A hundred turns of insulated copper wire are | vedclass

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(A) The total charge $Q$ that flows through the circuit is given by the formula $Q = \frac{\Delta \phi}{R}$,where $\Delta \phi$ is the change in magne

Vedclass · Accepted Answer

$2 	imes 10^{-2} \ C$

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(A) The total charge $Q$ that flows through the circuit is given by the formula $Q = \frac{\Delta \phi}{R}$,where $\Delta \phi$ is the change in magnetic flux and $R$ is the total resistance.Given:Number of turns $N = 100$Area $A = 1 	imes 10^{-3} \ m^2$Resistance $R = 10 \ \Omega$Initial magnetic induction $B_i = 1 \ Wb/m^2$Final magnetic induction $B_f = -1 \ Wb/m^2$ (opposite direction)Change in magnetic induction $\Delta B = |B_f - B_i| = |(-1) - 1| = 2 \ Wb/m^2$.The change in flux is $\Delta \phi = N \cdot A \cdot \Delta B$.Substituting the values: $\Delta \phi = 100 	imes (1 	imes 10^{-3}) 	imes 2 = 0.2 \ Wb$.Now,calculating the charge: $Q = \frac{0.2}{10} = 0.02 \ C = 2 	imes 10^{-2} \ C$.

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