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)
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