Import Question JSON

Current Question (ID: 16968)

Question:

$\text{A rectangular loop of wire shown below is coplanar with a long wire carrying current, } I.$ $\text{The loop is pulled to the right as indicated. What are the directions of the induced current in the loop and the magnetic forces on the left and right sides of the loop?}$

Options:

1. $\text{counterclockwise, to the left, to the right}$
2. $\text{clockwise, to the left, to the right}$
3. $\text{counterclockwise, to the right, to the left}$
4. $\text{clockwise, to the right, to the left}$

Solution:

$\text{Hint: Flux is inward and it is decreasing as the loop is going away from the wire.}$ $\text{Step 1: Find the direction of the current induced.}$ $\text{The magnetic field decreases away from the loop, so the inward flux decreases as it is moved away.}$ $\text{The direction of the induced current will be such that the magnetic field due to the induced current will be inwards.}$ $\text{Therefore the direction of the induced current is clockwise.}$ $\text{Step 2: Find the direction of force acting on the loop.}$ $\text{Two current-carrying wires carrying current in the same direction will attract each other and in opposite directions, current-carrying wires will repel each other.}$ $\text{Therefore force on the left side of the loop is towards the left and the force on the right side of the loop is towards the right.}$ $\text{Hence, option (2) is the correct answer.}$

Import JSON File

Upload a JSON file containing LaTeX/MathJax formatted question, options, and solution.

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
}