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Current Question (ID: 20362)

Question:

$\text{A square loop of side } 2a \text{ and carrying current } I \text{ is kept in xz plane}$ $\text{with its centre at origin. A long wire carrying the same current } I \text{ is}$ $\text{placed parallel to z-axis and passing through point } (0, b, 0), (b \gg a).$ $\text{The magnitude of torque on the loop about z-axis will be:}$

Options:

1. $\frac{2\mu_0 I^2 a^2 b}{\pi (a^2 - b^2)}$
2. $\frac{\mu_0 I^2 a^2 b}{\pi (a^2 - b^2)}$
3. $\frac{\mu_0 I^2 a^2}{2\pi ab}$
4. $\frac{2\mu_0 I^2 a^2}{\pi b}$

Solution:

$\tau = \vec{r} \times \vec{F}$ $F = B I 2a = \frac{\mu_0 I}{2\pi r} I \times 2a$ $F = \frac{\mu_0 I^2 a}{\pi \sqrt{b^2 + a^2}}$ $\tau = F \cos \theta \times 2a$ $\frac{\mu_0 I^2 a}{\pi \sqrt{b^2 + a^2}} \times \frac{b}{\sqrt{b^2 + a^2}} \times 2a$ $\tau = \frac{2\mu_0 I^2 a^2 b}{\pi (a^2 + b^2)}$ $\text{If } b \gg a \text{ then } \tau = \frac{2\mu_0 I^2 a^2}{\pi b}$

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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
}