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  • 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).

ODBAC is a fixed rectangular conductor of negligible resistance (CO is not connected) and OP is a conductor that 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 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 180^{o}.

Answers (1)

Explanation:-

 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 the 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<t<\frac{\pi }{4\omega }\; or\; 0<t<\frac{T}{8} 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 }<t<\frac{3\pi }{4\omega }\; or \; \frac{T}{8}<t<\frac{3T}{8} 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 intervals \frac{3\pi }{4\omega }<t<\frac{\pi }{\omega }or\frac{3T}{8}<t<\frac{T}{2}, 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

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.

Posted by

infoexpert23

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