# 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.

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I've added the homework tag. In the future, when posting to ask for help on assigned school work, please use the tag. – Ben Crowell Jun 12 '13 at 18:31
Will do! Didn't know about that rule. hope someone could help me with that though.Thanks! – Matheus Jun 14 '13 at 22:14