$A$ fixed rectangular conductor $ODBAC$ has negligible resistance (where $CO$ is not connected). $A$ conductor $OP$ rotates clockwise with an angular velocity $\omega$ as shown in the figure. The entire system is in a uniform magnetic field $B$ directed along the normal to the surface of the rectangular conductor $ABDC$. The conductor $OP$ is in electric contact with $ABDC$. The rotating conductor has a resistance of $\lambda$ per unit length. Find the current in the rotating conductor as it rotates by $180^{\circ}$.

(N/A) Let the wire $OP$ be at an angle $\theta = \omega t$ with the horizontal $OD$. The wire $OP$ intersects the vertical side $BD$ at point $Q$. In the right-angled triangle $\triangle ODQ$,the length $OQ = x = \frac{l}{\cos \theta}$.
The area of the triangle $\triangle ODQ$ is $A = \frac{1}{2} \times OD \times QD = \frac{1}{2} \times l \times (l \tan \theta) = \frac{1}{2} l^2 \tan \theta$.
The magnetic flux linked with $\triangle ODQ$ is $\phi = B \cdot A = \frac{1}{2} B l^2 \tan(\omega t)$.
The induced $emf$ is $\varepsilon = \frac{d\phi}{dt} = \frac{d}{dt} \left( \frac{1}{2} B l^2 \tan(\omega t) \right) = \frac{1}{2} B l^2 \omega \sec^2(\omega t)$.
The resistance of the portion of the wire $OP$ inside the loop is $R = \lambda x = \frac{\lambda l}{\cos(\omega t)}$.
The induced current $I$ is given by $I = \frac{\varepsilon}{R} = \frac{\frac{1}{2} B l^2 \omega \sec^2(\omega t)}{\frac{\lambda l}{\cos(\omega t)}} = \frac{B l \omega}{2 \lambda \cos^3(\omega t)}$.