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Current Question (ID: 16837)

Question:

$\text{If a square loop } ABCD \text{ carrying a current } i \text{ is placed near and coplanar with a long straight conductor } XY \text{ carrying a current } I, \text{ what will be the net force on the loop?}$

Options:

1. $\frac{\mu_0 I i}{2 \pi}$
2. $\frac{2 \mu_0 I i L}{3 \pi}$
3. $\frac{3 \pi}{\mu_0 I i L}$
4. $\frac{2 \pi}{2 \mu_0 I i}$

Solution:

$\text{Hint: The net force on the loop will be the vector sum of the forces on the sides of the loop.}$ $\text{Step 1: Find the force on the arm } BA \text{ and the arm } CD.$ $F_{BA} = \frac{\mu_0 I i L}{2 \pi \frac{L}{2}}$ $F_{CD} = \frac{\mu_0 I i L}{2 \pi \frac{3L}{2}}$ $\text{Step 2: Find the net force on the loop.}$ $F_{\text{loop}} = F_{BA} - F_{CD}$ $F_{\text{loop}} = \frac{\mu_0 I i L}{2 \pi \frac{L}{2}} - \frac{\mu_0 I i L}{2 \pi \frac{3L}{2}}$ $\Rightarrow F_{\text{loop}} = \frac{2 \mu_0 I i}{3 \pi}$ $\text{Hence, option (4) is the correct answer.}$

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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
}