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Current Question (ID: 16809)

Question:

$\text{Three infinitely long conductors carrying currents } I_1, I_2 \text{ and } I_3 \text{ lie perpendicular to the plane of the paper as shown below.}$ $\text{If the value of the integral } \oint \vec{B} \cdot \vec{dl} \text{ for the loops } C_1, C_2, \text{ and } C_3 \text{ are } 2\mu_0, 4\mu_0 \text{ and } \mu_0 \text{ in the units of N/A, respectively, then:}$

Options:

1. $I_1 = 3\text{A into the paper}$
2. $I_2 = 3\text{A out of the paper}$
3. $I_3 = 0$
4. $I_3 = 1\text{A out of the paper}$

Solution:

$\text{Hint: Use Ampere's circuital law}$ $\text{Step: Find the value of currents flowing through infinitely long conductors.}$ $\text{By using Ampere's circuital law;} \oint \vec{B} \cdot \vec{dl} = \mu_0 (I_1 + I_2)$ $2\mu_0 = (I_1 + I_2) \mu_0$ $I_1 + I_2 = 2 \quad \cdots (1)$ $4\mu_0 = \mu_0 (I_2 + I_3)$ $I_2 + I_3 = 4 \quad \cdots (2)$ $\mu_0 = (I_1 + I_2 + I_3) \mu_0$ $1 = (I_1 + I_2 + I_3)$ $\text{From the equations (1) and (2) we get;}$ $\Rightarrow I_1 = -3 \text{ A i.e. 3 A into the paper.}$ $\text{Hence, option (1) is the correct answer.}$

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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
}