# Magnetic levitation frog example

I'm trying to figure out the strength of the magnetic field needed to levitate the frog, assuming it's spherical, has mass $5$kg and diameter $20$cm. I also assume $\vec{B}$ is uniform. Since it's mostly water, I can treat the frog as a diamagnetic material, so it will be repelled away from the field.

I want to find the magnetic dipole moment of this frog in terms of the applied field, but I'm not sure how to go about doing it. My thoughts are, that the sphere will become uniformly magnetized in the field, and will have a magnetic dipole moment $\vec{m}=\frac{4}{3}\pi R^3\vec{M}$, where $\vec{M}$ is the magnetic dipole per unit volume and $R$ is the frog's radius. Then using the relation $\vec{H}=\frac{1}{\mu_0}\vec{B}-\vec{M}$, we'd have $\vec{m}=\frac{4}{3}\pi R^3(\frac{1}{\mu_0}\vec{B}-\vec{H})$, but by Ampere's law, $\oint\vec{H}\cdot\vec{dl}=$ enclosed free current $=0$, we'd just have $\vec{m}=\frac{4}{3}\pi R^3(\frac{1}{\mu_0}\vec{B})$. Is this correct?

Then to find the strenght of the field, I would have $mg\hat{z}=-\vec{F}=\nabla(\vec{m}\cdot\vec{B})$?

• Why a frog? Could you not use a plastic bladder filled with water of similar mass and shape? – Selene Routley Oct 29 '14 at 7:54
• Important mistake: A diamagnet does not get repelled in a uniform field. You need a field gradient. – mikuszefski Oct 29 '14 at 10:02
• @WetSavannaAnimalakaRodVance: I think he's trying to find an answer for a question from the opening question of a chapter from Fundamentals of Physics (HRW) 6th ed. The text only gives an explanation of the phenomenon that goes on, not the actual calculation as to 'how' it floats. – Gaurav Oct 31 '14 at 13:33

Yes, the frog will be repelled from the field if the field is non-uniform. Its magnetic moment is given through the susceptibility of Water

$\vec \mu = V \chi \vec H$ and $\chi_\mathrm{water}\approx -9\times 10^{-6}$

Lets say the average field over the Frog is $\mu_0 H = 10\;\mathrm{T}$. This results in an induced moment of $\mu = -0.3 \mathrm{Am}^2$

Now lets check the force due to a field gradient. With this volume the frog will have 4 Kg. That results in a force due to gravity of about $F\approx 40\;\mathrm{N}$ The force due to the field gradient (assumed to be in $z$-direction) will be

$F= \mu \; \mathrm{d}B/\mathrm{d}z$

We therefore need a field gradient of $133\;\mathrm{T/m}$. Note, this is the reason why the volume in this levitation experiments are so confined (and why you cannot try this at home). In reality the field will be larger (16 or more), such that the required gradient reduces accordingly. This gradient also results in a "stable" height. If the frog goes down, field strength increases. Therefore the magnetic moment increases, resulting in an increased force upwards. If it goes up the opposite takes place and there is only one height where gravity is exactly compensated (assuming a constant gradient of course and neglecting movement and variations in $x$ and $y$)
Long story short: you need

$B\;\mathrm{d}B/\mathrm{d}z >\frac{\mu_0 \rho g}{\chi}$

• Nice answer. And since magnetic levitation of frogs seems to typically be done at the mouths of solenoids, it might be useful to know B(z) along the central axis of a solenoid, extending past its mouth, so that the average B and dB/dz along the frog's body can be found--the equation for B along the axis of a solenoid can be found here. – Hypnosifl Oct 29 '14 at 13:13