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Current Question (ID: 20364)

Question:

$\text{Magnetic fields at two points on the axis of a circular coil at a distance of } 0.05 \text{ m and } 0.2 \text{ m from the centre are in the ratio } 8 : 1. \text{ The radius of the coil is:}$

Options:

1. $0.2 \text{ m}$
2. $0.1 \text{ m}$
3. $0.15 \text{ m}$
4. $1.0 \text{ m}$

Solution:

$\text{Hint: } \textbf{B} = \frac{\mu_0}{4\pi} \frac{2NIA}{(R^2 + x^2)^{3/2}}$ $\text{We know, the magnetic field on the axis of a current carrying circular ring is given by}$ $\textbf{B} = \frac{\mu_0}{4\pi} \frac{2NIA}{(R^2 + x^2)^{3/2}}$ $\therefore \frac{B_1}{B_2} = \frac{8}{1} = \left[ \frac{R^2 + (0.2)^2}{R^2 + (0.05)^2} \right]^{3/2}$ $4 \left[ R^2 + (0.05)^2 \right] = \left[ R^2 + (0.2)^2 \right]$ $4R^2 - R^2 = (0.2)^2 - 4 \times (0.05)^2$ $4R^2 - R^2 = (0.2)^2 - (0.1)^2$ $3R^2 = 0.3 \times 0.1$ $R^2 = (0.1)^2 \Rightarrow R = 0.1$

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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
}