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Q no 2

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Asked by manasvijha 25th March 2019, 7:40 AM
Answered by Expert
Answer:
when a proton is released from rest, intitially it is accelerated  twowards west direction due to electric field which is acting westwards
 
begin mathsize 12px style stack F subscript e with rightwards arrow on top space equals space e space E with rightwards arrow on top space equals space m left parenthesis stack a subscript 0 with rightwards arrow on top space right parenthesis space space space space o r space space open vertical bar E with rightwards arrow on top close vertical bar space equals space fraction numerator space m space a subscript 0 over denominator e end fraction space space end style ( in west direction )  ..............................(1)
when the proton is projected in North direction with velocity v0 , proton is subjected to electric field and magnetic field.
 
Proton is moving towards west with acceleration 3a0 . Since begin mathsize 12px style v with rightwards arrow on top space cross times B with rightwards arrow on top space end styleforce is acting in west direction and its velocity is towards north ,
magnetic field is acting downward direction ( see figure )
 
since it is known that electric field gives an acceleration ao , additional acceleration 2ais given by magnetic field
 
begin mathsize 12px style stack F subscript m with rightwards arrow on top space equals e left parenthesis space v with rightwards arrow on top space cross times B with rightwards arrow on top right parenthesis space equals space m open parentheses 2 stack a subscript o with rightwards arrow on top close parentheses space space space space space o r space space space space space space space open vertical bar B with rightwards arrow on top close vertical bar space equals space fraction numerator 2 m space a subscript 0 over denominator e space v subscript o end fraction end style  ( downward direction )
 
Hence in the list of options given in the answer,  (b) is the correct one
Answered by Expert 25th March 2019, 2:16 PM
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