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Current Question (ID: 20454)

Question:
$\text{An element } \Delta l = \Delta x \ \hat{i} \text{ is placed at the origin and carries a large current } I = 10 \ \text{A. The magnetic field on the y-axis at a distance of } 0.5 \ \text{m from the elements } \Delta x \text{ of } 1 \ \text{cm length is:}$
Options:
  • 1. $8 \times 10^{-8} \ \text{T}$
  • 2. $10 \times 10^{-8} \ \text{T}$
  • 3. $4 \times 10^{-8} \ \text{T}$
  • 4. $12 \times 10^{-8} \ \text{T}$
Solution:
$\text{Using the Biot-Savart law, the magnetic field } B \text{ at a distance } r \text{ from a current element is given by:}$ $B = \frac{\mu_0}{4\pi} \cdot \frac{I \cdot \Delta l \cdot \sin \theta}{r^2}$ $\text{Here, } \Delta l = 0.01 \ \text{m, } I = 10 \ \text{A, } r = 0.5 \ \text{m, and } \theta = 90^\circ \text{ (since the field is perpendicular to the current element).}$ $B = \frac{4\pi \times 10^{-7}}{4\pi} \cdot \frac{10 \cdot 0.01 \cdot 1}{0.5^2}$ $= 2 \times 10^{-8} \ \text{T}$ $\text{Therefore, the correct answer is option 3: } 4 \times 10^{-8} \ \text{T.}$

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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
}