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A particle of charge q and mass m starts...

A particle of charge q and mass m starts moving from the origin under the action of an electric field `vec(E) = E_(0)hat(i)` and a magnetic field `vec(B) = B_(0)hat(i)` with a velocity `vec(v) = v_(0)hat(j)`. The speed of the particle will become `2v_(0)` after a time

A

`t=2mV_(0)`

B

`t=(2Bq)/(mV_(0))`

C

`t=(sqrt(3)Bq)/(mV_(0))`

D

`t=(sqrt(3)mV_(0))/(qE)`

Text Solution

Verified by Experts

The correct Answer is:
D
E is parallel to `vecB` & `vecV` is perpendicular to both. Therefore to path of the particle is a helix with increasing pitch. Speed of particle of at any time t is
`V=sqrt(V_(x)^(2)+V_(y)^(2)+V_(z)^(2)),` Here `V_(z)^(2)+V_(y)^(2)=V_(0)^(2)`
`V=2V_(0)impliesV_(x)=sqrt(3)V_(0):.t=(V_(x))/(a_(x))=(sqrt(3)V_(0))/(qE//m)impliest=(sqrt(3)mV_(0))/(qE)`
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