A particle having charge 10⁻⁹ C moving in x-y plane in fields of 0.4ȷ̂ N/C and 4×10⁻³k̂ T experiences a force of (4ı̂+2ȷ̂)×10⁻¹⁰ N. The velocity of the particle at that instant is m/s.

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Accepted Answer

(1) 50ı̂+100ȷ̂. qE⃗=10⁻⁹×0.4ȷ̂=4×10⁻¹⁰ȷ̂ N, so the magnetic part is qv⃗×B⃗=(4ı̂-2ȷ̂)×10⁻¹⁰. With v⃗=vₓı̂+v_yȷ̂: v⃗× Bk̂=B(v_yı̂-vₓȷ̂). qBv_y=4×10⁻¹⁰⇒ v_y=(4×10⁻¹⁰)/(4×10⁻¹²)=100; qBvₓ=2×10⁻¹⁰⇒ vₓ=50. v⃗=50ı̂+100ȷ̂ m/s

A particle having charge 10⁻⁹ C moving in x-y plane in fields of 0.4ȷ̂ N/C and 4×10⁻³k̂ T experiences a force of (4ı̂+2ȷ̂)×10⁻¹⁰ N. The velocity of the particle at that instant is ______ m/s.

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