A rectangular coil of area $5.0 imes 10^{-4}$ m$^2$ and 60 turns ...

Question

A rectangular coil of area $5.0 	imes 10^{-4}$ m$^2$ and 60 turns is pivoted about one of its vertical sides in a radial horizontal magnetic field of 90 G. If a current of 0.20 mA produces an angular deflection of 18°, find the torsional constant $k$ of the hair spring.

Accepted Answer

Given: $A = 5.0 	imes 10^{-4}$ m$^2$, $n = 60$, $B = 90$ G $= 90 	imes 10^{-4}$ T $= 9 	imes 10^{-3}$ T,
$I = 0.20$ mA $= 0.20 	imes 10^{-3}$ A, $	heta = 18° = 18 	imes \pi/180 = \pi/10$ rad

At equilibrium: $nIAB = k	heta$
$$k = \frac{nIAB}{	heta} = \frac{60 	imes 0.20 	imes 10^{-3} 	imes 5.0 	imes 10^{-4} 	imes 9 	imes 10^{-3}}{\pi/10}$$
$$k = \frac{60 	imes 0.20 	imes 10^{-3} 	imes 4.5 	imes 10^{-6}}{\pi/10}$$
$$k = \frac{5.4 	imes 10^{-8}}{\pi/10} = \frac{5.4 	imes 10^{-7}}{\pi} \approx 1.72 	imes 10^{-7} 	ext{ N·m/rad}$$

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