A flexible wire bent in the form of a circle | vedclass

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(C) The magnetic flux $\phi$ through the coil is given by $\phi = B A = B \pi r^2$.Since $B$ is constant,$\phi \propto r^2$,so $\phi = k r^2$ where $k

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(C) The magnetic flux $\phi$ through the coil is given by $\phi = B A = B \pi r^2$.Since $B$ is constant,$\phi \propto r^2$,so $\phi = k r^2$ where $k = B \pi$ is a constant.The induced emf $e$ is given by Faraday's law: $e = -\frac{d\phi}{dt} = -\frac{d}{dt}(k r^2) = -2 k r \frac{dr}{dt}$.$1$. From $t = 0$ to $t = 1$,the radius $r$ is constant,so $\frac{dr}{dt} = 0$,which implies $e = 0$.$2$. From $t = 1$ to $t = 2$,the radius $r$ increases linearly with time,i.e.,$r = \alpha t + \beta$. Thus,$\frac{dr}{dt} = \alpha$ (a constant). The induced emf $e = -2 k r \alpha$. Since $r$ is increasing linearly with time,the magnitude of the induced emf $|e|$ increases linearly with time.$3$. From $t > 2$,the radius $r$ is constant again,so $\frac{dr}{dt} = 0$,which implies $e = 0$.Comparing this with the given options,the graph that shows zero emf,then a linear increase in magnitude,and then zero emf is represented by graph $C$.

$A$ flexible wire bent in the form of a circle is placed in a uniform magnetic field perpendicular to the plane of the coil. The radius of the coil changes as shown in the figure. The graph of induced emf in the coil is represented by

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