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A circular metal plate of radius $R$ is rotating with a uniform angular velocity $\omega$ with its plane perpendicular to a uniform magnetic field $B$. Then the emf developed between the centre and the rim of the plate is
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The correct answer is:
$\omega B R^2 / 2$
We write $d \in=\operatorname{Blv}(\epsilon \mathrm{HF})$
Now l $=\mathrm{dx} \mathrm{v}=\omega \mathrm{x}$
$\int \mathrm{d} \in=\mathrm{B} \omega \int_0^r \mathrm{x} d x$
$\in=\frac{\mathrm{B} \omega \mathrm{r}^2}{2}$
Now l $=\mathrm{dx} \mathrm{v}=\omega \mathrm{x}$
$\int \mathrm{d} \in=\mathrm{B} \omega \int_0^r \mathrm{x} d x$
$\in=\frac{\mathrm{B} \omega \mathrm{r}^2}{2}$
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