## torque in an MRI scanner

There are (at least) two issues regarding the presence of ferromagnetic materials around an MRI scanner (of strength e.g. 3 Tesla).
(i) that something (e.g. scissors in a health worker's pocket) becomes a projectile, and
(ii) that an implant may experience a twisting force (e.g. ferromagnetic shrapnel or bullets or a needle or other implanted medical device with ferromagnetic parts).

I am trying to determine the equation relating the torque τ experienced by an object, to
the magnetic susceptibility of a material χ
the static magnetic field strength B
the volume of the material V
the angle it makes with B, θ

I am stuck because my derivation doesn't agree with the textbooks I have. My working is:

The magnetisation of a sample which is not a permanent magnet is given by M=χH, where χ is the (dimensionless) magnetic susceptibility of the sample. We may also write B=μ$_{0}$(1+χ)=μ$_{0}$μ$_{r}$H=μH, where μ0 is the (constant) permeability of free space (units of henries per metre, or newtons per ampere squared), μ$_{r}$ is the (dimensionless) relative permeability of the sample to μ$_{0}$, μ is the magnetic permeability of the sample, B is the magnetic field strength as flux density (Tesla) and H is magnetic field strength given in ampere-turns per metre. We may write, then, M=χB/μ.

So, torque (τ) is given by τ = m x B (where m denotes the magnetic dipole moment, which equals the magnetisation of a sample (M) multiplied by the volume of the sample (V)). Thus, |τ| = mBsinθ where θ is the angle between the magnetic moment and the applied magnetic field. In terms of the magnetisability of a sample (χ), |τ| = χVB$^{2}$sinθ/μ (using M=χB/μ and m=MV).

One textbook says |τ| $\propto$ χ$^{2}$VB$^{2}$.
It seems that either I am missing one “χ” or my textbook has one “χ” too many.
Another says F$_{rot}$ $\approx$ χ$^{2}$VB$^{2}$/2Lμ$_{0}$, where L is the length of the object.

Thanks for any pointers.

Edit: sorry, probably this should be in the homework forum. I'd move it if I could.