Question 98 of 153

A metallic rod of length $l$ is tied to a string of length $2l$ and made to rotate with angular speed $\omega$ on a horizontal table with one end of the string fixed. If there is a vertical magnetic field B in the region, the e.m.f. induced across the ends of the rod is:

A$\dfrac{4\text{B}\omega {l}^{2}}{2}$B$\dfrac{5\text{B}\omega {l}^{2}}{2}$C$\dfrac{2\text{B}\omega {l}^{2}}{2}$D$\dfrac{3\text{B}\omega {l}^{2}}{2}$

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Answer: (B) $\dfrac{5\text{B}\omega {l}^{2}}{2}$

$e=\int _{2l}^{3l}BVdx$

$\text{e}=\int _{2\ell }^{3\ell }$B(ωx)dx
$e=\omega B\int _{2l}^{3l}xdx$

$e=\omega B{\left[\dfrac{{x}^{2}}{2}\right]}_{2l}^{3l}$

$e=\dfrac{\omega B}{2}{\left[{\left(3l\right)}^{2}-{\left(2l\right)}^{2}\right]}_{2l}^{3l}$

$e=\dfrac{5}{2}\omega {Bl}^{2}$