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Current Question (ID: 20458)

Question:

$\text{A rectangular loop with sides } 10 \text{ cm, carrying a current } I = 12 \text{ A, is placed in various orientations as shown in the figures. The loop is subjected to a uniform magnetic field of } 0.3 \text{ T in the positive } z\text{-direction.}$ $\text{In which orientations is the loop in (i) stable equilibrium and (ii) unstable equilibrium?}$

Options:

1. $(a) \text{ and } (b), \text{ respectively}$
2. $(a) \text{ and } (c), \text{ respectively}$
3. $(b) \text{ and } (d), \text{ respectively}$
4. $(b) \text{ and } (c), \text{ respectively}$

Solution:

$\text{Hint: } U = \vec{M} \cdot \vec{B}$ $\text{(a) } \vec{M} = M \hat{i}, \quad \vec{B} = B \hat{k}$ $\therefore \vec{M} \perp \vec{B}$ $\therefore \text{Torque } \neq 0$ $\therefore \text{It is not in equilibrium position}$ $\text{(b) } \vec{M} = M \hat{k}, \quad \vec{B} = B \hat{k}$ $\therefore \vec{M} \parallel \vec{B}$ $\therefore U = -MB \cos 0^\circ = -MN \text{ (min. potential energy)}$ $\therefore \text{loop will be in stable equilibrium}$ $\text{(c) } \vec{M} = M (-\hat{j}), \quad \vec{B} = B \hat{k}$ $\therefore \vec{M} \perp \vec{B}$ $\therefore \text{Torque } \neq 0$ $\therefore \text{It is not in equilibrium position}$ $\text{(d) } \vec{M} = M (-\hat{k}), \quad \vec{B} = B \hat{k}$ $\therefore U = -MB \cos 180^\circ = +MB \text{ (max. potential energy)}$ $\therefore \text{It is in unstable equilibrium}$

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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
}