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Current Question (ID: 16782)

Question:

$\text{When a charged particle with velocity } \vec{v} \text{ is subjected to an induction magnetic field } \vec{B}, \text{ the force on it is non-zero. What does this imply?}$

Options:

1. $\text{Angle between } \vec{v} \text{ and } \vec{B} \text{ is necessarily } 90^\circ.$
2. $\text{Angle between } \vec{v} \text{ and } \vec{B} \text{ can have any value other than } 90^\circ.$
3. $\text{Angle between } \vec{v} \text{ and } \vec{B} \text{ can have any value other than zero and } 180^\circ.$
4. $\text{Angle between } \vec{v} \text{ and } \vec{B} \text{ is either zero or } 180^\circ.$

Solution:

$\text{Hint: The magnetic force is zero when the charged particle's velocity is parallel or anti-parallel to the magnetic field.}$ $\text{Step 1: Find the formula of the magnetic force.}$ $\text{The magnetic force is given by,}$ $F = qvB \sin \theta.$ $\text{Step 2: Find the magnetic force for } \theta = 0^\circ \text{ or } 180^\circ.$ $\text{If, } \theta = 0^\circ \text{ or } 180^\circ, \text{ then } \sin \theta = 0$ $\therefore F = qvB \sin \theta = 0$ $\text{Since the force on a charged particle is non-zero, so the angle between } \vec{v} \text{ and } \vec{B} \text{ can have any value other than zero and } 180^\circ.$ $\text{Note: The force experienced by the charged particle is Lorentz force.}$

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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
}