Two tangent galvanometers A and B have coils | vedclass

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(B) The current $I$ through a tangent galvanometer is given by $I = \frac{2r B_H}{\mu_0 N} 	an 	heta$, where $r$ is the radius, $N$ is the number of

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$12$

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(B) The current $I$ through a tangent galvanometer is given by $I = \frac{2r B_H}{\mu_0 N} 	an 	heta$, where $r$ is the radius, $N$ is the number of turns, and $	heta$ is the deflection.Since the galvanometers are connected in parallel to a cell of emf $V$, the potential difference across each is the same $(V_A = V_B = V)$.Given $V = IR$, we have $I = V/R$. Since $R$ is the same for both $(8 \Omega)$, the currents $I_A$ and $I_B$ are equal.Thus, $\frac{2 r_A B_H}{\mu_0 N_A} 	an 	heta_A = \frac{2 r_B B_H}{\mu_0 N_B} 	an 	heta_B$.Simplifying, we get $\frac{r_A 	an 	heta_A}{N_A} = \frac{r_B 	an 	heta_B}{N_B}$.Substituting the given values: $r_A = 8 	ext{ cm}$, $r_B = 16 	ext{ cm}$, $N_A = 2$, $	heta_A = 30^{\circ}$, $	heta_B = 60^{\circ}$.$\frac{8 	an 30^{\circ}}{2} = \frac{16 	an 60^{\circ}}{N_B}$.$4 	imes \frac{1}{\sqrt{3}} = \frac{16 	imes \sqrt{3}}{N_B}$.$N_B = \frac{16 	imes \sqrt{3} 	imes \sqrt{3}}{4} = \frac{16 	imes 3}{4} = 12$ turns.

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