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Current Question (ID: 16763)

Question:
$\text{If a wire in the form of a square with a side } a \text{ carries a current } i, \text{ then the magnetic induction at the centre of the square wire will be:}$ $\text{(Magnetic permeability of free space } = \mu_0)$
Options:
  • 1. $\frac{\mu_0 i}{2 \pi a}$
  • 2. $\frac{\mu_0 i \sqrt{2}}{\pi a}$
  • 3. $\frac{2 \sqrt{2} \mu_0 i}{\pi a}$
  • 4. $\frac{\mu_0 i}{\sqrt{2} \pi a}$
Solution:
$B_0 = 4 \times \frac{\mu_0}{4 \pi} \times \frac{i}{(a/2)} (\sin(45^\circ) + \sin(45^\circ))$ $= 4 \times \frac{\mu_0}{4 \pi} \times \frac{2i}{a} \times \frac{2}{\sqrt{2}}$ $= \frac{\mu_0 i 2 \sqrt{2}}{\pi a}$

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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
}