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A metal rod of length ' 1 ' rotates about one of its ends in a plane perpendicular to a magnetic field of induction ' $B$ '. If the e.m.f. induced between the ends of the rod is ' $\mathrm{e}$ ' then the number of revolutions made by the rod per second is
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The correct answer is:
$\frac{\mathrm{e}}{\mathrm{B} \pi \ell^2}$
The induced emf e $=\frac{\mathrm{d} \phi}{\mathrm{dt}}=\frac{\mathrm{d}}{\mathrm{dt}}(\mathrm{AB})=\mathrm{B} \frac{\mathrm{dA}}{\mathrm{dt}}$
In one rotation the rod traces out a circle of radius 1 or an area $\mathrm{A}=\pi \mathrm{l}^2$
So, in $n$ rotations it will trace out an area of $n \pi l^2$
$\begin{aligned} & \therefore \mathrm{e}=\mathrm{B} \frac{\Delta \mathrm{A}}{\Delta \mathrm{t}}=\mathrm{B} \cdot \frac{\mathrm{n} \pi \mathrm{l}^2}{\Delta \mathrm{t}} \\ & \text { But } \frac{\mathrm{n}}{\Delta \mathrm{t}}=\frac{\text { No.of rev.performed }}{\text { time }}=\text { frequency } \mathrm{f} \\ & \therefore \mathrm{e}=\mathrm{Bf} \pi \mathrm{l}^2=\text { Bf.A } \\ & \therefore \mathrm{f}=\frac{\mathrm{e}}{\mathrm{B} \pi \mathrm{l}^2}\end{aligned}$
In one rotation the rod traces out a circle of radius 1 or an area $\mathrm{A}=\pi \mathrm{l}^2$
So, in $n$ rotations it will trace out an area of $n \pi l^2$
$\begin{aligned} & \therefore \mathrm{e}=\mathrm{B} \frac{\Delta \mathrm{A}}{\Delta \mathrm{t}}=\mathrm{B} \cdot \frac{\mathrm{n} \pi \mathrm{l}^2}{\Delta \mathrm{t}} \\ & \text { But } \frac{\mathrm{n}}{\Delta \mathrm{t}}=\frac{\text { No.of rev.performed }}{\text { time }}=\text { frequency } \mathrm{f} \\ & \therefore \mathrm{e}=\mathrm{Bf} \pi \mathrm{l}^2=\text { Bf.A } \\ & \therefore \mathrm{f}=\frac{\mathrm{e}}{\mathrm{B} \pi \mathrm{l}^2}\end{aligned}$
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