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Current Question (ID: 17399)

Question:

$\oint \vec{B} \cdot \vec{dl} = \mu_0 \left( i + \varepsilon_0 \cdot \frac{d(\phi)E}{dt} \right)$ $\text{is a statement of:}$ $1. \text{Faraday's law of induction}$ $2. \text{Modified Ampere's law}$ $3. \text{Gauss's law of electricity}$ $4. \text{Gauss's law of magnetism}$

Options:

1. $\text{Faraday's law of induction}$
2. $\text{Modified Ampere's law}$
3. $\text{Gauss's law of electricity}$
4. $\text{Gauss's law of magnetism}$

Solution:

$\text{Hint: The given Maxwell's equation is the modified Ampère's law, which includes the displacement current term to account for changing electric fields.}$ $\text{Step: Identify Maxwell's equation.}$ $\text{The given Maxwell's equation:}$ $\oint \vec{B} \cdot \vec{dl} = \mu_0 \left( i + \varepsilon_0 \frac{d\phi_E}{dt} \right)$ $\text{This equation is a form of Ampère's law modified by Maxwell to include the displacement current term } (\varepsilon_0 \frac{d\phi_E}{dt}), \text{ which accounts for the changing electric field.}$ $\text{The term } i \text{ represents the conduction current. The term } \varepsilon_0 \frac{d\phi_E}{dt} \text{ represents the displacement current, which arises due to a time-varying electric field.}$ $\text{This modification was introduced by Maxwell to complete the symmetry in Maxwell's equations and ensure consistency with the continuity of current.}$ $\text{Therefore, the equation is a modified form of Ampère's law.}$ $\text{Hence, option (2) is the correct answer.}$

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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
}