Import Question JSON

« Back to Question List Edit Question »

Current Question (ID: 20613)

Question:
$\text{The electric field of a plane electromagnetic wave propagating, along the } x \text{ direction is vacuum is;} \quad \vec{E} = E_0 \cos(\omega t - kx) \hat{j}. \text{ The magnetic field } \vec{B} \text{ at the moment } t = 0 \text{ s is:}$
Options:
  • 1. $\vec{B} = \frac{E_0}{\sqrt{\mu_0 \epsilon_0}} \cos(kx) \hat{j}$
  • 2. $\vec{B} = E_0 \sqrt{\mu_0 \epsilon_0} \cos(kx) \hat{j}$
  • 3. $\vec{B} = \frac{E_0}{\sqrt{\mu_0 \epsilon_0}} \cos(kx) \hat{k}$
  • 4. $\vec{B} = E_0 \sqrt{\mu_0 \epsilon_0} \cos(kx) \hat{k}$
Solution:
$\text{Hint: } B_0 = \frac{E_0}{c}$ $\text{Step: Find the magnetic field } \vec{B} \text{ at the moment } t = 0 \text{ s.}$ $\text{For a plane electromagnetic wave propagating in the } x \text{ direction with an electric field is given by;} \Rightarrow \vec{E} = E_0 \cos(\omega t - kx) \hat{j},$ $\text{the magnetic field } \vec{B} \text{ is related to the electric field } \vec{E} \text{ using the following properties of electromagnetic waves:}$ $\text{The electric and magnetic fields are perpendicular to each other and to the direction of propagation.}$ $\text{The magnitude of the magnetic field is related to the electric field is given by the equation;} \Rightarrow B_0 = \frac{E_0}{c} = \frac{E_0}{\sqrt{\mu_0 \epsilon_0}} \text{ where } c \text{ is the speed of light in vacuum.}$ $\text{At } t = 0, \text{ the electric field is given by;} \Rightarrow \vec{E} = E_0 \cos(kx) \hat{j}$ $\text{For the magnetic field } \vec{B} \text{ since the wave propagates in the } x \text{ direction and the electric field is along the } y\text{-axis, the magnetic field will be along the } z\text{-axis } (\hat{k}), \text{ and its magnitude will be:}$ $\Rightarrow \vec{B} = \frac{E_0}{\sqrt{\mu_0 \epsilon_0}} \cos(kx) \hat{k}$ $\text{Hence, option (3) is the correct answer.}$

Import JSON File

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
}