Conducting rod of length l with one end pivoted is rotated with a uniform angular speed omega in vertical plane normal to a uniform magnetic field B. deduce an expression for emf induced in this rod if the resistance of the rod is R . what is the current induced in it?

Hint:-Here first we will use expression for magnetic flux in terms of magnetic field and area vector to find induced emf then equate with resistance with the help of ohm’s law.
Formula used
$d\phi = BdA$
Where $d\phi $ is magnetic flux, B is magnetic field and dA is area vector$d\phi = B\dfrac{1}{2}{l^2}\theta $
$\dfrac{{d\theta }}{{dt}} = \omega $
$\omega $is angular velocity , $\theta $is angular displacement
$i = e/R$
Where i is induced current, e is induced emf, R is resistance

Complete step-by-step solution:We know that magnetic flux is the number of magnetic field lines passing through a given closed surface.
ie; $\phi = BdA$ (1)
now here area will be equal to $dA = \dfrac{\theta }{{2\pi }}\pi {l^2} = \dfrac{1}{2}{l^2}\theta $
then in equation (1)
\[\phi = B\dfrac{1}{2}{l^2}\theta \]
Differentiating above equation w.r to time
$\dfrac{{d\phi }}{{dt}} = B\dfrac{1}{2}{l^2}\dfrac{{d\theta }}{{dt}}$ (2)
From Faraday's law of electromagnetic induction we know that Ie; $e = - \dfrac{{d\phi }}{{dt}}$
And we have know that angular velocity is time rate at which an object rotates, or revolves, about an axis, or at which the angular displacement between two bodies change
Ie; $\dfrac{{d\theta }}{{dt}} = \omega $
Then in equation (2)
$ - e = \dfrac{{d\phi }}{{dt}} = B\dfrac{1}{2}{l^2}\omega $ (3)
From ohm’s law
$i = e/R$
From equation (3) induced current is
$i = - \dfrac{{B\dfrac{1}{2}{l^2}\omega }}{R}$

Note:- Faraday's law of electromagnetic induction, also known as Faraday's law is the basic law of electromagnetism which helps us to predict how a magnetic field would interact with an electric circuit to produce an electromotive force (EMF). This phenomenon is known as electromagnetic induction.