A moving coil galvanometer has been fitted with a rectangular coi… - Vidyadip

Question

A moving coil galvanometer has been fitted with a rectangular coil having $50$ turns and dimensions $5\ cm \times 3\ cm$. The radial magnetic field in which the coil is suspended is of $0.05 Wb / m ^2$. The torsional constant of the spring is $1.5$ $\times 10^{-9} Nm /$ degree. Obtain the current required to be passed through the galvanometer so as to produce a deflection of $30^{\circ}$.

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Answer

$\text { Data }: N =50, C=1.5 \times 10^{-9} Nm / \text { degree, }$
$A = lb =5 cm \times 3 cm =15 cm ^2=15 \times 10^{-4} m ^2,$
$B =0.05 Wb / m ^2, \theta=30^{\circ}$
$NIAB = C \theta$
$\therefore \text { The current through the coil, } I =\frac{C \theta}{N A B}$
$=\frac{1.5 \times 10^{-9} \times 30}{50 \times 15 \times 10^{-4} \times 0.05}=\frac{3 \times 10^{-5}}{5 \times 0.5}$
$=1.2 \times 10^{-5} A $

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