Import Question JSON

Current Question (ID: 16759)

Question:

$\text{Two identical long conducting wires } (AOB) \text{ and } (COD) \text{ are placed at a right angle to each other, with one above the other such that } 'O' \text{ is the common point for the two. The wires carry } I_1 \text{ and } I_2 \text{ currents, respectively. The point } 'P' \text{ is lying at a distance } 'd' \text{ from } 'O' \text{ along a direction perpendicular to the plane containing the wires. What will be the magnetic field at the point } P?$

Options:

1. $\frac{\mu_0}{2\pi d} \left( \frac{I_1}{I_2} \right)$
2. $\frac{\mu_0}{2\pi d} [I_1 + I_2]$
3. $\frac{\mu_0}{2\pi d} \left[ I_1^2 + I_2^2 \right]$
4. $\frac{\mu_0}{2\pi d} \sqrt{I_1^2 + I_2^2}$

Solution:

$\text{Hint: The magnetic field at the point } P \text{ will be the vector sum of the magnetic fields due to both wires.}$ $\text{Step 1: Find the magnetic field due to each wire.}$ $B_1 = \frac{\mu_0 I_1}{2\pi d} \hat{i}$ $B_2 = \frac{\mu_0 I_2}{2\pi d} \hat{j}$ $B_{\text{net}} = \sqrt{B_1^2 + B_2^2}$ $B_{\text{net}} = \sqrt{\left( \frac{\mu_0 I_1}{2\pi d} \right)^2 + \left( \frac{\mu_0 I_2}{2\pi d} \right)^2}$ $\Rightarrow B_{\text{net}} = \frac{\mu_0}{2\pi d} \sqrt{I_1^2 + I_2^2}$ $\text{Hence, option (4) is the correct answer.}$

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
}