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# A particle of charge +q and mass m moving under the influence of a uniform electric field E^i and uniform magnetic field B^k follows a trajectory from P to Q as shown in figure. The velocities at P and Q are v^i and −2v^j. Among the following the correct statements is/are :rate of work done by the electric field at P is 34(mv2a)E=34(mv2qa)rate of work done by electric field at P is zerorate of work done by both the fields at Q is zero

A
E=34(mv2qa)
B
rate of work done by the electric field at P is 34(mv2a)
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
Solution
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#### Net work done = Change in Kinetic EnergyTherefore, net work done from P to Q=32mv2And all work is done by electric field therefore equatingqE(2a)=32mv2⇒E=3mv24qaTherefore, option A is correct.Option B is wrong. Using Dimensional Analysis of given option is of force not of power.Option D is correct.Because at Q, electric force is perpendicular to velocity vector.

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