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Current Question (ID: 16841)

Question:

$\text{A square loop with a side } l \text{ is held in a uniform magnetic field } B, \text{ such that its plane making an angle } \alpha \text{ with } B. \text{ A current } i \text{ flows through the loop. What will be the torque experienced by the loop in this position?}$

Options:

1. $Bil^2$
2. $Bil^2 \sin \alpha$
3. $Bil^2 \cos \alpha$
4. $\text{zero}$

Solution:

$\text{The torque } \tau \text{ experienced by a current loop in a magnetic field is given by } \tau = nBiA \sin \alpha. \text{ For a square loop, } A = l^2 \text{ and } n = 1. \text{ Therefore, } \tau = Bil^2 \sin \alpha.$

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Upload a JSON file containing LaTeX/MathJax formatted question, options, and solution.

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
}