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Current Question (ID: 20373)

Question:

$\text{A wire is first coiled in } n \text{ circular turns and current } I \text{ is run through it.}$ $\text{Now the same wire is coiled in } N \text{ circular turns and the same current } I \text{ is run through it.}$ $\text{If } B_1 \text{ and } B_2 \text{ are the magnetic field at centre of two coils respectively then } \frac{B_1}{B_2} \text{ is equal to:}$

Options:

1. $\sqrt{\frac{n}{N}}$
2. $\left(\frac{n}{N}\right)^2$
3. $\frac{n}{N}$
4. $\frac{n^3}{N^3}$

Solution:

$\text{Let the length of wire is } l,$ $\text{Radius of the first coil } R_1 = \frac{l}{2\pi n}$ $\text{Radius of the second coil } R_2 = \frac{l}{2\pi N}$ $B_1 = \frac{\mu_0 n I}{2 R_1} = \frac{\mu_0 n I}{\frac{2l}{2\pi n}} = \frac{\mu_0 \pi n^2 I}{l}$ $B_2 = \frac{\mu_0 N I}{2 R_2} = \frac{\mu_0 N I}{\frac{2l}{2\pi N}} = \frac{\mu_0 \pi N^2 I}{l}$ $\frac{B_1}{B_2} = \left(\frac{n}{N}\right)^2$

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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
}