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Current Question (ID: 16764)

Question:

$\text{A current loop consists of two identical semicircular parts each of radius } R, \text{ one lying in the } x-y \text{ plane, and the other in the } x-z \text{ plane. If the current in the loop is } i, \text{ what will be the resultant magnetic field due to the two semicircular parts at their common centre?}$

Options:

1. $\frac{\mu_0 i}{2\sqrt{2}R}$
2. $\frac{\mu_0 i}{2R}$
3. $\frac{\mu_0 i}{4R}$
4. $\frac{\mu_0 i}{\sqrt{2}R}$

Solution:

$\text{Hint: } B = \frac{\mu_0 I}{4R}$ $\text{Step: Find the resultant magnetic field due to the two semicircular parts at their common centre.}$ $\text{The magnetic field at centre } O \text{ for each semicircular part each of radius } R, \text{ is given by:}$ $\vec{B}_1 = \frac{\mu_0 I}{4R} (-\hat{k})$ $\vec{B}_2 = \frac{\mu_0 I}{4R} (-\hat{j})$ $\text{The net magnetic field at their common centre is given by:}$ $\vec{B}_{\text{net}} = \vec{B}_1 + \vec{B}_2$ $B = \sqrt{B_1^2 + B_2^2}$ $B = \sqrt{\left(\frac{\mu_0 I}{4R}\right)^2 + \left(\frac{\mu_0 I}{4R}\right)^2}$ $\Rightarrow B = \frac{\mu_0 I}{2\sqrt{2}R}$ $\text{Hence, option (1) is the correct answer.}$

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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
}