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Current Question (ID: 16814)

Question:

$\text{A rectangular loop carrying a current } I_1, \text{ is situated near a long straight wire carrying a steady current } I_2. \text{ If the wire is parallel to one of the sides of the loop and is in the plane of the loop as shown in the figure, then the current loop will:}$

Options:

1. $\text{move away from the wire.}$
2. $\text{move towards the wire.}$
3. $\text{remain stationary.}$
4. $\text{rotate about an axis parallel to the wire.}$

Solution:

$\text{Hint: } \vec{F} = i(\vec{l} \times \vec{B}) = ilB$ $\text{Step: Find the magnetic forces on each part of the square loop.}$ $\text{The magnetic field is due to the straight wire is given by:}$ $B = \frac{\mu_0 i}{2 \pi r}$ $\text{The magnetic force on the loop is given by:}$ $\vec{F} = i(\vec{l} \times \vec{B}) = ilB$ $\text{Here force on } DC \text{ and } BA \text{ parts are equal and opposite so its resultant will be zero.}$ $\text{Now as } AD \text{ is closer to the straight wire than } CB \text{ so the magnitude of the force on } AD \text{ will be more than that on } CD.$ $i.e., F_1 > F_2$ $\text{Therefore the net force on the loop will be towards the straight wire.}$ $\text{Hence, option (2) is the correct answer.}$

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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
}