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Current Question (ID: 20386)

Question:
$\text{Consider the following current carrying structure. Then the magnetic field at the centre is:}$ $\text{Given that } (r_1 = 2\pi \text{ units and } r_2 = 4\pi \text{ units.})$
Options:
  • 1. $1 \times 10^{-8} \text{ T}$
  • 2. $5 \times 10^{-8} \text{ T}$
  • 3. $3 \times 10^{-7} \text{ T}$
  • 4. $4 \times 10^{-7} \text{ T}$
Solution:
$\text{Hint: The magnetic field at the centre, } B = \frac{\mu_0 i}{2R}$ $B = |B_1 - B_2|$ $= \left| \frac{\mu_0 \left( \frac{i}{2} \right)}{4r_1} - \frac{\mu_0 \left( \frac{i}{2} \right)}{4r_2} \right|$ $= \frac{\mu_0 i}{2 \times 4} \left[ \frac{1}{4\pi} \right]$ $= \frac{4}{8} \times 10^{-7} \text{ T} = 5 \times 10^{-8} \text{ T}$

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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
}