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

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 P is `(3)/(4) (mv^(3))/(a)`

C

Rate of work done by electric field at p is zero

D

Rate of work done by both the fields at Q is zero

Text Solution

Verified by Experts

The correct Answer is:
A, B, C
In going from P to Q increase in kinetic energy
`= (1)/(2) m (2v)^(2) - (1)/(2) mv^(2) = (1)/(2) m (3v^(2))` = work done by electric field
or `(3)/(2) mv^(2) = Eq xx 2a` `implies` `E = (3)/(4) (mv^(2))/(qa)`
The rate of work done by E at P = force due to E X velocity
`= qEv = q ((3mv^(2))/(4qa)) V = (3mv^(3))/(4a)`
At q. `vec(v)` is a perpendicular to `vec(E )` and `vec(B)` therefore no work is done by the either field
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