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Current Question (ID: 16834)

Question:

$\text{A circular loop with a radius of } 20 \text{ cm is placed in a uniform magnetic field } B = 2 \text{ T in the } XY \text{ plane as shown in the figure. If the loop carries a current of } i = 1 \text{ A, then the magnitude of torque acting on the loop will be:}$

Options:

1. $0.25 \text{ N-m}$
2. $5.2 \text{ N-m}$
3. $2.5 \text{ N-m}$
4. $0.52 \text{ N-m}$

Solution:

$\text{Hint: } |\tau| = MB \sin \theta$ $\text{Step: Find the magnitude of torque acting on the loop.}$ $\text{The magnetic moment is given by:}$ $M = I \cdot A = I \cdot \pi R^2$ $\therefore M = 1 \times \pi \times (20 \times 10^{-2})^2$ $\Rightarrow M = 4\pi \times 10^{-2} \text{ A.m}^2 \text{ (Perpendicular to plane)}$ $\text{The magnitude of the torque is given by:}$ $|\tau| = MB \sin \theta$ $|\tau| = MB \quad (\theta = 90^\circ, \sin 90 = 1)$ $\Rightarrow |\tau| = 4\pi \times 10^{-2} \times 2 = 0.25 \text{ N/m}$ $\text{Hence, option (1) is the correct answer.}$

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
}