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Current Question (ID: 20347)

Question:

$\text{A thin ring of } 10 \text{ cm radius carries a uniformly distributed charge.}$ $\text{The ring rotates at a constant angular speed of } 40\pi \text{ rad s}^{-1} \text{ about}$ $\text{its axis, perpendicular to its plane. If the magnetic field at its centre is}$ $3.8 \times 10^{-9} \text{ T, then the charge carried by the ring is close to:}$ $\left( \mu_0 = 4\pi \times 10^{-7} \text{ N/A}^2 \right)$

Options:

1. $7 \times 10^{-6} \text{ C}$
2. $4 \times 10^{-5} \text{ C}$
3. $2 \times 10^{-6} \text{ C}$
4. $3 \times 10^{-5} \text{ C}$

Solution:

$\text{Magnetic field at the centre of the ring} = \frac{\mu_0 I}{2R}$ $R = 10 \text{ cm} = 10^{-1} \text{ m}$ $\omega = 40 \pi \text{ rad/s}$ $B_0 = 3.8 \times 10^{-9} \text{ T}$ $I = \frac{\Delta Q}{\Delta t} = \frac{Q}{\left[ \frac{2\pi}{\omega} \right]} = \mu_0 \left[ \frac{Q}{\left( \frac{2\pi}{\omega} \right)} \right] \frac{1}{2R}$ $B_0 = \frac{\mu_0 I}{2R} = \frac{\mu_0 Q \omega}{4\pi R}$ $B_0 4\pi R = \mu_0 \omega Q$ $Q = \frac{B_0 4\pi R}{\mu_0 \omega}$ $= \frac{(3.8 \times 10^{-9}) (4\pi) (10^{-1})}{(4\pi \times 10^{-7}) (40\pi)}$ $= \frac{380 \times 10^{-2} \times 10^{-9} \times 10^{-1}}{40 \times 3.14 \times 10^{-7}}$ $= \frac{380}{125.6} \times 10^{-12+7} = 3 \times 10^{-5} \text{ C}$

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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
}