Answer:
[tex]EMF = 0.236\times 10^{-3}[/tex]
Explanation:
As we know that induced EMF in rotating rod is given as
[tex]EMF = \frac{1}{2}B\omega L^2[/tex]
here we know that
[tex]B = 2 \times 10^{-4} T[/tex]
L = 0.50 m
[tex]\omega = 2\pi f[/tex]
[tex]\omega = 2\pi (\frac{90}{60})[/tex]
[tex]\omega = 3\pi[/tex]
now we have
[tex]EMF = \frac{1}{2}(2\times 10^{-4})(3\pi)(0.50)^2)[/tex]
[tex]EMF = 0.236\times 10^{-3}[/tex]
