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A metallic rod of $$1\text{ m }$$  length is rotated with a frequency of $$50\text{ rev/s }$$ . with one end hinged at the centre and the other end at the circumference of a circular metallic ring of radius $$1\text{ m },$$  about an axis passing through the centre and perpendicular to the plane of the ring .A constant and uniform magnetic field of $$1T$$  parallel to the axis is present everywhere. What is the emf between the centre and the metallic ring?
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## QuestionPhysicsClass 12

A metallic rod of $$1\text{ m }$$  length is rotated with a frequency of $$50\text{ rev/s }$$ . with one end hinged at the centre and the other end at the circumference of a circular metallic ring of radius $$1\text{ m },$$  about an axis passing through the centre and perpendicular to the plane of the ring .A constant and uniform magnetic field of $$1T$$  parallel to the axis is present everywhere. What is the emf between the centre and the metallic ring?

Angular speed of the rod $$\left(\omega \right)=50\frac{\text{ rev }}{\text{ sec }}=50\times 2\pi =100\pi \text{ rad/sec}$$
Radius of the rod $$\left(R\right)=1m$$
Magnetic field $$(B)=1T$$
Let us consider a small elemental length $$dl$$  of the metallic rod at a distance $$l$$  from the centre moving at a linear speed $$v$$  in a magnetic field $$B$$  perpendicular to the length and speed.
The emf induced across the elemental length $$\mathit{dl}$$
$$\mathit{d\varepsilon }=\mathit{vBdl}$$
Since the wheel is rotating at an angular speed $$\omega$$ , we have
$$v=\mathit{\omega l}$$
$$\Rightarrow \mathit{d\varepsilon }=\mathit{\omega lBdl}$$
The total emf induced across the metallic rod is given by:
$${\therefore}\varepsilon =\int _0^R\mathit{d\varepsilon }=\int _0^R\mathit{\omega lBdl}$$
$$=\mathit{\omega B}\int _0^R\mathit{ldl}$$
= $$\frac 1 2\mathit{\omega B}R^2$$
$$=\frac 1 2\times 100\pi \times 1\times (1)^2$$
$$=157V$$