Two tangent galvanometer coils of the same ra | vedclass

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(B) For a tangent galvanometer,the current $I$ is given by $I = K 	an 	heta$,where $K$ is the reduction factor.Since the coils are connected in seri

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$\frac{\sqrt{3}}{1}$

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(B) For a tangent galvanometer,the current $I$ is given by $I = K 	an 	heta$,where $K$ is the reduction factor.Since the coils are connected in series,the current $I$ is the same for both.Therefore,$K_1 	an 	heta_1 = K_2 	an 	heta_2$.Given $	heta_1 = 60^o$ and $	heta_2 = 45^o$,we have $K_1 	an 60^o = K_2 	an 45^o$.$K_1 (\sqrt{3}) = K_2 (1) \Rightarrow \frac{K_1}{K_2} = \frac{1}{\sqrt{3}}$.The reduction factor $K$ is given by $K = \frac{2 R B_H}{\mu_0 N}$,where $R$ is the radius and $N$ is the number of turns.Since $R$ is the same for both,$K \propto \frac{1}{N}$,which implies $\frac{K_1}{K_2} = \frac{N_2}{N_1}$.Thus,$\frac{N_2}{N_1} = \frac{1}{\sqrt{3}}$,which means $\frac{N_1}{N_2} = \frac{\sqrt{3}}{1}$.

Two tangent galvanometer coils of the same radius are connected in series. The current flowing produces deflections of $60^o$ and $45^o$. The ratio of the number of turns in the coils is:

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