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Current Question (ID: 16750)

Question:

$\text{A straight wire carrying a current of } 12 \text{ A is bent into a semi-circular arc of radius } 2.0 \text{ cm as shown in the figure. Considering the magnetic field } B \text{ at the centre of the arc, what will be the magnetic field due to the straight segments?}$

Options:

1. $0$
2. $1.2 \times 10^{-4} \text{ T}$
3. $2.1 \times 10^{-4} \text{ T}$
4. $\text{None of these}$

Solution:

$\text{Hint: } d\vec{B} = \frac{\mu_0}{4\pi} \frac{I(d\vec{l} \times \vec{r})}{r^3}$ $\text{Explanation: Find the magnetic field due to the straight segments.}$ $\text{If we assume a small segment of wire } dl \text{ and find the magnetic field at the middle point of the semi-circular arc, we get;}$ $\Rightarrow d\vec{B} = \frac{\mu_0}{4\pi} \frac{I(d\vec{l} \times \vec{r})}{r^3}$ $d\vec{l} \text{ and } \vec{r} \text{ for each element of the straight segments are parallel. By the law of vectors, } I d\vec{l} \times \vec{r} = \vec{0}$ $\text{Therefore, } d\vec{B} = \frac{\mu_0}{4\pi} \frac{I(d\vec{l} \times \vec{r})}{r^3} \text{ so straight segments do not contribute to } \vec{B}.$ $\text{Hence, option (1) is the correct answer.}$

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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
}