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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 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)/(qE)`

B

`t=(2Bq)/(mv_0)`

C

`t=(sqrt3Bq)/(mv_0)`

D

`t=(sqrt2mv_0)/(qE)`

Text Solution

Verified by Experts

The correct Answer is:
d
`vecE` is parallel to `vecB` and `vecv` is perpendicular to both. Therefore, path of the particle is a helix with increasing pitch. Speed of particle at any time t is
`v=sqrt(v_x^2+v_y^2+v_z^2)....(i)`
Here, `v_y^2+v_z^2=v_0^2`
and `v=2v_0 implies v_x=sqrt3v_0`
`v_xa_xt implies sqrt3v_0=(qE)/mt implies t=(sqrt3mv_0)/(qE)`
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Knowledge Check

  • 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 vec B = B_0 hat i with a velocity vec v = v_0 hatj . The speed of the particle will becomes (sqrt(5))/(2) v_0 after a time

    A
    `(mv_(0))/(qE_(0))`
    B
    `(mv_(0))/(2qE_(0))`
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    `(sqrt5mv_(0))/(2qE_(0))`

  • The electric field acts positive x -axis. A charged particle of charge q and mass m is released from origin and moves with velocity vec(v)=v_(0)hatj under the action of electric field and magnetic field, vec(B)=B_(0)hati . The velocity of particle becomes 2v_(0) after time (sqrt(3)mv_(0))/(sqrt(2)qE_(0)) . find the electric field:

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  • Charge Q is given a displacement vec r = a hat i + b hat j in an electric field vec E = E_1 hat i + E_2 hat j . The work done is.

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    `Q sqrt((E_1^2 + E_2^2)^2) sqrt (a^2 + b^2)`

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