A particle of charge 10⁻⁹ C moves in the x-y plane. At a certain instant, the electric field is E = 0.4ĵ N·C⁻¹ and the magnetic field is B = 4×10⁻³ k̂ T. The total electromagnetic force on the particle is F = (4î+2ĵ)×10⁻¹⁰ N. What is the velocity of the particle at that instant?
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Question

A particle of charge 10⁻⁹ C moves in the x-y plane. At a certain instant, the electric field is E = 0.4ĵ N·C⁻¹ and the magnetic field is B = 4×10⁻³ k̂ T. The total electromagnetic force on the particle is F = (4î+2ĵ)×10⁻¹⁰ N. What is the velocity of the particle at that instant?

Options

G1 Physics · Accepted Answer

Use the Lorentz-force equation: F = q(E+v×B). Let v = vₓî+vᵧĵ. Given, q = 10⁻⁹ C, E = 0.4ĵ N·C⁻¹, B = 4×10⁻³ k̂ T. Now, v×B = (vₓî+vᵧĵ)×(4×10⁻³ k̂) = 4×10⁻³[vᵧî-vₓĵ]. Therefore, F ÷ q = 4×10⁻³vᵧî + [0.4-4×10⁻³vₓ]ĵ. From the given force, F ÷ q = [(4î+2ĵ)×10⁻¹⁰] ÷ 10⁻⁹ = 0.4î+0.2ĵ. Compare the î components: 4×10⁻³vᵧ = 0.4, so vᵧ = 100 m·s⁻¹. Compare the ĵ components: 0.4-4×10⁻³vₓ = 0.2. Thus, 4×10⁻³vₓ = 0.2, vₓ = 50 m·s⁻¹. Hence, v = 50î+100ĵ m·s⁻¹. Correct option: A.