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Current Question (ID: 20515)

Question:

$\text{A conducting circular loop made of a thin wire has area } 3.5 \times 10^{-3} \text{ m}^2 \text{ and resistance } 10 \, \Omega. \text{ It is placed perpendicular to a time-dependent magnetic field } B(t) = 0.4 \sin(50 \pi t) \, \text{T. The field is uniform in space. Then the net charge flowing through the loop during } t = 0 \, \text{s and } t = 10 \, \text{ms is close to:}$

Options:

1. $0.14 \, \text{mC}$
2. $0.7 \, \text{mC}$
3. $0.21 \, \text{mC}$
4. $0.6 \, \text{mC}$

Solution:

$\text{Hint: } q = \frac{\Delta \phi}{R}$

Import JSON File

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
}