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Current Question (ID: 16943)

Question:

$\text{The radius of a loop as shown in the figure is } 10 \text{ cm.}$ $\text{If the magnetic field is uniform and has a value } 10^{-2} \text{ T, then the flux through the loop will be:}$

Options:

1. $2\pi \times 10^{-2} \text{ Wb}$
2. $3\pi \times 10^{-4} \text{ Wb}$
3. $5\pi \times 10^{-5} \text{ Wb}$
4. $5\pi \times 10^{-4} \text{ Wb}$

Solution:

$\text{Hint: } \phi = BA \cos \theta$ $\text{Step: Find the flux through the loop.}$ $\text{For the magnetic flux through the loop is given by;}$ $\phi = BA \cos \theta$ $\text{The angle between the magnetic field and the normal to the surface of the loop } \theta = 60^\circ.$ $\text{The area of a circular loop is given by;}$ $\Rightarrow A = \pi r^2 = \pi (0.1)^2 = \pi \times 0.01 = 0.01\pi \text{ m}^2$ $\text{The magnetic flux through the loop is given by;}$ $\phi = BA \cos \theta$ $\phi = 10^{-2} \times \pi (0.1)^2 \cos 60^\circ$ $\phi = \frac{1}{2} \times \pi \times 10^{-4}$ $\Rightarrow \phi = 5\pi \times 10^{-5} \text{ Wb}$ $\text{Hence, option (3) is the correct answer.}$

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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
}