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

A particle of charge +q and mass m moving under the influence of uniform electric field `Ehati` and uniform magnetic field `BhatK` follows a trajectory from P to Q as as shown in figure. The velocities at P and Q are `vhati` and `-2vhatj` of the following statements is/are correct ?

A

`E=3/4 [(m upsilon^2)/(qa)]`

B

Rate of work done by the electric field at P is `E=3/4 [(m v^3)/a]`

C

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

D

Rate of work done by both the feilds at Q is zero.

Text Solution

Verified by Experts

The correct Answer is:
(a,b,d)
In going from P and Q, increase in kinetic energy
`=1/2 m(2v)^2-1/2mv^2=1/2m(3v^2)`
=work done by the electric filed
`3/2 mv^2=Eqxx2a or E=3/4((mv^2)/(qa))`
The rate of work done by E at P =force due to E x velocity
`=(qE)v=qv[3/4((mv^2)/(qa))]=3/4((mv^3)/a)`
At Q, `vecv` is perpendicular to `vecE` hence no power or rate of work
done by electric field. Rate of work done by magnetic field is
always zero.
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