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This question has been asked in AIEEE 2007 . Please answer this question with explanation.

Naina Singh , 7 Years ago
Grade 12th pass
anser 1 Answers
deepak

Last Activity: 7 Years ago

consider an setup like \overrightarrow{E} along the +ve x direction and \overrightarrow{B} −y axis and the velocity \overrightarrow{v} along the positive direction
 
force due to \overrightarrow{E} on the particle will be in the +ve x direction =Eq
 
force due to \overrightarrow{B} on the particle will be in the -ve x direction =qvB sin(90)
 
if there is no alteration in the path of the particle then Fdue to B= Fdue to E    .....they can cancle as they are in the opposite directions
 
=> qvB=Eq
=> v = E/B = EB/B2
 we also know that the velocity is in the direction of \overrightarrow{E} × \overrightarrow{B} (vectorially by right hand rule)
 
hence \overrightarrow{v} = [\overrightarrow{E} * \overrightarrow{B}] / B^{2}
 
hence option d is correct
 
hope this was helpfull XD
 

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