Source: High school physics (Chinese)
Problem Sets:
Problem
The coil of a moving-coil galvanometer is $2.0$ cm long and $1.0$ cm wide, has $N = 250$ turns, and lies in a magnetic field $B = 0.2$ T (in the radial gap between the poles). When a current $I = 0.10$ mA flows, the coil deflects by $\varphi = 30^{\circ}$.
- Find the magnetic torque acting on the coil.
- Find the torsion constant of the spiral spring.
(1) $\tau = 1.0\times 10^{-6}$ N$\cdot$m.
(2) $k = \dfrac{6}{\pi}\times 10^{-6} \approx 1.9\times 10^{-6}$ N$\cdot$m/rad.
In a moving-coil galvanometer the magnetic field in the gap is radial, so the plane of the coil always contains $\vec{B}$. The torque on the coil is therefore
$$\tau = NBIA,$$independent of the deflection angle, where $A$ is the area of one turn.
(1) $A = 2.0\times 10^{-2}\times 1.0\times 10^{-2} = 2.0\times 10^{-4}\ \text{m}^2$; $I = 0.10\ \text{mA} = 1.0\times 10^{-4}$ A.
$$\tau = 250 \times 0.2 \times 1.0\times 10^{-4}\times 2.0\times 10^{-4} = 1.0\times 10^{-6}\ \text{N}\cdot\text{m}.$$(2) At equilibrium, the spring's restoring torque $k\varphi$ (with $\varphi$ in radians) balances the magnetic torque. Convert $\varphi = 30^{\circ} = \pi/6$ rad:
$$k = \frac{\tau}{\varphi} = \frac{1.0\times 10^{-6}}{\pi/6} = \frac{6}{\pi}\times 10^{-6} \approx 1.9\times 10^{-6}\ \text{N}\cdot\text{m}/\text{rad}.$$