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Current Question (ID: 16884)

Question:
$\text{Assume that the magnetic field of the earth is due to a small magnetic dipole,}$ $\text{placed at the centre of the earth (radius: } R).$ $\text{The magnetic field at the equator}$ $\text{is } B_e. \text{ The dipole moment of the dipole is:}$
Options:
  • 1. $\frac{\pi}{\mu_0} \left( B_e R^3 \right)$
  • 2. $\frac{2\pi}{\mu_0} \left( B_e R^3 \right)$
  • 3. $\frac{4\pi}{\mu_0} \left( B_e R^3 \right)$
  • 4. $\frac{2}{\mu_0} \left( B_e R^3 \right)$
Solution:
$\text{Hint: } B_e = \frac{\mu_0 m}{4\pi r^3}$ $\text{Step: Find the magnetic moment of the earth.}$ $B_e = \frac{\mu_0 m}{4\pi r^3}$ $\Rightarrow m = \frac{4\pi}{\mu_0} \left( B_e R^3 \right)$

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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
}