A metal disc of radius ' $\mathrm{R}$ ' rotates with an angular velocity ' $\omega$ ' about an axis…

A metal disc of radius ' $\mathrm{R}$ ' rotates with an angular velocity ' $\omega$ ' about an axis perpendicular to its plane passing through its centre in a magnetic field of induction ' $\mathrm{B}$ ' acting perpendicular to the plane of the disc. The induced e.m.f. between the rim and axis of the disc is (magnitude only)
  1. $\frac{B \omega^2 R^2}{2}$
  2. $\frac{\mathrm{B} \omega \mathrm{R}}{2}$
  3. $\frac{B \omega^2 R}{2}$
  4. $\frac{\mathrm{B} \omega \mathrm{R}^2}{2}$

Solution

The correct option is (D). We can imagine the disc to be a collection of thin rods connected in parallel between the center of the disc and the rim. So if we calculate the induced emf on a thin rod rotating about its axis then this should be equal to that of the disc. The tiny motional emf developed across the element dr can be written as: $\mathrm{dE}=\mathrm{Bvdr}$ Taking the velocity $\mathrm{v}=\omega \mathrm{r}$, $\mathrm{dE}=\mathrm{B} \omega \mathrm{rdr}$ On integrating across the rod, $\left.\mathrm{E}=\int_0^{\mathrm{R}} \mathrm{B} \omega \mathrm{rdr}=\frac{\mathrm{B} \omega \mathrm{r}^2}{2}\right]_0^{\mathrm{R}}=\frac{\mathrm{B} \omega \mathrm{R}^2}{2}$

Asked in: MHT CET 2022 (05 Aug Shift 1)

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