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Magnetics question

2020 · 3 Sep · Shift 1 · Q61
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Magnetics question

2020 · 3 Sep · Shift 1 · Q61

JEE MainPhysicsMagneticsMCQ+4 / −1
A charged particle carrying charge 1 μ\muμ C is moving with velocity (2i^+3j^+4k^)\left( {2\widehat i + 3\widehat j + 4\widehat k} \right)(2i+3j​+4k) ms–1. If an external magnetic field of (5i^+3j^−6k^)\left( {5\widehat i + 3\widehat j - 6\widehat k} \right)(5i+3j​−6k)× 10–3 T exists in the region where the particle is moving then the force on the particle is F→\overrightarrow FF× 10–9 N. The vector F→\overrightarrow FF is :
  1. A
    −0.30i^+0.32j^−0.09k^{ - 0.30\widehat i + 0.32\widehat j - 0.09\widehat k}−0.30i+0.32j​−0.09k
  2. B
    −300i^+320j^−90k^{ - 300\widehat i + 320\widehat j - 90\widehat k}−300i+320j​−90k
  3. C
    −30i^+32j^−9k^{ - 30\widehat i + 32\widehat j - 9\widehat k}−30i+32j​−9k
  4. D
    −3.0i^+3.2j^−0.9k^{ - 3.0\widehat i + 3.2\widehat j - 0.9\widehat k}−3.0i+3.2j​−0.9k
View written solutionFree

Correct answer: C

  1. The magnetic force on a moving charge is F⃗=q(v⃗×B⃗).\vec F = q(\vec v \times \vec B).F=q(v×B).

  2. Given: q=1 μC=10−6 Cq = 1\,\mu C = 10^{-6}\,Cq=1μC=10−6C v⃗=2i^+3j^+4k^\vec v = 2\hat i + 3\hat j + 4\hat kv=2i^+3j^​+4k^ B⃗=(5i^+3j^−6k^)×10−3 T\vec B = (5\hat i + 3\hat j - 6\hat k)\times 10^{-3}\,TB=(5i^+3j^​−6k^)×10−3T

  3. First compute the cross product v⃗×B⃗\vec v \times \vec Bv×B.

    Since B⃗=10−3(5i^+3j^−6k^)\vec B = 10^{-3}(5\hat i + 3\hat j - 6\hat k)B=10−3(5i^+3j^​−6k^), v⃗×B⃗=10−3[(2i^+3j^+4k^)×(5i^+3j^−6k^)].\vec v \times \vec B = 10^{-3}\left[(2\hat i+3\hat j+4\hat k)\times(5\hat i+3\hat j-6\hat k)\right].v×B=10−3[(2i^+3j^​+4k^)×(5i^+3j^​−6k^)].

  4. Now evaluate:

    \begin{vmatrix} \hat i & \hat j & \hat k \\ 2 & 3 & 4 \\ 5 & 3 & -6 \end{vmatrix}$$ Expanding, $$= 10^{-3}\left[\hat i(3\cdot(-6)-4\cdot3)-\hat j(2\cdot(-6)-4\cdot5)+\hat k(2\cdot3-3\cdot5)\right]$$ $$= 10^{-3}\left[\hat i(-18-12)-\hat j(-12-20)+\hat k(6-15)\right]$$ $$= 10^{-3}\left[-30\hat i+32\hat j-9\hat k\right].$$

  5. Multiply by charge q=10−6q=10^{-6}q=10−6: F⃗=10−6×10−3(−30i^+32j^−9k^)\vec F = 10^{-6}\times 10^{-3}\left(-30\hat i+32\hat j-9\hat k\right)F=10−6×10−3(−30i^+32j^​−9k^) F⃗=(−30i^+32j^−9k^)×10−9 N.\vec F = \left(-30\hat i+32\hat j-9\hat k\right)\times 10^{-9}\,N.F=(−30i^+32j^​−9k^)×10−9N.

  6. The question says force is F⃗×10−9 N\vec F\times 10^{-9}\,NF×10−9N. Therefore, F⃗=−30i^+32j^−9k^.\vec F = -30\hat i + 32\hat j - 9\hat k.F=−30i^+32j^​−9k^.

  7. Comparing with options, this matches: C\boxed{\text{C}}C​

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