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Magnetic field exists in the region between lines PQ and RS as shown in the figure. A particle having charge to mass ratio of a enters the magnetic field as shown in the figure with a speed v. Lines PQ and RS are parallel to each other. Find the value of d such that the charged particle exits the magnetic field region with velocity perpendicular to line RS.


Option 1 -

(√3v)/(2aB)

Option 2 -

v/(2aB)

Option 3 -

v/(aB)

Option 4 -

v/(√3aB)

0 7 Views | Posted a month ago
Asked by Shiksha User

  • 1 Answer

  • V

    Answered by

    Vishal Baghel | Contributor-Level 10

    a month ago
    Correct Option - 1


    Detailed Solution:

    d = Rsin 60°


    ⇒ d = (mv√3)/ (qB 2) = (√3v)/ (2aB)

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