text: • # question_answer  Calculate the rate at which the flux linked with the generated area changes with time when a rod of length x is   (i) translated  (ii) rotated in a uniform field of induction B as shown in figure.    Or An inductor, L of reactance, ${{X}_{L}}$ is connected in series with a bulb B to an AC source as shown in the figure. Briefly explain how does the brightness of the bulb change, when AC source (i) number of turns of the inductor is reduced (ii) a capacitor of reactance, ${{X}_{C}}={{X}_{L}}$ is included in series in the same circuit?
  
  (i) Component of velocity perpendicular to the rod $=v\sin \theta .$ Therefore, in time t, area transversed (A) is,             $=x\times v\sin \theta \times t$ Since,    $\phi =B\cdot \,\,A=BA\cos \theta$             Here,     $\theta =0{}^\circ$             $\therefore$      $\phi =B(x\times v\sin \theta \times t)\cos 0{}^\circ =Bxvt\sin \theta$             Also,     $e=\frac{d\phi }{dt}=Bxv\sin \theta$ (ii) In time t, if $\theta$ is the angle traced by the free end, then $\therefore$ Area swept, $A=\pi {{x}^{2}}\times \left( \frac{\theta }{2\pi } \right)=\frac{1}{2}{{x}^{2}}\theta$ Also, $\phi =B\left( \frac{1}{2}{{x}^{2}}\theta \right)\cos 0{}^\circ =\frac{1}{2}B{{x}^{2}}\theta$ $\therefore$      $e=\frac{d\phi }{dt}$             $=\frac{1}{2}B{{x}^{2}}\frac{d\theta }{dt}=\frac{1}{2}B{{x}^{2}}\omega$ Or (i) When the number of turns of the inductor is reduced, its inductance, L decreases. A decrease in inductance causes ${{X}_{L}}$ to decrease and hence current in the circuit increases therefore brightness of the bulb increases. (ii) When the capacitor is connected in series in the circuit, then its impedance is given by             $Z=\sqrt{{{R}^{2}}+{{({{X}_{L}}-{{X}_{C}})}^{2}}}$ Here,     ${{X}_{L}}={{X}_{C}}$ $\therefore$      Z = R Thus, impedance of the circuit becomes minimum and is equal to the resistance and hence, the brightness of the bulb increases to maximum.\
explanation: The document contains a math and physics question along with its detailed answer. The question involves calculating the rate of change of magnetic flux linked with a generated area when a rod is either translated or rotated in a uniform magnetic field. Additionally, it asks about the effect of changes in an inductor and the addition of a capacitor on the brightness of a bulb in an AC circuit. The answer provides step-by-step solutions for both parts of the question.
distilabel_metadata.content: Evaluate the user given text extract and determine if it contains a math question and optionally and answer to this question.
  
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  1. Briefly describe the document and mention if it contains a question or a question and answer.
  2. Finalize the output with: Label: no-question,question,question-and-answer,multiple-question,multiple-question-and-answer
distilabel_metadata.role: system
distilabel_metadata.content: • # question_answer  Calculate the rate at which the flux linked with the generated area changes with time when a rod of length x is   (i) translated  (ii) rotated in a uniform field of induction B as shown in figure.    Or An inductor, L of reactance, ${{X}_{L}}$ is connected in series with a bulb B to an AC source as shown in the figure. Briefly explain how does the brightness of the bulb change, when AC source (i) number of turns of the inductor is reduced (ii) a capacitor of reactance, ${{X}_{C}}={{X}_{L}}$ is included in series in the same circuit?
  
  (i) Component of velocity perpendicular to the rod $=v\sin \theta .$ Therefore, in time t, area transversed (A) is,             $=x\times v\sin \theta \times t$ Since,    $\phi =B\cdot \,\,A=BA\cos \theta$             Here,     $\theta =0{}^\circ$             $\therefore$      $\phi =B(x\times v\sin \theta \times t)\cos 0{}^\circ =Bxvt\sin \theta$             Also,     $e=\frac{d\phi }{dt}=Bxv\sin \theta$ (ii) In time t, if $\theta$ is the angle traced by the free end, then $\therefore$ Area swept, $A=\pi {{x}^{2}}\times \left( \frac{\theta }{2\pi } \right)=\frac{1}{2}{{x}^{2}}\theta$ Also, $\phi =B\left( \frac{1}{2}{{x}^{2}}\theta \right)\cos 0{}^\circ =\frac{1}{2}B{{x}^{2}}\theta$ $\therefore$      $e=\frac{d\phi }{dt}$             $=\frac{1}{2}B{{x}^{2}}\frac{d\theta }{dt}=\frac{1}{2}B{{x}^{2}}\omega$ Or (i) When the number of turns of the inductor is reduced, its inductance, L decreases. A decrease in inductance causes ${{X}_{L}}$ to decrease and hence current in the circuit increases therefore brightness of the bulb increases. (ii) When the capacitor is connected in series in the circuit, then its impedance is given by             $Z=\sqrt{{{R}^{2}}+{{({{X}_{L}}-{{X}_{C}})}^{2}}}$ Here,     ${{X}_{L}}={{X}_{C}}$ $\therefore$      Z = R Thus, impedance of the circuit becomes minimum and is equal to the resistance and hence, the brightness of the bulb increases to maximum.\
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The document contains a math and physics question along with its detailed answer. The question involves calculating the rate of change of magnetic flux linked with a generated area when a rod is either translated or rotated in a uniform magnetic field. Additionally, it asks about the effect of changes in an inductor and the addition of a capacitor on the brightness of a bulb in an AC circuit. The answer provides step-by-step solutions for both parts of the question.
  
  Label: question-and-answer