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Current Question (ID: 20344)

Question:

$\text{A rigid square loop of side } a, \text{ carrying a current } I_2, \text{ lies on a horizontal surface near a long, straight wire carrying a current } I_1. \text{ Both the loop and the wire are in the same plane, as shown in the figure. The net force acting on the loop due to the current in the wire is:}$

Options:

1. $\text{zero}$
2. $\text{repulsive and equal to } \frac{\mu_0 I_1 I_2}{4\pi}$
3. $\text{repulsive and equal to } \frac{\mu_0 I_1 I_2}{2\pi}$
4. $\text{attractive and equal to } \frac{\mu_0 I_1 I_2}{3\pi}$

Solution:

$\text{Hint: } \frac{F}{l} = \frac{\mu_0 I_1 I_2}{2\pi r}$ $\text{Step: Find the net force acting on the loop due to the current in the wire.}$ $\text{The force on the } AB \text{ wire is given by:}$ $F_1 = I_2 B_1 a$ $\Rightarrow F_1 = I_2 \left( \frac{\mu_0 I_1}{2\pi a} \right) a$ $\text{The force on the } CD \text{ wire is given by:}$ $F_2 = I_2 B_2 a$ $\Rightarrow F_2 = I_2 \left( \frac{\mu_0 I_1}{2\pi (2a)} \right) a$ $\text{The net force acting on the wire is given by:}$ $F_{\text{net}} = F_1 - F_2$ $\Rightarrow F_{\text{net}} = \frac{\mu_0 I_1 I_2}{2\pi} - \frac{\mu_0 I_1 I_2}{4\pi}$ $\Rightarrow F_{\text{net}} = \frac{\mu_0 I_1 I_2}{4\pi}$ $\text{As the direction of the current in the wires is opposite.}$ $\text{Therefore, the force between the wires is repulsive and equal to } \frac{\mu_0 I_1 I_2}{4\pi}. \text{Hence, option (2) is the correct answer.}$

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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
}