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Current Question (ID: 16762)

Question:

$\text{A circular coil is in the } y-z \text{ plane with its centre at the origin. The coil carries a constant current. Assuming the direction of the magnetic field at } x = -25 \text{ cm to be positive, which of the following graphs shows the variation of the magnetic field along the } x\text{-axis?}$

Options:

1. $\text{Graph 1}$
2. $\text{Graph 2}$
3. $\text{Graph 3}$
4. $\text{Graph 4}$

Solution:

$\text{Hint: } B = \frac{\mu_0 NIR^2}{2(R^2 + x^2)^{3/2}}$ $\text{Explanation: The direction of the magnetic field at every point on the axis of a current-carrying coil remains the same though the magnitude varies.}$ $\text{Hence, magnetic induction for the whole } x\text{-axis will remain positive.}$ $\text{Therefore, (3) and (4) are wrong.}$ $\text{The magnitude of the magnetic field will vary with } x \text{ according to the law is given by;}$ $B = \frac{\mu_0 NIR^2}{2(R^2 + x^2)^{3/2}}$ $\text{Hence, at } x = 0,$ $B = \frac{\mu_0 NI}{2R} \text{ and when } x \rightarrow \infty, B \rightarrow 0.$ $\text{The slope of the graph will be;}$ $\frac{dB}{dx} = \frac{-3\mu_0 NIR^2 \cdot x}{2(R^2 + x^2)^{5/2}}$ $\text{It means, that at } x = 0, \text{ the slope is equal to zero or tangent to the graph at } x = 0, \text{ and must be parallel to the } x\text{-axis.}$ $\text{Therefore, (2) is correct and (1) is wrong.}$ $\text{Hence, option (2) is the correct answer.}$

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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
}