The arm PQ of a rectangular conductor is movi | vedclass

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(A) $1$. Magnetic Flux $(\phi)$: As the rod moves from $x=0$ to $x=b$,the area inside the magnetic field increases linearly,so flux increases. From $x

Vedclass · Accepted Answer

$A-$Flux,$B-$EMF,$C-$Power dissipated

Vedclass · Answer

(A) $1$. Magnetic Flux $(\phi)$: As the rod moves from $x=0$ to $x=b$,the area inside the magnetic field increases linearly,so flux increases. From $x=b$ to $x=2b$,the rod is outside the magnetic field,so flux remains constant. On the return journey from $x=2b$ to $x=b$,flux remains constant,and from $x=b$ to $x=0$,it decreases linearly. This corresponds to curve $A$.$2$. Induced $EMF$ $(e)$: The induced $EMF$ is given by $e = -\frac{d\phi}{dt} = -Blv$. As the rod moves from $x=0$ to $x=b$,$e$ is constant and negative. From $x=b$ to $x=2b$,$e=0$. On the return journey from $x=b$ to $x=0$,the velocity is reversed,so $e$ becomes positive and constant. This corresponds to curve $B$.$3$. Power Dissipated $(P)$: Power dissipated is given by $P = \frac{e^2}{R}$. Since $P \propto e^2$,it is non-zero only when the rod is in the magnetic field ($0$ to $b$). This corresponds to curve $C$.

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