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Current Question (ID: 16822)

Question:

$\text{If a charged particle (charge } q \text{) is moving in a circle of radius } R \text{ at a uniform speed } v, \text{ then the value of its associated magnetic moment } \mu \text{ will be:}$

Options:

1. $\frac{qvR}{2}$
2. $qvR^2$
3. $\frac{qvR^2}{2}$
4. $qvR$

Solution:

$\text{Hint: The magnetic moment is given by, } \mu = I A.$ $\text{Step 1: Find the current produced by the particle.}$ $\text{As the revolving charge is equivalent to a current,}$ $I = qxf \times \frac{\omega}{2\pi}$ $\text{But } \omega = \frac{v}{R}$ $\text{Therefore, } I = \frac{qv}{2\pi R}$ $\text{Step 2: Find the magnetic moment.}$ $\text{Now, the magnetic moment is given by,}$ $\mu = I A = I \times \pi R^2$ $\text{Or } \mu = \frac{qv}{2\pi R} \times \pi R^2 = \frac{qvR}{2}$

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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
}