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Current Question (ID: 16752)

Question:

$\text{A long wire carrying a steady current is bent into a circular loop of one turn.}$ $\text{The magnetic field at the centre of the loop is } B. \text{ It is then bent into a circular coil of } n \text{ turns.}$ $\text{What will the magnetic field be at the centre of this } n\text{-turn coil?}$

Options:

1. $nB$
2. $n^2B$
3. $2nB$
4. $2n^2B$

Solution:

$\text{Hint - } B_{\text{loop}} = \frac{\mu_0 I}{2r}$ $\text{Step 1: Find the radius of new coil}$ $2\pi R = n(2\pi r)$ $r = \frac{R}{n}$ $\text{Step 2: Find the initial magnetic field}$ $B = \frac{\mu_0 i}{2R}$ $\text{Step 3: Find the final magnetic field}$ $B_f = n\left(\frac{\mu_0 i}{2r}\right)$ $B_f = n^2\left(\frac{\mu_0 i}{2R}\right)$ $B_f = n^2B$

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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
}