A particle having charge 10-9 C moving in x-y plane in fields of 0.4 j N/C and 4 10-3 k T experiences a force of (4 i+2 j ) 10-10 N. The velocity of the particle at that instant is m/s.
- 50 i+100 j
- 100 i+50 j
- -50 i+100 j
- 50 i-100 j
Answer
(1) 50 i+100 j
Solution
Lorentz force: F = q E + q( v B)
Electric part: q E = 10-9 0.4 j = 4 10-10 j N
∴ Magnetic part: q( v B) = (4 i + 2 j) 10-10 - 4 10-10 j = (4 i - 2 j) 10-10 N
Let v = vx i + vy j (motion in the x-y plane), B = B k:
v B = B(vy i - vx j)
qB = 10-9 4 10-3 = 4 10-12
⇒ 4 10-12 vy = 4 10-10 ⇒ vy = 100 m/s
⇒ -4 10-12 vx = -2 10-10 ⇒ vx = 50 m/s
∴ v = 50 i + 100 j m/s
Answer: (1)