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Updated on : 2022-09-05

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Correct option is D)

$v=v_{0}j^ ⊥E$

$v⊥B$

Hence, the path of the particle is a helix with speed remaining constant for the circular path in the $yz$ plane.

Magnitude of velocity of particle at any time $t:$ $v=v_{x}+v_{y}+v_{z} =(a_{x}t)_{2}+v_{0} $ where $v_{y}+v_{z}=v_{0}$

Given $v=2v_{0}$

$⇒(2v_{0})_{2}=v_{x}+v_{0}$

$⇒(v_{x})_{2}=3(v_{0})_{2}$

$⇒v_{x}=3 v_{0}$

$⇒a_{x}t=3 v_{0}$

$⇒(mqE )t=3 v_{0}$

$⇒t=qE3 v_{0}m $

Solve any question of Moving Charges and Magnetism with:-

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