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Current Question (ID: 16756)

Question:

$\text{Two similar coils of radius } R \text{ are lying concentrically with their planes at right angles to each other. The currents flowing in them are } I \text{ and } 2I, \text{ respectively. What will be the resultant magnetic field induction at the centre?}$

Options:

1. $\frac{\sqrt{5} \mu_0 I}{2R}$
2. $\frac{3 \mu_0 I}{2R}$
3. $\frac{\mu_0 I}{2R}$
4. $\frac{\mu_0 I}{R}$

Solution:

$\text{The magnetic field due to the first coil is } B_1 = \frac{\mu_0 I}{2R}.$ $\text{The magnetic field due to the second coil is } B_2 = \frac{\mu_0 (2I)}{2R} = \frac{\mu_0 I}{R}.$ $\text{Since the coils are at right angles, the resultant magnetic field is } B = \sqrt{B_1^2 + B_2^2} = \sqrt{\left(\frac{\mu_0 I}{2R}\right)^2 + \left(\frac{\mu_0 I}{R}\right)^2} = \frac{\sqrt{5} \mu_0 I}{2R}.$

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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
}