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Current Question (ID: 16830)

Question:

$\text{A square current-carrying loop is suspended in a uniform magnetic field acting in the plane of the loop. If the force on one arm of the loop is } \vec{F}, \text{ what will be the net force on the remaining three arms of the loop?}$

Options:

1. $3\vec{F}$
2. $-\vec{F}$
3. $-3\vec{F}$
4. $\vec{F}$

Solution:

$\text{Hint: The net force on the loop will be equal to zero.}$ $\text{Step 1: Find the net force on the loop.}$ $\text{The net force on the loop} = \vec{F}_1 + \vec{F}_2 + \vec{F}_3 + \vec{F}_4$ $\text{Step 2: Find the net force on the other three arms.}$ $\text{As the net force on a closed-loop} = 0,$ $\vec{F}_1 + \vec{F}_2 + \vec{F}_3 + \vec{F}_4 = 0$ $\vec{F}_2 + \vec{F}_3 + \vec{F}_4 = -\vec{F}_1$

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
}