The deflection in a moving coil galvanometer | vedclass

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(A) Given: Resistance of galvanometer $G = 45 \Omega$,specific resistance $ho = 5 	imes 10^{-7} \Omega m$,area $A = 4 	imes 10^{-7} m^2$.When the

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$4$

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(A) Given: Resistance of galvanometer $G = 45 \Omega$,specific resistance $ho = 5 	imes 10^{-7} \Omega m$,area $A = 4 	imes 10^{-7} m^2$.When the deflection falls from $30$ divisions to $3$ divisions,the current through the galvanometer $I_g$ becomes $I_g = \frac{3}{30} I = \frac{1}{10} I$.The shunt resistance $S$ is given by the formula $S = \frac{I_g G}{I - I_g}$.Substituting the values: $S = \frac{(\frac{1}{10} I) 	imes 45}{I - \frac{1}{10} I} = \frac{4.5 I}{0.9 I} = 5 \Omega$.Since $S = \frac{ho L}{A}$,we have $L = \frac{S A}{ho}$.$L = \frac{5 	imes 4 	imes 10^{-7}}{5 	imes 10^{-7}} = 4 \ m$.

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