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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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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?

Answer

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:
\({\therefore}\varepsilon =\int _0^R\mathit{d\varepsilon }=\int _0^R\mathit{\omega lBdl}\)  
\(=\mathit{\omega B}\int _0^R\mathit{ldl}\)  
= \(\frac 1 2\mathit{\omega B}R^2\)  
\(=\frac 1 2\times 100\pi \times 1\times (1)^2\)  
\(=157V\)  
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