Given \(: \ell=1\,m , B =5 \times 10^{-3}\,Wb / m ^2\)
\(f =\frac{1800}{60}=30 \text { rotations } / sec\)
in one rotation, the moving rod of the metal races a circle of radius \(r=\ell\)
\(\therefore\) Area swept in one rotation \(=\pi r ^2\)
\(\frac{ d \phi}{ dt }=\frac{ d }{ dt }( BA )= B \cdot \frac{ dA }{ dt }=\frac{ B \pi r ^2}{ T }\)
\(= B f \pi r ^2=\left(5 \times 10^{-3}\right) \times 3.14 \times 30 \times 1=0.471 \,V\)