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Magnetics question

2024 · 29 Jan · Shift 2 · Q89
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  5. /2024 · 29 Jan · Shift 2 · Q89

Magnetics question

2024 · 29 Jan · Shift 2 · Q89

JEE MainPhysicsMagneticsNumerical+4 / −1
A charge of 4.0μC4.0 \mu \mathrm{C}4.0μC is moving with a velocity of 4.0×106 ms−14.0 \times 10^6 \mathrm{~ms}^{-1}4.0×106 ms−1 along the positive yyy axis under a magnetic field B⃗\vec{B}B of strength (2k^)T(2 \hat{k}) \mathrm{T}(2k^)T. The force acting on the charge is xi^Nx \hat{i} Nxi^N. The value of xxx is ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 32

  1. Given data
  • Charge: q=4.0 μC=4.0×10−6 Cq = 4.0\,\mu\text{C} = 4.0 \times 10^{-6}\,\text{C}q=4.0μC=4.0×10−6C
  • Velocity: v⃗=4.0×106 j^ m s−1\vec v = 4.0 \times 10^6 \, \hat{j}\,\text{m s}^{-1}v=4.0×106j^​m s−1
  • Magnetic field: B⃗=2k^ T\vec B = 2\hat{k}\,\text{T}B=2k^T

We need the magnetic force:

F⃗=q(v⃗×B⃗)\vec F = q(\vec v \times \vec B)F=q(v×B)
  1. Find the cross product
v⃗×B⃗=(4.0×106j^)×(2k^)\vec v \times \vec B = (4.0 \times 10^6\hat{j}) \times (2\hat{k})v×B=(4.0×106j^​)×(2k^)

Taking constants out:

v⃗×B⃗=8.0×106(j^×k^)\vec v \times \vec B = 8.0 \times 10^6 (\hat{j} \times \hat{k})v×B=8.0×106(j^​×k^)

Using unit vector relation:

j^×k^=i^\hat{j} \times \hat{k} = \hat{i}j^​×k^=i^

So,

v⃗×B⃗=8.0×106i^\vec v \times \vec B = 8.0 \times 10^6 \hat{i}v×B=8.0×106i^
  1. Multiply by charge
F⃗=q(v⃗×B⃗)\vec F = q(\vec v \times \vec B)F=q(v×B) F⃗=(4.0×10−6)(8.0×106)i^\vec F = (4.0 \times 10^{-6})(8.0 \times 10^6)\hat{i}F=(4.0×10−6)(8.0×106)i^ F⃗=32i^ N\vec F = 32\hat{i}\,\text{N}F=32i^N
  1. Compare with the given form

Given that the force is xi^ Nx\hat{i}\,\text{N}xi^N, we get

x=32x = 32x=32
  1. Comparison with stored correct answer

Stored correct answer = 323232

This matches the derived answer.

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