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A positively charged particle of specific charge α, accelerated by a potential difference V moves through a uniform transverse magnetic field B. The field occupies a region of width l. The angle θ through which the charged particle deviates from initial direction of its motion is

Option 1 -

sin⁻¹[Bl√(α/V)]

Option 2 -

sin⁻¹[2Bl√(α/V)]

Option 3 -

cos⁻¹[2Bl√(α/3V)]

Option 4 -

sin⁻¹[Bl√(α/2V)]

0 4 Views | Posted a month ago
Asked by Shiksha User

  • 1 Answer

  • V

    Answered by

    Vishal Baghel | Contributor-Level 10

    a month ago
    Correct Option - 4


    Detailed Solution:

    p²/2m = Vq => p = √2mVq
    θ = sin? ¹ (l/R') = l/R = mV/qB = p/qB

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