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Current Question (ID: 20633)

Question:

$\text{The magnetic field of an electromagnetic wave is given by:}$ $\vec{B} = 1.6 \times 10^{-6} \cos \left(2 \times 10^{-7}z + 6 \times 10^{15}t \right) \left(2\hat{i} + \hat{j} \right) \text{ Wb/m}^2$ $\text{The associated electric field will be:}$

Options:

1. $\vec{E} = 4.8 \times 10^2 \cos \left(2 \times 10^7z + 6 \times 10^{15}t \right) \left(-\hat{i} + 2\hat{j} \right) \text{ V/m}$
2. $\vec{E} = 4.8 \times 10^2 \cos \left(2 \times 10^7z - 6 \times 10^{15}t \right) \left(2\hat{i} + 2\hat{j} \right) \text{ V/m}$
3. $\vec{E} = 4.8 \times 10^2 \cos \left(2 \times 10^7z - 6 \times 10^{15}t \right) \left(\hat{i} - 2\hat{j} \right) \text{ V/m}$
4. $\vec{E} = 4.8 \times 10^2 \cos \left(2 \times 10^7z - 6 \times 10^{15}t \right) \left(-2\hat{i} + \hat{j} \right) \text{ V/m}$

Solution:

$\text{Hint: } E_0 = cB_0$ $E_0 = cB_0$ $= 3 \times 10^8 \times 1.6 \times 10^{-6}$ $= 4.8 \times 10^2$

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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
}