Q.

A particle having charge 10-9 C moving in x-y plane in fields of 0.4j^ N/C and 4×10-3k^ T experiences a force of (4i^+2j^)×10-10 N. The velocity of the particle at that instant is ______ m/s.                      [2026]

1 50i^+100j^  
2 100i^+50j^  
3 -50i^+100j^  
4 50i^-100j^  

Ans.

(1)

F→=qE→+q(v→×B→)

(4i^+2j^)×10-10=10-9[0.4j^+(vxi^+vyj^)×(4×10-3)k^]

(4i^+2j^)×10-10=10-9(0.4j^-4×10-3vxj^+4×10-3vyi^)

On comparing: vy=100, vx=50

v→=(50i^+100j^) ms-1