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Current Question (ID: 16998)

Question:

$\text{A square loop of side } 5 \text{ cm enters a magnetic field with } 1 \text{ cm s}^{-1}. \text{ If the front edge enters the magnetic field at } t = 0, \text{ then which graph best depicts emf?}$

Options:

1. $\begin{array}{c} V \\ 3 \times 10^{-4} \\ 0 \\ 5 \\ 15 \\ 20 \\ t (s) \end{array}$
2. $\begin{array}{c} V \\ 3 \times 10^{-4} \\ 0 \\ 5 \\ 15 \\ 20 \\ t (s) \end{array}$
3. $\begin{array}{c} V \\ 3 \times 10^{-4} \\ 0 \\ 5 \\ 15 \\ 20 \\ t (s) \end{array}$
4. $\begin{array}{c} V \\ 3 \times 10^{-4} \\ 0 \\ 5 \\ 15 \\ 20 \\ t (s) \end{array}$

Solution:

$\text{(3) When the loop is entering the field, the magnetic flux (i.e. } x \text{) linked with the loop increases, so induced emf in it } e = -Bvl = -0.6 \times 10^{-2} \times 5 \times 10^{-2} = -3 \times 10^{-4} V \text{ (negative).}$ $\text{When the loop completely entered the field (after 5 sec), the flux linked with the loop remains constant, so } e = 0.$ $\text{After 15 sec, the loop begins to exit out, the linked magnetic flux decreases, so induced emf, } e = +3 \times 10^{-4} V \text{ (Positive).}$

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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
}