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A metallic rod of 1 m length is rotated with a frequency of 100 rev/s, with one end hinged at the centre and the other end at the circumference of a circular metallic ring of radius 1 m, about an axis passing through the centre and perpendicular to the plane of the ring (see fig). A constant and uniform magnetic field of 1 T parallel to the axis is present everywhere. What is the emf between the centre and the metallic ring?

Picture $\begin{array}{1 1} 157V \\ 314V \\ 471V \\ 78.5\;V\end{array}$

1 Answer

Free electrons in the rod move towards the outer end due to Lorentz force when the rod is rotated.As they get distributed over the ring, the resulting separation of charges produces an emf across the ends of the rod. At a certain value of emf, there is no more flow of electrons and a steady state is reached. The magnitude of the emf generated across the rod as it moves at right angles to the magnetic field is given by: $d\varepsilon = Bv \;dr$Hence $\varepsilon = \int d\varepsilon = \int_{R}^{0} Bv\; dr = \int_{R}^{0} B \omega dr = \large\frac{B\omega R^2}{2}$Given that $v = \omega r$, we get:$\varepsilon = \large\frac{1}{2}$$ \times 1.0 \times 2\;\pi \times 100 \times 1^2$$\varepsilon = 314 V$
answered Mar 10, 2014 by balaji.thirumalai

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