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Current Question (ID: 20502)

Question:

$\text{A small square loop of wire of side } \ell \text{ is placed inside a large square loop of wire } L \ (L \gg \ell).$ $\text{Both loops are coplanar and their centres coincide at point } O \text{ as shown in the figure.}$ $\text{The mutual inductance of the system is:}$

Options:

1. $\frac{2\sqrt{2}\mu_0L^2}{\pi \ell}$
2. $\frac{\mu_0\ell^2}{2\sqrt{2}\pi L}$
3. $\frac{2\sqrt{2}\mu_0\ell^2}{\pi L}$
4. $\frac{\mu_0L^2}{2\sqrt{2}\pi \ell}$

Solution:

$\text{Hint: } \phi = MI$ $\text{Step 1: Find the magnetic field due to the outer loop.}$ $\text{The magnetic field at the centre due to the outer loop is given by:}$ $B = \frac{2\sqrt{2}\mu_0i}{\pi L}$ $\text{The flux linking through the square loop is given by:}$ $\phi_{21} = B\ell^2 = \frac{2\sqrt{2}\mu_0i}{\pi L} \ell^2$ $\text{Step 2: Find the mutual inductance of the system.}$ $\text{The mutual inductance of the system is given by:}$ $M = \frac{\phi_{21}}{i}$ $\Rightarrow M = \frac{2\sqrt{2}\mu_0\ell^2}{\pi L}$ $\text{Hence, option (3) is the correct answer.}$

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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
}