Question

ODBAC is a fixed rectangular conductor of negilible resistance (CO is not connnected) and OP is a conductor which rotates clockwise with an angular velocity ω (Figure). The entire system is in a uniform magnetic field B whose direction is 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 λ per unit length. Find the current in the rotating conductor, as it rotates by 180°.

Answer 1

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

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

t = 0 to ${\displaystyle t=\frac{\pi }{4\omega }}$ (or 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 ${\displaystyle 0<t<\frac{\pi }{4\omega }}$ or ${\displaystyle \omega <t<\frac{T}{8}}$ be x.

The flux through the area ODQ is ${\displaystyle {\varphi }_m=BA=B(\frac{1}{2}\times QD\times OD)=B(\frac{1}{2}\times l\tan \theta xl)}$

⇒ ${\displaystyle {\varphi }_m=\frac{1}{2}B{l}^2\tan \theta }$, where θ = ωt

By applying Faraday's law of EMI,

Thus, the magnitude of the emf induced is ${\displaystyle |\epsilon |=\left|\frac{d\varphi }{dt}\right|=\frac{1}{2}B{l}^2\omega {\sec }^2\omega t}$

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

Where, ${\displaystyle R \infty \lambda }$

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

∴ ${\displaystyle I=\frac{1}{2}\frac{B{l}^2\omega }{\lambda l}{\sec }^2\omega t\cos \omega t=\frac{Bl\omega }{2\lambda \cos \omega t}}$

(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 ${\displaystyle \frac{\pi }{4\omega }<t<\frac{3\pi }{4\omega }}$ or ${\displaystyle T/8<t<\frac{3T}{8}}$ be x.

The flux through the area OQBD is ${\displaystyle {\varphi }_m=(l^2+\frac{1}{2}\frac{l^2}{2\tan \theta })B}$

Where, θ = ωt

 Thus, the magnitude of emf induced in the loop is 

${\displaystyle |\epsilon |=\left|\frac{d\varphi }{dt}\right|=\frac{B{l}^2\omega {\sec }^2\omega t}{2{\tan }^2\omega t}}$

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

(iii) When the flux through OQABD = ${\displaystyle \varphi }$ = B.A

${\displaystyle \varphi =B.(2l^2+\frac{1}{2}ly)}$

${\displaystyle \varphi =B.(2l^2+\frac{l^2\tan \omega t}{2})}$  ......${\displaystyle \left[\begin{array}{c}\tan (180-\theta )=\frac{y}{l}\\ \begin{array}{c}y=l(-\tan \theta )\\ y=-l\tan \theta \end{array}\end{array}\right]}$

${\displaystyle \frac{d\varphi }{dt}=\frac{d}{dt}[2l^2+\frac{l^2}{2}\left(\tan \omega t\right)]B}$

${\displaystyle -\epsilon =0+\frac{B{l}^2}{2}+\frac{d}{dt}\tan \omega t}$

${\displaystyle -\epsilon =+\frac{B{l}^2\omega }{2}{\sec }^2\omega t=-\frac{B{l}^2\omega }{2{\cos }^2\omega t}}$

${\displaystyle I=\frac{\epsilon }{R}=\frac{\epsilon }{\lambda x}}$

${\displaystyle I=\frac{-B{l}^2\omega }{2{\cos }^2 \omega t}\frac{\cos \omega t}{\lambda (-1)}}$  ......${\displaystyle [\because \frac{l}{x}=\cos (180-\theta )\frac{1}{x}=-\cos \theta \Rightarrow x=\frac{-1}{\cos \omega t}]}$

${\displaystyle I=\frac{Bl\omega }{2\lambda \cos \omega t}}$