The circular coil of a galvanometer has 50 tu | vedclass

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(B) The deflection $\phi$ of a moving coil galvanometer is given by the formula: $\phi = \frac{N A B i}{k}$.Here, $N = 50$, $r = 1 \ cm = 0.01 \ m$, $

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$30$

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(B) The deflection $\phi$ of a moving coil galvanometer is given by the formula: $\phi = \frac{N A B i}{k}$.Here, $N = 50$, $r = 1 \ cm = 0.01 \ m$, $B = 0.010 \ T$, $i = 1.0 \ mA = 1.0 	imes 10^{-3} \ A$, and $k = 3 	imes 10^{-7} \ N-m-rad^{-1}$.The area $A = \pi r^2 = \pi 	imes (0.01)^2 = \pi 	imes 10^{-4} \ m^2$.Substituting these values into the formula:$\phi = \frac{50 	imes (\pi 	imes 10^{-4}) 	imes 0.010 	imes 1.0 	imes 10^{-3}}{3 	imes 10^{-7}}$$\phi = \frac{50 	imes \pi 	imes 10^{-4} 	imes 10^{-2} 	imes 10^{-3}}{3 	imes 10^{-7}} = \frac{50 	imes \pi 	imes 10^{-9}}{3 	imes 10^{-7}} = \frac{50 	imes \pi}{3} 	imes 10^{-2} = \frac{0.5 \pi}{3} \approx 0.5236 \ radians$.Converting radians to degrees: $\phi_{deg} = 0.5236 	imes \frac{180}{\pi} = 30^{\circ}$.

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