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Current Question (ID: 20345)

Question:

$\text{Two wires } A \text{ and } B \text{ are carrying currents } I_1 \text{ and } I_2 \text{ as shown in the figure.}$ $\text{The separation between them is } d. \text{ A third wire } C \text{ carrying a current } I$ $\text{is to be kept parallel to them at a distance } x \text{ from } A \text{ such that the net}$ $\text{force acting on it is zero. The possible values of } x \text{ are:}$

Options:

1. $x = \left( \frac{I_1}{I_1 - I_2} \right) d \text{ and } x = \frac{I_2}{(I_1 - I_2)} d$
2. $x = \frac{I_1 d}{I_1 - I_2} \text{ and } x = \frac{I_2}{(I_1 - I_2)} d$
3. $x = \left( \frac{I_1}{I_1 - I_2} \right) d \text{ and } x = \frac{I_2}{(1 - I_2)} d$
4. $x = \left( \frac{I_2}{I_1 - I_2} \right) d \text{ and } x = \frac{I_2}{(1 - I_2)} d$

Solution:

$\text{Hint: } F_{\text{net}} = \frac{\mu_0 i_1 i_2}{2 \pi r}$ $F_{\text{net}} \text{ on } C = \frac{\mu_0 I_1 I}{2 \pi x} + \frac{\mu_0 I_2 I}{2 \pi (d-x)} = 0$ $\frac{I_1}{x} = \frac{I_2}{x-d}$ $x = \frac{I_1 d}{I_1 - I_2}$

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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
}