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Magnetics question

2019 · 11 Jan · Shift 2 · Q54
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Magnetics question

2019 · 11 Jan · Shift 2 · Q54

JEE MainPhysicsMagneticsMCQ+4 / −1
A particle of mass m and charge q is in an electric and magnetic field given by E→=2i^+3j^;   B→=4j^+6k^.\overrightarrow E = 2\widehat i + 3\widehat j;\,\,\,\overrightarrow B = 4\widehat j + 6\widehat k.E=2i+3j​;B=4j​+6k. The charged particle is shifted from he origin to the point P(x = 1; y = 1) along a straight path. The magnitude of the total work done is :
  1. A
    (2.5) q
  2. B
    (0.35) q
  3. C
    (0.15) q
  4. D
    5 q
View written solutionFree

Correct answer: D

  1. Given fields

    E⃗=2i^+3j^,B⃗=4j^+6k^\vec E = 2\hat i + 3\hat j, \qquad \vec B = 4\hat j + 6\hat kE=2i^+3j^​,B=4j^​+6k^

    The particle is moved from the origin O(0,0)O(0,0)O(0,0) to point P(1,1)P(1,1)P(1,1) along a straight line.

  2. Work done by magnetic field

    Magnetic force is always perpendicular to velocity:

    F⃗B=q(v⃗×B⃗)\vec F_B = q(\vec v \times \vec B)FB​=q(v×B)

    Hence,

    WB=∫F⃗B⋅dr⃗=0W_B = \int \vec F_B \cdot d\vec r = 0WB​=∫FB​⋅dr=0

    So only the electric field does work.

  3. Work done by electric field

    Electric force:

    F⃗E=qE⃗=q(2i^+3j^)\vec F_E = q\vec E = q(2\hat i + 3\hat j)FE​=qE=q(2i^+3j^​)

    Displacement from origin to (1,1)(1,1)(1,1) is

    Δr⃗=i^+j^\Delta \vec r = \hat i + \hat jΔr=i^+j^​

    Therefore work done:

    W=F⃗E⋅Δr⃗=q(2i^+3j^)⋅(i^+j^)W = \vec F_E \cdot \Delta \vec r = q(2\hat i + 3\hat j) \cdot (\hat i + \hat j)W=FE​⋅Δr=q(2i^+3j^​)⋅(i^+j^​)

    W=q(2⋅1+3⋅1)=5qW = q(2\cdot 1 + 3\cdot 1) = 5qW=q(2⋅1+3⋅1)=5q

  4. Magnitude of total work done

    ∣W∣=5q|W| = 5q∣W∣=5q

  5. Option check

    • A: (2.5)q(2.5)q(2.5)q ❌
    • B: (0.35)q(0.35)q(0.35)q ❌
    • C: (0.15)q(0.15)q(0.15)q ❌
    • D: 5q5q5q ✅

Therefore, the correct answer is D.

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