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Current Question (ID: 17031)

Question:

$\text{A long solenoid of diameter } 0.1 \text{ m has } 2 \times 10^4 \text{ turns per meter. At the centre of the solenoid, a coil of } 100 \text{ turns and a radius of } 0.01 \text{ m is placed with its axis coinciding with the solenoid's axis. The current in the solenoid reduces at a constant rate from } 0 \text{ A to } 4 \text{ A in } 0.05 \text{ s. If the resistance of the coil is } 10\pi^2 \, \Omega, \text{ the total charge flowing through the coil during this time is:}$

Options:

1. $32\pi \, \mu\text{C}$
2. $16 \, \mu\text{C}$
3. $32 \, \mu\text{C}$
4. $16\pi \, \mu\text{C}$

Solution:

$\text{(c) Current induced in the coil given by}$ $i = \frac{1}{R} \left( \frac{d\phi}{dt} \right)$ $\Rightarrow \frac{\Delta q}{\Delta t} = \frac{1}{R} \left( \frac{\Delta \phi}{\Delta t} \right)$ $\text{Given, the resistance of the solenoid,}$ $R = 10\pi^2 \, \Omega$ $\text{The radius of the second coil } r = 10^{-2}$ $\Delta t = 0.05 \, \text{s}, \quad \Delta i = 4 - 0 = 4 \, \text{A}$ $\text{The charge flowing through the coil is given by}$ $\Delta q = \left( \frac{\Delta \phi}{\Delta t} \right) \frac{1}{R} \left( \Delta t \right)$ $= \mu_0 N_1 N_2 \pi r^2 \left( \frac{\Delta i}{\Delta t} \right) \frac{1}{R} \Delta t$ $= 4\pi \times 10^{-7} \times 2 \times 10^4 \times 100 \times \pi$ $\times (10^{-2})^2 \times \left( \frac{4}{0.05} \right) \times \frac{1}{10\pi^2} \times 0.05$ $= 32 \times 10^{-6} \text{C} = 32 \, \mu\text{C}$

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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
}