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A charged particle with charge q enters ...

A charged particle with charge q enters a region of constant, uniform and mutually orthogonal fields `vecE and vecB` with a velocity `vecv` perpendicular toboth `vecE and vecB`, and comes out without any change in its magnitude or direction. Then

A

`vecv=vecExx(vecB)/(B^(2))`

B

`vecv=vecBxx(vecE)/(B^(2))`

C

`vecv=vecExx(vecB)/(E^(2))`

D

`vecv=vecBxx(vecE)/(E^(2))`

Text Solution

Verified by Experts

The correct Answer is:
A
As v of charged particle is remaining constant, it means force acting on charged particle is zero:
`f=qE+q(vxx B)=0, qE-(vxxB)=0`
So `q(vxxB)=qEimpliesvxxB=Eimpliesv=(ExxB)/(B^(2))`
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