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Current Question (ID: 16969)

Question:

$\text{A conducting wireframe is placed in a magnetic field that is directed into the paper.}$ $\text{The magnetic field is increasing at a constant rate.}$ $\text{The directions of induced current in wires } AB \text{ and } CD \text{ are:}$

Options:

1. $B \text{ to } A \text{ and } D \text{ to } C$
2. $A \text{ to } B \text{ and } C \text{ to } D$
3. $A \text{ to } B \text{ and } D \text{ to } C$
4. $B \text{ to } A \text{ and } C \text{ to } D$

Solution:

$\text{(1) Inward magnetic field (x) increasing.}$ $\text{Therefore, induced current in both the loops should be anticlockwise.}$ $\text{But as the area of loop on right side is more, induced emf in this will be more compared to the left side loop}$ $\left(e = -\frac{d\phi}{dt} = -A \cdot \frac{dB}{dt}\right).$ $\text{Therefore, net current in the complete loop will be in a direction shown below.}$ $\text{Hence only option (1) is correct.}$

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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
}