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Electromagnetic Induction — Physics STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 SciencePhysicsElectromagnetic Induction5 Marks
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 $\omega$ (Fig). 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 $\lambda$ per unit length. Find the current in the rotating conductor, as it rotates by 180º.
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Answer


Key concept: When the conductor OP is rotated, then the rate of change if area hence the rate of change of flux can be considered uniform from $0<\theta <\frac{\pi}{4};\frac{\pi}{4}<\theta<\frac{3\pi}{4}\text{ and }\frac{3\pi}{4}<\theta<\frac{\pi}{2}$.
  1. Let us first assume the position of rotating conductor at time interval
$\text{t}=0 \text{ to }\text{t}=\frac{\pi}{4\omega}\Big(\text{or}\frac{\text{T}}{8}\Big)$

The rod OP will make contact with the side BD. Let the length OQ of the contact after some time interval t such that $0<\text{t}<\frac{\pi}{4\omega}$ or $0<\text{t}<\frac{\text{T}}{8}$ be x. The flux through the area ODQ is,
$\phi_\text{m}=\text{BA}=\text{B}\Big(\frac{1}{2}\times\text{QD}\times\text{OD}\Big)=\Big(\frac{1}{2}\times\text{l}\tan\theta\times\text{l}\Big)$
$\Rightarrow\ \phi_\text{m}=\frac{1}{2}\text{Bl}^2\tan\theta,\text{where}\theta=\omega\text{t}$
By applying Faraday's law of EMI,
Thus, the magnitude of the emf induced is $|\in|=\Big|\frac{\text{d}\phi}{\text{dt}}\Big|=\frac{1}{2}\text{Bl}^2\omega\sec^2\omega\text{t}$
The current induced in the circuit will be $\text{I}=\frac{\in}{\text{R}}$ where, R is the where $\text{R}\propto\lambda$.
$\text{R}=\lambda\text{x}=\frac{\lambda\text{l}}{\cos\omega\text{t}}$
$\therefore\ \text{I}=\frac{1}{2}\frac{\text{Bl}^2\omega}{\lambda\text{l}}\sec^2\omega\text{t}\cos\omega\text{t}=\frac{\text{Bl}\omega}{2\lambda\cos\omega\text{t}}$
  1. 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 $\frac{\pi}{4\omega}<\text{t}<\frac{3\phi}{4\omega}$ or $\frac{\text{T}}{8}<\text{t}<\frac{3\text{t}}{8}$ be x. The flux though the area OQBD is,
$\phi_\text{m}=\Big(\text{l}^2+\frac{1}{2}\frac{\text{l}^2}{\tan\theta}\Big)\text{B}$
where, $\theta=\omega\text{t}$
Thus, the magnitude of emf indueced in loop is
$|\in|=\Big|\frac{\text{d}\phi}{\text{dt}}\Big|=\frac{\text{Bl}^2\omega\sec^2\omega\text{t}}{2\tan^2\omega\text{t}}$
The current induced in the circuit in the circuit is $\text{I}=\frac{\in}{\text{R}}=\frac{\in}{\lambda\text{x}}=\frac{\in\sin\omega\text{t}}{\lambda\text{l}}=\frac{1}{3}\frac{\text{Bl}\omega}{\lambda\sin\omega\text{t}}$
  1. Similarly, for time interval $\frac{3\pi}{4\omega}<\text{t}<\frac{\pi}{\omega}\text{ or }\frac{3\text{T}}{8}<\text{t}<\frac{\text{T}}{2}$, the rod will be in touch with AC.

The flux through OQABD is given by
$\phi_\text{m}=\bigg(2\text{l}^2-\frac{\text{l}^2}{1\tan\omega\text{t}}\bigg)\text{B}$
And the magnitude of emf generated in loop is given by
$\in=\frac{\text{d}\phi}{\text{dt}}=\frac{\text{B}\omega\text{l}^2\sec^2\omega\text{t}}{1\tan^2\omega\text{t}}$
$\text{I}=\frac{\in}{\text{R}}=\frac{\in}{\lambda\text{x}}=\frac{1}{2}\frac{\text{BI}\omega}{\lambda\sin\omega\text{t}}$
These are the required expressions.

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