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Current Question (ID: 20460)

Question:

$\text{Two magnetic dipoles, } X \text{ and } Y, \text{ are separated by a distance } d, \text{ with their axes oriented perpendicular to each other.}$ $\text{The dipole moment of } Y \text{ is twice that of } X. \text{A charged particle with charge } q \text{ moves with velocity } v \text{ through their midpoint } P, \text{ which makes an angle } \theta = 45^\circ \text{ with the horizontal axis, as shown in the diagram.}$ $\text{Assuming } d \text{ is much larger than the dimensions of the dipoles, the magnitude of the force acting on the charged particle at this instant is:}$

Options:

2. $\left( \frac{\mu_0}{4\pi} \right) \left( \frac{M}{\left( \frac{d}{2} \right)^3} \right) \times qv$
3. $\sqrt{2} \left( \frac{\mu_0}{4\pi} \right) \left( \frac{M}{\left( \frac{d}{2} \right)^3} \right) \times qv$
4. $\left( \frac{\mu_0}{4\pi} \right) \left( \frac{2M}{\left( \frac{d}{2} \right)^3} \right) \times qv$

Solution:

$\text{Hint: } B = \frac{\mu_0}{4\pi} \frac{2M}{r^3}$ $\text{Step: Find the magnetic field due to both dipoles.}$ $\text{The magnetic field due to both dipoles is shown in the figure below;}$ $B_X = \frac{\mu_0}{4\pi} \frac{2M}{r^3} \hat{i} \quad \left[ r = \frac{d}{2} \right]$ $B_Y = \frac{\mu_0}{4\pi} \frac{2M}{r^3} \hat{j} \quad \left[ r = \frac{d}{2} \right]$ $B_{\text{net}} = \sqrt{B_X^2 + B_Y^2}$ $\text{Step 2: Find the magnitude of the force acting on the charged particle at this instant.}$ $F = q(\vec{v} \times \vec{B})$ $\text{As the velocity vector is along the magnetic field, then the net magnetic force acting on the charged particle is zero.}$ $\text{Hence, option (1) is the correct answer.}$

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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
}