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Current Question (ID: 16987)

Question:
$\text{When a conducting wire } XY \text{ is moved towards the right, a current flows in the anti-clockwise direction.}$ $\text{Direction of magnetic field at point } O \text{ is:}$
Options:
  • 1. $\text{parallel to the motion of wire.}$
  • 2. $\text{along with } XY.$
  • 3. $\text{perpendicular outside the paper.}$
  • 4. $\text{perpendicular inside the paper.}$
Solution:
$\text{Hint: Recall Lenz's law.}$ $\text{Step 1: Find flux through the loop}$ $\phi = B_0 A$ $\text{As the area of the loop decreases, so the flux through the loop will also decrease.}$ $\text{Step 2: Find the direction of field at point O.}$ $\text{According to Lenz's Law direction of induced current should oppose the decrease in flux so induced current should produce field in the same direction as that of original field to support original field.}$ $\text{Here the direction of the field due to the current in the anti-clockwise direction is perpendicular to and outside the paper.}$ $\text{Thus, the direction of original field at point O will be perpendicular outside the paper.}$

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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
}