A 10 Omega coil of 180 turns and diameter 4 t | vedclass

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Question

(C) Given: Resistance of the coil, $R = 10 \Omega$. Number of turns, $N = 180$. Diameter of the coil, $d = 4 	ext{ cm} = 4 	imes 10^{-2} 	ext{ m}$.

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$1$

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(C) Given: Resistance of the coil, $R = 10 \Omega$. Number of turns, $N = 180$. Diameter of the coil, $d = 4 	ext{ cm} = 4 	imes 10^{-2} 	ext{ m}$. Radius, $r = 2 	imes 10^{-2} 	ext{ m}$. Resistance of the galvanometer, $R_g = 618 \Omega$. Total resistance, $R_{eq} = R + R_g = 10 + 618 = 628 \Omega$. Charge, $q = 360 \mu 	ext{C} = 360 	imes 10^{-6} 	ext{ C}$.The magnetic flux is maximum, so $\phi = BA$.The induced charge is given by $q = \frac{N \Delta \phi}{R_{eq}}$.Since the field is removed, $\Delta \phi = BA - 0 = BA$.Area $A = \pi r^2 = \pi (2 	imes 10^{-2})^2 = 4 \pi 	imes 10^{-4} 	ext{ m}^2$.Substituting the values: $360 	imes 10^{-6} = \frac{180 	imes B 	imes 4 \pi 	imes 10^{-4}}{628}$.$B = \frac{360 	imes 10^{-6} 	imes 628}{180 	imes 4 	imes 3.14 	imes 10^{-4}} = \frac{2 	imes 10^{-6} 	imes 628}{12.56 	imes 10^{-4}} = \frac{1256 	imes 10^{-6}}{1256 	imes 10^{-4}} = 1 	ext{ T}$.

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