A conducting circular loop made of a thin wir | vedclass

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Question

(A) The magnetic flux $\Phi(t)$ through the loop is given by $\Phi(t) = B(t) \cdot A = A \cdot B_0 \sin(50 \pi t)$.The induced electromotive force $(E

Vedclass · Accepted Answer

$14$

Vedclass · Answer

(A) The magnetic flux $\Phi(t)$ through the loop is given by $\Phi(t) = B(t) \cdot A = A \cdot B_0 \sin(50 \pi t)$.The induced electromotive force $(EMF)$ is $\varepsilon = -\frac{d\Phi}{dt} = -A \cdot B_0 \cdot (50 \pi) \cos(50 \pi t)$.The current in the loop is $I(t) = \frac{\varepsilon}{R} = -\frac{A \cdot B_0 \cdot 50 \pi}{R} \cos(50 \pi t)$.The charge $q$ flowing through the loop is the integral of current: $q = \int_{0}^{t} I(t) dt = \frac{1}{R} [\Phi(0) - \Phi(t)]$.At $t = 0$, $\Phi(0) = A \cdot B_0 \sin(0) = 0$.At $t = 10 \, ms = 0.01 \, s$, $\Phi(0.01) = A \cdot B_0 \sin(50 \pi 	imes 0.01) = A \cdot B_0 \sin(0.5 \pi) = A \cdot B_0$.Substituting the values: $A = 3.5 	imes 10^{-3} \, m^2$, $B_0 = 0.4 \, T$, $R = 10 \, \Omega$.$q = \frac{1}{10} |0 - (3.5 	imes 10^{-3} 	imes 0.4)| = \frac{1.4 	imes 10^{-3}}{10} = 0.14 	imes 10^{-3} \, C = 0.14 \, mC$.Wait, re-evaluating the calculation: $q = \frac{A \cdot B_0}{R} = \frac{3.5 	imes 10^{-3} 	imes 0.4}{10} = 0.14 	imes 10^{-3} \, C = 0.14 \, mC$. Given the options, there might be a scale factor error in the provided options. However, based on the standard formula, the result is $0.14 \, mC$. If we assume the question implies $B(t) = 0.4 \sin(50 \pi t)$ and the flux change is calculated, the magnitude is $0.14 \, mC$. Given the options, $0.14$ is closest to $0.14$ (if units were different) or the calculation $140 \, \mu C$ is intended.

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