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Current Question (ID: 20492)

Question:

$\text{An elliptical loop with resistance } R, \text{ semi-major axis } a, \text{ and semi-minor axis } b \text{ is placed in a magnetic field as shown in the figure. If the loop is rotated about the } x\text{-axis with an angular frequency } \omega, \text{ the average power loss in the loop due to Joule heating is:}$

Options:

1. $\text{zero}$
2. $\frac{\pi^2 a^2 b^2 B^2 \omega^2}{R}$
3. $\frac{\pi^2 a^2 b^2 B^2 \omega^2}{2R}$
4. $\frac{\pi a b B \omega}{R}$

Solution:

$\text{Hint: } P_{\text{avg}} = \left\langle \frac{E^2}{R} \right\rangle$ $\text{Step 1: Find the induced emf in the elliptical loop.}$ $\text{The emf induced in the loop is given by:}$ $\varepsilon = -\frac{d\phi}{dt}$ $\Rightarrow \varepsilon = -\frac{d(BA \cos \omega t)}{dt}$ $\Rightarrow \varepsilon = -BA \omega (-\sin \omega t) = B \pi a b \omega \sin \omega t$ $\text{Step 2: Find the average power loss in the loop.}$ $P_{\text{avg}} = V_{\text{rms}} I_{\text{rms}} \cos \phi$ $\Rightarrow P_{\text{avg}} = \frac{V_{\text{rms}}^2}{R} \left[ \cos \phi = 1 \right]$ $\Rightarrow P_{\text{avg}} = \frac{(B \pi a b \omega)^2}{2R}$ $\Rightarrow P_{\text{avg}} = \frac{B^2 \pi^2 a^2 b^2 \omega^2}{2R}$ $\text{Hence, option (3) is the correct answer.}$

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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
}