The deflection produced in a tangent galvanom | vedclass

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(B) For a tangent galvanometer,the current $I$ is given by $I = \frac{2r B_H 	an 	heta}{\mu_0 N}$.Given: Resistance $R = 9 \ \Omega$,Potential diffe

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$10 \sqrt{3} 	imes 10^{-2} \ m$

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(B) For a tangent galvanometer,the current $I$ is given by $I = \frac{2r B_H 	an 	heta}{\mu_0 N}$.Given: Resistance $R = 9 \ \Omega$,Potential difference $V = 4.5 \ V$,Number of turns $N = 10$,Deflection $	heta = 30^{\circ}$,Horizontal component of Earth's magnetic field $B_H = 3.14 	imes 10^{-5} \ T$.First,calculate the current $I$ using Ohm's law: $I = \frac{V}{R} = \frac{4.5}{9} = 0.5 \ A$.Rearranging the tangent galvanometer formula for radius $r$: $r = \frac{\mu_0 N I}{2 B_H 	an 	heta}$.Substituting the values: $r = \frac{4 \pi 	imes 10^{-7} 	imes 10 	imes 0.5}{2 	imes 3.14 	imes 10^{-5} 	imes 	an 30^{\circ}}$.Since $\mu_0 = 4 \pi 	imes 10^{-7} \approx 4 	imes 3.14 	imes 10^{-7}$,we have $r = \frac{4 	imes 3.14 	imes 10^{-7} 	imes 5}{2 	imes 3.14 	imes 10^{-5} 	imes (1/\sqrt{3})}$.$r = \frac{2 	imes 10^{-7} 	imes 5}{10^{-5} 	imes (1/\sqrt{3})} = 10 	imes 10^{-2} 	imes \sqrt{3} \ m = 10 \sqrt{3} 	imes 10^{-2} \ m$.

The deflection produced in a tangent galvanometer,whose coil has a resistance of $9 \Omega$,is $30^{\circ}$. The potential difference across the coil is $4.5 \ V$. If the number of turns in the coil is $10$,the radius of the coil is (Given,$B_{H} = 3.14 \times 10^{-5} \ T$)

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