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Current Question (ID: 20411)

Question:

$\text{An iron rod of volume } 10^{-3} \text{ m}^3 \text{ and relative permeability } 1000 \text{ is placed as core in a solenoid with } 10 \text{ turns/cm. If a current of } 0.5 \text{ A is passed through the solenoid, then the magnetic moment of the rod will be:}$

Options:

1. $0.5 \times 10^2 \text{ Am}^2$
2. $50 \times 10^2 \text{ Am}^2$
3. $500 \times 10^2 \text{ Am}^2$
4. $5 \times 10^2 \text{ Am}^2$

Solution:

$\text{Hint: } M = \mu_r N i A$ $M = \mu_r N i A$ $\text{Here } \mu_r = \text{Relative permeability}$ $N = \text{No. of turns}$ $i = \text{Current}$ $A = \text{Area of cross section}$ $M = \mu_r N i A = \mu_r n l i$ $M = \mu_r n i V = 1000 (1000) 0.5 (10^{-3})$ $= 500 = 5 \times 10^2 \text{ Am}^2$

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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
}