A particle of charge +q and mass m moving under the influence of a uniform electric field Ei^ and uniform magnetic field Bk^ follows a trajectory from P to Q as shown in fig. The velocities at P and Q are vi^ and −2vj^ which of the following statement(s) is/are correct?
This question has multiple correct options
Rate of work done by the electric field at P is 43[amv3]
Rate of work done by the electric field at P is zero
Rate or work done by both the fields at Q is zero
Open in App
Updated on : 2022-09-05
Verified by Toppr
Correct options are A) , B) and D)
From the work- energy theorem W=△K.E. ⇒qE.2a=21m[4υ2−υ2] ⇒E=43(qamυ2) At P rate of work done by E field will be =F.υ=(qE)(υ)cos0° =43(amυ3) At Q rate of work done =F.υ=(qE)(2υ)cos90°=0 And work done by magnetic field is always zero.
Solve any question of Moving Charges and Magnetism with:-