= Moving-coil galvanometer torque
{title2=$\tau_{\mathrm m}=NIAB$}
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 balance
$$
\tau_{\mathrm m}=NIAB,\qquad \tau_{\mathrm s}=-k\theta,\qquad NIAB-k\theta=0,\qquad \theta=\frac{NAB}{k}I.
$$
Here $B$ 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 $NIAB|\sin\alpha|$. <Physics Part I, Class XII/Moving Charges and Magnetism>[§4.9.1, equations (4.20)–(4.24), and §4.10, equation (4.26): fixed-field versus radial-field torque]{ref}
For an original illustrative design, choose $N=120$, $A=3.00\times10^{-4}\,\mathrm{m^2}$, $B=0.250\,\mathrm T$, and $k=3.00\times10^{-6}\,\mathrm{N\,m/rad}$. Then $NAB=9.00\times10^{-3}\,\mathrm{N\,m/A}$. At $I=+40.0\,\mathrm{\mu A}$, the magnetic <torque> is $+3.60\times10^{-7}\,\mathrm{N\,m}$ 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>.
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