The coil in a moving coil galvanometer experi | vedclass

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Question

(A) The torque on the coil is given by $	au = NIAB \sin 	heta$. For small deflections,$	au = NIAB = C 	heta$,where $C$ is the torsional constant.G

Vedclass · Accepted Answer

$Q\sqrt {\frac{{\pi NAB}}{{2iI}}} $

Vedclass · Answer

(A) The torque on the coil is given by $	au = NIAB \sin 	heta$. For small deflections,$	au = NIAB = C 	heta$,where $C$ is the torsional constant.Given that for current $i$,the deflection is $	heta = 90^o = \frac{\pi}{2}$ radians.So,$NiAB = C \frac{\pi}{2} \Rightarrow C = \frac{2NiAB}{\pi}$ ........$(1)$When a charge $Q$ is passed suddenly,the impulse of the torque is equal to the change in angular momentum:$\int 	au dt = \int (NIAB) dt = NAB \int i dt = NABQ$Since $\int 	au dt = \Delta L = I \omega$,we have $I \omega = NABQ$ ........$(2)$By the law of conservation of energy,the initial kinetic energy is converted into potential energy of the spring at maximum deflection $	heta_{max}$:$\frac{1}{2} I \omega^2 = \frac{1}{2} C 	heta_{max}^2$$	heta_{max} = \omega \sqrt{\frac{I}{C}} = \frac{I \omega}{\sqrt{IC}} = \frac{NABQ}{\sqrt{IC}}$Substituting $C$ from $(1)$:$	heta_{max} = \frac{NABQ}{\sqrt{I \cdot \frac{2NiAB}{\pi}}} = NABQ \sqrt{\frac{\pi}{2NiABI}} = Q \sqrt{\frac{\pi NAB}{2iI}}$

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