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Current Question (ID: 16761)

Question:

$\text{A ring of radius } R \text{ carries a linear charge density } \lambda. \text{ It is rotating with angular}$ $\text{speed } \omega \text{ about an axis passing through the centre and perpendicular to the}$ $\text{plane. What is the magnetic field at its centre?}$

Options:

1. $\frac{3\mu_0\lambda\omega}{2}$
2. $\frac{\mu_0\lambda\omega}{2}$
3. $\frac{\mu_0\lambda\omega}{\pi}$
4. $\mu_0\lambda\omega$

Solution:

$\text{Hint: The current in the loop = Q/T.}$ $\text{Step 1: Find the current in the loop.}$ $Q = \lambda \times 2\pi R$ $i_{\text{loop}} = \frac{Q}{T} = \frac{\lambda \times 2\pi R}{T} = \lambda \omega R$ $\text{Step 2: Find the magnetic field at the centre.}$ $B_C = \frac{\mu_0}{2} \times \frac{i_{\text{loop}}}{R} = \frac{\mu_0}{2} \times \frac{\lambda \omega R}{R}$

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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
}