Import Question JSON

Current Question (ID: 16985)

Question:

$\text{Consider the situation shown in the figure. The wire } AB \text{ is sliding on the fixed rails with a constant velocity.}$ $\text{If the wire } AB \text{ is replaced by a semicircular wire, the magnitude of the induced current will:}$

Options:

1. $\text{increase.}$
2. $\text{remain the same.}$
3. $\text{decrease.}$
4. $\text{increase or decrease depending on whether the semicircle bulges towards the resistance or away from it.}$

Solution:

$\text{Hint: The effective length between } A \text{ and } B \text{ remains the same.}$ $\text{Explanation: For a semicircular wire: Even though the wire has a different shape, the way it sweeps through the magnetic field still leads to the same rate of change of the magnetic flux.}$ $\text{The changing area, in this case, is still proportional to the motion of the wire, and since the velocity of motion is constant, the rate of change of the magnetic flux remains the same.}$ $\text{The induced current is given by } I = \frac{\varepsilon}{R} , \text{ where } \varepsilon \text{ is the induced EMF and } R \text{ is the resistance.}$ $\text{Since the rate of change of the magnetic flux (and thus the EMF) remains the same and the resistance doesn't change significantly between the straight and semicircular wire, the magnitude of the induced current remains the same.}$ $\text{Hence, option (2) is the correct answer.}$

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
}