 Moving-coil galvanometer torque

For a rectangular winding with $N$ turns, active side length $\ell$, side separation $w$, and per-turn area $A=w\ell$, the magnetic force on a current-carrying wire gives equal and opposite side forces of magnitude $N|I|\ell B$. Their force couple has lever arm $w$. Choosing the positive rotation to match positive electric current in the illustration gives the signed torque and spring balanceHere $B$ is the working-gap magnetic flux density, $k>0$ is the torsional spring constant in $\mathrm{N}m/rad$, and $\theta$ is the deflection angle in radians from the no-current zero. The balance holds after motion has settled, within the radial-field and linear elasticity ranges. Do not identify $\theta$ with the angle between a magnetic dipole moment and a fixed uniform magnetic field: that different geometry has torque magnitude $NIAB|\sin \alpha |$. ^[ref]

For an original illustrative design, choose $N=120$, $A=3.00\times 10^{-4}{\mathrm{m}}^2$, $B=0.250\mathrm{T}$, and $k=3.00\times 10^{-6}\mathrm{N}m/rad$. Then $NAB=9.00\times 10^{-3}\mathrm{N}m/A$. At $I=+40.0\mu \mathrm{A}$, the magnetic torque is $+3.60\times 10^{-7}\mathrm{N}\mathrm{m}$ and $\theta =+0.120\mathrm{rad}\simeq +6.88^{\circ }$. Reversing the electric current reverses both the torque and the deflection, giving $-6.88^{\circ }$ without changing the magnetic field.