#### ODBAC is a fixed rectangular conductor of negligible resistance ($CO$ is not connected) and OP is a conductor which rotates clockwise with an angular velocity $\omega$ (figure). The entire system is in a uniform magnetic field B whose direction is along the normal to the surface of the rectangular conductor A’BDC. 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 $180^{o}$.

When the conductor OP is rotated, then the rate of change of area and hence the rate of change of flux can be considered uniform from

$0<\theta <\frac{\pi }{4};\frac{\pi }{4}<\theta <\frac{3\pi }{4}\; and \frac{3\pi }{4}<\theta <\frac{\pi }{2}.$

(i) Let us first assume the position of rotating conductor at the time interval

$t=0\; to\; t=\frac{\pi }{4\omega }(or \frac{T}{8})$

The rod OP will make contact with the side BD. Let the length OQ of the contact after some time interval t such that $0 be x. The flux through the area OQD is

$\phi _{m}=BA=B\left ( \frac{1}{2}\times QD\times OD \right )=B\left ( \frac{1}{2}\times l \tan \theta \times l \right )$

$\Rightarrow \phi _{m}=\frac{1}{2}Bl^{2}\tan \theta ,$ where $\theta =\omega t$

By applying Faraday's law of EMI,

Thus, the magnitude of the emf induced is $\left | \varepsilon \right |=\left | \frac{d\phi }{dt} \right |=\frac{1}{2}Bl^{2}\omega \sec^{2 } \omega t$

The current induced in the circuit will be $I=\frac{\varepsilon }{R}$ where, R is the resistance of the rod in contact.

where, $R\infty \lambda$

$R=\lambda x=\frac{\lambda l}{\cos \omega t}$

$\therefore I=\frac{1}{2}\frac{Bl^{2}\omega }{\lambda l}\sec ^{2}\omega t\; \cos\; \omega t =\frac{Bl\omega }{2\lambda \; \cos\; \omega t}$

(ii) Now let the rod OP will make contact with the side AB. And the length of OQ of the contact after some time interval t such that $\frac{\pi }{4\omega } be x. The flux through the area OQBD is

$\phi _{m}=\left ( l^{2}+\frac{1}{2}\frac{l^{2}}{\tan \theta } \right )B$

Where, $\theta =\omega t$

Thus, the magnitude of emf induced in the loop is

$\left | \varepsilon \right |=\left | \frac{d\phi }{dt} \right |=\frac{Bl^{2}\omega \; \sec ^{2}\omega t}{2\; \tan ^{2}\omega t}$

The current induced in the circuit is $I=\frac{\varepsilon }{R}=\frac{\varepsilon }{\lambda x}=\frac{\varepsilon \; \sin \omega t}{\lambda l}=\frac{1}{2}\frac{Bl\omega }{\lambda \sin \omega t}$

(iii) Similarly, for time interval $\frac{3\pi }{4\omega } the rod will be in touch with AC.

The flux through OQABD is given by

$\phi _{m}=\left ( 2l^{2}-\frac{l^{2}}{2\; \tan \omega t} \right )B$

And the magnitude of emf generated in the loop is given by

$\varepsilon =\frac{d\phi }{dt}=\frac{B\omega l^{2}\; \sec ^{2}\omega t}{2\; \tan ^{2}\omega t}$

$I=\frac{\varepsilon }{R}=\frac{\varepsilon }{\lambda x}=\frac{1}{2}\frac{Bl\omega }{\lambda \sin \omega t}$

These are the required expressions.