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A particle of charge +q and mass m moving under the influence of a uniform electric field E and a magnetic field B  enters in I quadrant of a coordinate system at a point (0, a) with initial velocity v  and leaves the quadrant at a point (2a, 0) with velocity - 2v . Find
Rate of work done by both the fields at (2a, 0)
    Correct answer is '0'. Can you explain this answer?
    Verified Answer
    A particle of charge +q and mass m moving under the influence of a uni...
    From the work- energy theorem
    W=△K.E.
    ⇒qE.2a=(1/2)​m[4υ2−υ2]
    ⇒E=(3/4)​( mυ2​/qa)
    At P rate of work done by E field will be 
    =F.υ=(qE)(υ)cos0°
    =(3/4)​( mυ3/a​)
    At Q rate of work done
    =F.υ=(qE)(2υ)cos90°=0
    And work done by the magnetic field is always zero.
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    A particle of charge +q and mass m moving under the influence of a uniform electric field E and a magnetic field Benters in I quadrant of a coordinate system at a point (0, a) with initial velocity vand leaves the quadrant at a point (2a, 0) with velocity -2v . FindRate of work done by both the fields at (2a, 0)Correct answer is '0'. Can you explain this answer? for Class 12 2024 is part of Class 12 preparation. The Question and answers have been prepared according to the Class 12 exam syllabus. Information about A particle of charge +q and mass m moving under the influence of a uniform electric field E and a magnetic field Benters in I quadrant of a coordinate system at a point (0, a) with initial velocity vand leaves the quadrant at a point (2a, 0) with velocity -2v . FindRate of work done by both the fields at (2a, 0)Correct answer is '0'. Can you explain this answer? covers all topics & solutions for Class 12 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A particle of charge +q and mass m moving under the influence of a uniform electric field E and a magnetic field Benters in I quadrant of a coordinate system at a point (0, a) with initial velocity vand leaves the quadrant at a point (2a, 0) with velocity -2v . FindRate of work done by both the fields at (2a, 0)Correct answer is '0'. Can you explain this answer?.
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