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Current Question (ID: 16807)

Question:

$\text{A long straight wire of radius } a \text{ carries a steady current } I. \text{ The current is uniformly distributed over its cross-section. The ratio of the magnetic fields } B \text{ and } B' \text{ at radial distances } \frac{a}{2} \text{ and } 2a \text{ respectively, from the axis of the wire, is:}$

Options:

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

Solution:

$\text{Consider two Amperian loops of radius } \frac{a}{2} \text{ and } 2a \text{ as shown in the diagram.}$ $\text{Applying Ampere's circuital law for these loops we get,}$ $\oint B \cdot dL = \mu_0 I_{\text{enclosed}}$ $\text{For the smaller loop}$ $\Rightarrow B \times 2\pi \frac{a}{2} = \mu_0 \times \frac{I}{\pi a^2} \times \pi \left(\frac{a}{2}\right)^2$ $= \mu_0 I \times \frac{1}{4} = \frac{\mu_0 I}{4}$ $\Rightarrow B_1 = \frac{\mu_0 I}{4\pi a}, \text{ at distance } \frac{a}{2} \text{ from the axis of the wire.}$ $\text{Similarly, for bigger Amperian loop.}$ $B' \times 2\pi (2a) = \mu_0 I$ $\text{(total current enclosed by Amperian loop is 2)}$ $\Rightarrow B' = \frac{\mu_0 I}{4\pi a}$ $\text{at distance } 2a \text{ from the axis of the wire.}$ $\text{So, ratio of, } \frac{B}{B'} = \frac{\mu_0 I}{4\pi a} \times \frac{4\pi a}{\mu_0 I} = 1$

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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
}