99 Percentile Qs Bank for JEE MainPhysicsWork, Power and Energy
A charged particle of mass ‘ m ’ and charge ‘ q ’ moving under the influence of uniform electric field E   → i ^ and a uniform magnetic field B   → k ^ follows a trajectory from point P to Q as shown in figure. The velocities at P and Q are respectively, v i → and - 2 v j → . Then which of the following statements (A, B, C, D) are the correct? (Trajectory show
Options
- AA , C , D
- BB , C , D
- CA , B , C
- DA , B , C , D
Correct answer
C. A , B , C
Step-by-step solution
( A ) by work energy theorem W mag + W ele = 1 2 m ( 2 v ) 2 - 1 2 m ( v ) 2 0 + qE 0 2 a = 3 2 mv 2 E 0 = 3 4 mv 2 qa ( B ) Rate of work done at A = power of electric force = qE 0 V = 3 4 mv 3 a ( C ) at Q , dw dt = 0 for both forces D Δ L → = ( - m 2  v  2 a k ^ ) - ( -  mvak  ) | Δ L → | = 3 mva