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Current Question (ID: 17385)

Question:

$\text{In AC (alternating current) generator, a coil of wire rotates in a magnetic field.}$ $\text{Which of the following would change the potential difference measured by the voltmeter in the system above?}$ $\text{(a) use more turns of wire in the coil.}$ $\text{(b) use thicker wire.}$ $\text{(c) change the speed of rotation.}$ $\text{Choose the correct option from the given ones:}$

Options:

1. $(c) \text{ only}$
2. $(a), (b) \text{ and } (c)$
3. $(a) \text{ and } (b) \text{ only}$
4. $(a) \text{ and } (c) \text{ only}$

Solution:

$\text{Hint: } e = NBA\omega \sin(\omega t)$ $\text{Explanation: The induced emf is given by;}$ $\Rightarrow e = NBA\omega \sin(\omega t)$ $\text{It depends on the number of turns } N \text{ in the coil, and the speed of rotation } \omega.$ $\text{It does not depend on the resistance of the coil. Increasing } N \text{ increases the}$ $\text{induced emf directly because the emf is proportional to the number of turns.}$ $\text{Increasing the speed of rotation increases the rate of change of magnetic}$ $\text{flux. This will increase the potential difference. Therefore, more turns of wire}$ $\text{in the coil and change the speed of rotation would change the potential}$ $\text{difference measured by the voltmeter in the above system.}$ $\text{Hence, option (4) is the correct answer.}$

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Upload a JSON file containing LaTeX/MathJax formatted question, options, and solution.

Expected JSON Format:

{
  "question": "The mass of carbon present in 0.5 mole of $\\mathrm{K}_4[\\mathrm{Fe(CN)}_6]$ is:",
  "options": [
    {
      "id": 1,
      "text": "1.8 g"
    },
    {
      "id": 2,
      "text": "18 g"
    },
    {
      "id": 3,
      "text": "3.6 g"
    },
    {
      "id": 4,
      "text": "36 g"
    }
  ],
  "solution": "\\begin{align}\n&\\text{Hint: Mole concept}\\\\\n&1 \\text{ mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6] = 6 \\text{ moles of carbon atom}\\\\\n&0.5 \\text{ mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6] = 6 \\times 0.5 \\text{ mol} = 3 \\text{ mol}\\\\\n&1 \\text{ mol of carbon} = 12 \\text{ g}\\\\\n&3 \\text{ mol carbon} = 12 \\times 3 = 36 \\text{ g}\\\\\n&\\text{Hence, 36 g mass of carbon present in 0.5 mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6].\n\\end{align}",
  "correct_answer": 4
}