Two tangent galvanometers A and B have coils | vedclass

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(B) The current $I$ in a tangent galvanometer is given by $I = \frac{2 r H}{\mu_{0} N} 	an 	heta$, where $r$ is the radius, $H$ is the horizontal co

Vedclass · Accepted Answer

$12$

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(B) The current $I$ in a tangent galvanometer is given by $I = \frac{2 r H}{\mu_{0} N} 	an 	heta$, where $r$ is the radius, $H$ is the horizontal component of the Earth's magnetic field, $N$ is the number of turns, and $	heta$ is the deflection.Since the galvanometers are connected in parallel to a $4 \, V$ source, the voltage across each is $4 \, V$. The current in each galvanometer is $I_A = I_B = \frac{V}{R} = \frac{4 \, V}{8 \, \Omega} = 0.5 \, A$.From the formula, we have $\frac{N I}{r 	an 	heta} = \frac{2 H}{\mu_{0}} = 	ext{constant}$.Therefore, $\frac{N_A I_A}{r_A 	an 	heta_A} = \frac{N_B I_B}{r_B 	an 	heta_B}$.Since $I_A = I_B$, we have $\frac{N_A}{r_A 	an 	heta_A} = \frac{N_B}{r_B 	an 	heta_B}$.Substituting the values: $\frac{2}{8 \cdot 	an 30°} = \frac{N_B}{16 \cdot 	an 60°}$.$\frac{2}{8 \cdot (1/\sqrt{3})} = \frac{N_B}{16 \cdot \sqrt{3}}$.$\frac{2 \sqrt{3}}{8} = \frac{N_B}{16 \sqrt{3}}$.$N_B = \frac{2 \sqrt{3} \cdot 16 \sqrt{3}}{8} = \frac{32 \cdot 3}{8} = 4 \cdot 3 = 12$ turns.

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