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A particle of charge +q and mass m movi...

A particle of charge ` +q` and mass `m` moving under the influence of a uniform electric field ` E hat(i)` and uniform magnetic field ` B hat(k)` follows a trajectory from ` P to Q` as shown in fig. The velocities at `P and Q` are ` v hat(i)` and ` - 2v hat(j)` . which of the following statement(s) is/are correct ?

A

`E=(3)/(4)(mv^(2))/(qa)`

B

Rate of work done by electric field at `P` is `(3)/(4)(mv^(3))/(a)`

C

Rate of work done by electric field at `P` is zero

D

Rate of word down by both the fields at `Q` is zero

Text Solution

Verified by Experts

The correct Answer is:
1,2,3
Kinetic energy of the particle at point `P=(1)/(2)mv^(2)`
`K.E.` of the particle at point `Q=(1)/(2)m(2v)^(2)`
Increase in `K.E.=(3)/(2)mv^(2)`
It comes from the work done by the electric force `qE` on the particle as it covers a distance `2a` along the `x-` axis.
Thus `(3)/(2)mv^(2)=qExx2aimplies E=(3)/(4)(mv^(2))/(qa)`
The rate of work done by the electric field at `P=Fxxv=qExxv=(3)/(4)(mv^(2))/(a)`
At `A, vec(F)_(e)=qvec(E)` is along `x-` axis while velocity is along negative `y-` axis. Hence rate of work done by electric field `=vec(F)_(e).vec(v)=0 ( :. theta =90^(@))`. Similarly, according to equation `vec(F)_(m)=q(vec(v)xxvec(B))` force `vec(F)_(m)` is also perpendicular to velocity vector `vec(v)`.
Hence the rate of work done by the magnetic field `=0`.
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Knowledge Check

  • A particle of charge +q and mass m moving under the influnce of a uniform electric field E hati and a uniform magnetic field Bhatk follows trajectory from P to Q as shown in figure. The velocities at P and Q vhati and -2vhatj respectively. Which of the following statement(s) is/are correct

    A
    `E = 3/4 (mv^2)/(qa)`
    B
    Rate of work done by electric field at Pis `3/4 (mv^3)/(a)`
    C
    Rate of work done by electric field at P is zero
    D
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  • 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)`
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    `E = (3)/( 2) (( mv^(2))/(qa))`
    B
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    C
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