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Current Question (ID: 16767)

Question:

$\text{Two circular coils 1 and 2 are made from the same wire but the radius of the 1}^{\text{st}} \text{ coil is twice that of the 2}^{\text{nd}} \text{ coil. What is the ratio of the potential difference applied across them so that the magnetic field at their centres is the same?}$ $1.\ 3$ $2.\ 4$ $3.\ 6$ $4.\ 2$

Options:

1. $3$
2. $4$
3. $6$
4. $2$

Solution:

$\text{Hint: The magnetic field at the centre of the coil is given by, } B = \frac{\mu_0}{4\pi} \times \frac{2\pi i}{r}. $ $\text{Step 1: Put the magnetic field at the centre of the coils equal to each other.}$ $\text{As } B_1 = B_2, \frac{\mu_0}{4\pi} \frac{2\pi i_1}{r_1} = \frac{\mu_0}{4\pi} \frac{2\pi i_2}{r_2} \Rightarrow \frac{i_1}{r_1} = \frac{i_2}{r_2}.$ $\text{As } r_1 = 2r_2, \frac{i_1}{2r_2} = \frac{i_2}{r_2} \Rightarrow i_1 = 2i_2.$ $\text{Step 2: Find the ratio of the potential difference.}$ $\text{Now the ratio of the potential differences,}$ $\frac{V_2}{V_1} = \frac{i_2 R_2}{i_1 R_1} = \frac{i_2 r_2}{i_1 r_1} = \frac{i_2 r_2}{2i_2 \times 2r_2} = \frac{1}{4}$ $\text{(Both the coils are made of the same wire with the same cross-sectional area)}$

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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
}