\pi f \propto \gamma * B_T$ 
$\gamma$ is called the gyromagnetic ratio and is proportional to charge over mass. $B_T$ is the net field including the Earth's magnetic field, any stray field sources from magnets, and the field produced by the coil current. Your equation above is a form of $(1/2\pi)\gamma * B_T$, and/or the corresponding classical energy, proportional to $\omega^2$. So, when you cancel the earths magnetic field (plus any other stray magnetic field source), it will be observed when the precessional frequency vanishes or becomes zero, or the magnet's energy vanishes (ceases motion). Under these assumptions, your approach is correct :) Hopefully this explains a bit more as to why. So, well done, even though you are probably a Professor by now! 
