The magnetic flux through a circuit of resist | vedclass

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(D) According to Faraday's law of electromagnetic induction,the induced electromotive force $(e)$ is given by $e = -\frac{d\phi}{dt}$.Taking the magni

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$Q = \frac{\Delta \phi}{R}$

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(D) According to Faraday's law of electromagnetic induction,the induced electromotive force $(e)$ is given by $e = -\frac{d\phi}{dt}$.Taking the magnitude,we have $|e| = \frac{d\phi}{dt}$.Since $e = iR$,where $i$ is the induced current and $R$ is the resistance,we can write $iR = \frac{d\phi}{dt}$.We know that current $i = \frac{dq}{dt}$,where $dq$ is the small amount of charge passing in time $dt$.Substituting this into the equation: $\left(\frac{dq}{dt}\right)R = \frac{d\phi}{dt}$.Canceling $dt$ from both sides,we get $dq \cdot R = d\phi$,which implies $dq = \frac{d\phi}{R}$.Integrating both sides for the total change,the total charge $Q$ is given by $Q = \frac{\Delta \phi}{R}$.

Column-$I$	Column-$II$
$a$. Superior Vena Cava	$p$. Carries deoxygenated blood to the lungs
$b$. Inferior Vena Cava	$q$. Carries oxygenated blood to the heart
$c$. Pulmonary Artery	$r$. Brings deoxygenated blood from lower body parts to the right atrium
$d$. Pulmonary Vein	$s$. Brings deoxygenated blood from upper body parts to the right atrium

The magnetic flux through a circuit of resistance $R$ changes by an amount $\Delta \phi$ in time $\Delta t$. Then the total quantity of electric charge $Q$ passing during this time through any point of the circuit is given by:

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