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Current Question (ID: 20489)

Question:

$\text{The figure shows a square loop } L \text{ with a side length of } 5 \text{ cm, which is connected to a network of resistances.}$ $\text{The entire setup is moving to the right with a constant speed of } 1 \text{ cm/s.}$ $\text{At a certain instant, a part of the loop } L \text{ is in a uniform magnetic field of } 1 \text{ T, perpendicular to the plane of the loop.}$ $\text{If the resistance of the loop is } 1.7 \, \Omega, \text{ the current in the loop at that instant will be close to:}$

Options:

1. $115 \, \mu\text{A}$
2. $150 \, \mu\text{A}$
3. $170 \, \mu\text{A}$
4. $60 \, \mu\text{A}$

Solution:

$\text{Hint: } I = \frac{Bvl}{R_{\text{Total}}}$ $\text{Step 1: Find the induced emf in the loop.}$ $\text{The induced emf in the loop is given by:}$ $\varepsilon = vBl$ $\Rightarrow \varepsilon = 1 \times 10^{-2} \times 1 \times 5 \times 10^{-2}$ $\Rightarrow \varepsilon = 5 \times 10^{-4} \, \text{V}$ $\text{Step 2: Find the total resistance of the loop.}$ $\text{The total resistance of the loop is the series combination of the resistance of the loop and Wheatstone Bridge as shown in the diagram below:}$ $R_{\text{Wheatstone}} = \frac{4 \times 2}{4 + 2} = \frac{8}{6} \approx 1.3 \, \Omega$ $R_{\text{Total}} = 1.7 + 1.3 = 3 \, \Omega$ $\text{Step 3: Find the current in the loop.}$ $\text{The total current in the loop is given by:}$ $I = \frac{\varepsilon}{R_{\text{Total}}}$ $\Rightarrow I = \frac{5 \times 10^{-4}}{3} = 167 \, \mu\text{A} \approx 170 \, \mu\text{A}$ $\text{Hence, option (3) is the correct answer.}$

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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
}