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Current Question (ID: 16788)

Question:

$\text{A particle of charge } q \text{ and mass } m \text{ is moving along the } x\text{-axis with a velocity of } v \text{ and enters a region of electric field } E \text{ and magnetic field } B \text{ as shown in the figure below. For which figure is the net force on the charge zero?}$

Options:

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

Solution:

$\text{Hint: } E = vB$ $\text{Step: Find the figure for which force is zero.}$ $\text{Forces Acting on the Charge:}$ $\text{Electric Force } E = q(v \times B)$ $\text{The direction of } E \text{ is given by the Right-Hand Rule.}$ $\text{Point your fingers in the direction of velocity } (v). \text{ Curl them towards the magnetic field } (B).$ $\text{Your thumb shows the direction of the magnetic force (for a positive charge).}$ $\text{For a negative charge, the force direction is opposite.}$ $\text{For the net force to be zero, the electric force and magnetic force must be equal in magnitude and opposite in direction:}$ $qE = qvB \Rightarrow E = vB$ $\text{Thus, the correct figure will show } E \text{ and } B \text{ perpendicular to each other, such that the electric force cancels out the magnetic force. Therefore, figure } (2) \text{ has net force zero.}$ $\text{Hence, option } (2) \text{ is the correct answer.}$

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Upload a JSON file containing LaTeX/MathJax formatted question, options, and solution.

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
}