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Current Question (ID: 20491)

Question:
$\text{A circular coil of radius } 10 \text{ cm is placed in a uniform magnetic field of } 3.0 \times 10^{-5} \text{ T with its plane perpendicular to the field initially.}$ $\text{It is rotated at constant angular speed about an axis along the diameter of coil and perpendicular to magnetic field so that it undergoes half of rotation in } 0.2 \text{ s.}$ $\text{The maximum value of EMF induced (in } \mu \text{V) in the coil will be close to:}$
Options:
  • 1. $5$
  • 2. $10$
  • 3. $15$
  • 4. $20$
Solution:
$\text{Hint: } \phi = BA \cos \omega t$ $\text{Step: The maximum value of EMF induced in the coil.}$ $\text{Given: } r = 0.1 \text{ m, } B = 3 \times 10^{-5} \text{ T}$ $\text{The magnetic flux induced in the coil is given by:}$ $\phi = BA \cos \omega t$ $\text{The EMF induced in the coil is given by:}$ $|\varepsilon| = \left| \frac{d\phi}{dt} \right| = |BA \omega \sin \omega t|$ $\text{The maximum value of the EMF induced in the coil is given by:}$ $|\varepsilon| = BA \omega = BA \left( \frac{2\pi}{T} \right)$ $\Rightarrow |\varepsilon| = 3 \times 10^{-5} \times \pi \times (0.1)^2 \left( \frac{2\pi}{0.4} \right) \quad [\pi^2 \approx 10]$ $\Rightarrow \varepsilon = 1.5 \times 10^{-5} \text{ V} = 15 \mu \text{V}$ $\text{Hence, option (3) is the correct answer.}$

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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
}