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(B) The induced emf is given by Faraday's law: $\varepsilon = A \left| \frac{dB}{dt} \right|$.Since the area $A$ is constant,the induced emf $\varepsi

Vedclass · Accepted Answer

$b > (d = e) > (a = c)$

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(B) The induced emf is given by Faraday's law: $\varepsilon = A \left| \frac{dB}{dt} ight|$.Since the area $A$ is constant,the induced emf $\varepsilon$ is directly proportional to the magnitude of the slope of the $B-t$ graph,i.e.,$\varepsilon \propto \left| \frac{dB}{dt} ight|$.$1$. In region $b$,the slope is positive and steep,so the magnitude of the slope is high.$2$. In regions $d$ and $e$,the slope is negative but constant,and its magnitude is smaller than that in region $b$.$3$. In regions $a$ and $c$,the magnetic field $B$ is constant,so the slope is $0$,which means the induced emf is $0$.Comparing the magnitudes of the slopes: $	ext{slope}_b > 	ext{slope}_d = 	ext{slope}_e > 	ext{slope}_a = 	ext{slope}_c = 0$.Therefore,the correct ranking is $b > (d = e) > (a = c)$.

The graph shows the magnitude $B(t)$ of a uniform magnetic field that exists throughout a conducting loop,perpendicular to the plane of the loop. Rank the five regions of the graph according to the magnitude of the emf induced in the loop,greatest first.

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