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Current Question (ID: 20504)

Question:

$\text{A small circular loop of radius } r \text{ is placed in the plane of a square loop of side length } L \ (r \ll L).$ $\text{A circular loop is at the center of the square as shown in the figure.}$ $\text{The mutual inductance of the given system is:}$

Options:

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

Solution:

$\text{Hint: } \phi = MI$ $\text{Step 1: Find the magnetic field at the centre of the rectangular loop.}$ $\text{The magnetic field at any point, which is at a distance } a \text{ from the wire, is given by:}$ $\Rightarrow B = \frac{\mu_0 I}{4 \pi a} (\sin \alpha + \sin \beta)$ $\text{Then, the magnetic field at the centre of the rectangular loop is given by:}$ $\Rightarrow B = \frac{\mu_0 I}{4 \pi a} (\sin \alpha + \sin \beta)$ $\Rightarrow B_{\text{centre}} = 4 \times \frac{\mu_0 I}{4 \pi \left( \frac{L}{2} \right)} \left[ \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} \right]$ $\Rightarrow B_{\text{centre}} = \frac{4 \mu_0 I}{\sqrt{2} \pi L}$ $\Rightarrow B_{\text{centre}} = \frac{2 \sqrt{2} \mu_0 I}{\pi L}$ $\text{Step 2: Find the mutual inductance of the system.}$ $\text{The magnetic flux in a circular loop is given by:}$ $\Rightarrow (\phi) = B \cdot A = B_{\text{centre}} \cdot A_2 = B \times \pi r^2 \quad \cdots (1)$ $\text{The magnetic flux is also given in terms of mutual inductance as:}$ $\Rightarrow \phi = MI \quad \cdots (2)$ $\text{From (1) and (2) we get,}$ $\Rightarrow MI = \frac{2 \sqrt{2} \mu_0 I \times \pi r^2}{\pi L} \Rightarrow M = \frac{2 \sqrt{2} \mu_0 r^2}{L}$ $\text{Hence, option (3) is the correct answer.}$

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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
}