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Current Question (ID: 20365)

Question:

$\text{An infinitely long hollow conducting cylinder with radius } R \text{ carries a uniform current along its surface. The correct representation of magnetic field } (B) \text{ as a function of radial distance } (r) \text{ from the axis of the cylinder is:}$

Options:

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

Solution:

$\text{Hint:}$ $r < R, B_p = 0$ $r \geq R, B_p \propto \frac{1}{r}$ $\text{Step 1: Find the magnetic field inside the cylinder.}$ $\text{For a point inside the wire, the enclosed current within the Amperian loop is zero due to the uniform distribution of current along the wire's surface.}$ $B(2\pi r) = \mu_0 \times 0$ $\Rightarrow B = 0$ $\text{The magnetic field inside the cylinder is zero.}$ $\text{Step 2: Find the magnetic field outside the cable.}$ $\text{For an outside point, the current outside the ampere loop is } I.$ $B(2\pi r) = \mu_0 I$ $\Rightarrow B = \frac{\mu_0 I}{2\pi r}$ $\Rightarrow B \propto \frac{1}{r}$ $\text{The magnetic field is inversely proportional to the distance from outside the cylinder.}$ $\text{Therefore, the correct variation of the magnetic field as a function of radial distance } (r) \text{ is shown in the figure below;}$ $\text{Graph 4}$ $\text{Hence, option (4) is the correct answer.}$

Import JSON File

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
}