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Current Question (ID: 16760)

Question:

$\text{If the magnetic field at the centre of the circular coil is } B_0, \text{ then what is the distance on its axis from the centre of the coil where } B_x = \frac{B_0}{8}?$ $\left(R = \text{radius of the coil}\right)$

Options:

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

Solution:

$\text{Hint: } B_{\text{axis}} = \frac{\mu_0}{2} \frac{iR^2}{(x^2+R^2)^{3/2}}$ $\text{Step: Find the magnetic field on the axis of the coil.}$ $\text{The magnetic field on the axis of the coil is given by:}$ $B_{\text{axis}} = \frac{\mu_0}{2} \frac{iR^2}{(x^2+R^2)^{3/2}}$ $B_{\text{axis}} = \frac{B_0}{8} = \frac{\mu_0 i R^2}{2[x^2+R^2]^{3/2}}$ $\text{The magnetic field at the centre of the circular coil is given by:}$ $B_C = \frac{\mu_0 i}{2R}$ $\frac{\mu_0 i}{2R} \frac{1}{8} = \frac{\mu_0 i R^2}{2[x^2+R^2]^{3/2}}$ $8R^3 = [x^2 + R^2]^{3/2} 2R = (x^2 + R^2)^{1/2}$ $4R^2 = x^2 + R^2$ $\Rightarrow x = \sqrt{3}R$ $\text{Hence, option (2) is the correct answer.}$

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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
}