A particle of charge q and mass m is projected from the origin with initial velocity v = (v₀/√2)î+(v₀/√2)ĵ. The region 0 ≤ x ≤ L contains a uniform magnetic field B = B₀ k̂ and a position-dependent electric field E = E₀ exp(-λx) î, where λ is a positive constant. After the particle moves from x = 0 to x = L, what is the change in its kinetic energy?
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Question

A particle of charge q and mass m is projected from the origin with initial velocity v = (v₀/√2)î+(v₀/√2)ĵ. The region 0 ≤ x ≤ L contains a uniform magnetic field B = B₀ k̂ and a position-dependent electric field E = E₀ exp(-λx) î, where λ is a positive constant. After the particle moves from x = 0 to x = L, what is the change in its kinetic energy?

Options

G1 Physics · Accepted Answer

The change in kinetic energy equals the total work done on the particle. The magnetic force is always perpendicular to the instantaneous velocity, so it does no work: W(magnetic) = 0. Only the electric field changes the kinetic energy. Therefore, ΔK = q∫E·dr = q∫₀ᴸ E₀ exp(-λx) dx. Thus, ΔK = qE₀[-exp(-λx) ÷ λ]₀ᴸ = (qE₀ ÷ λ)[1-exp(-λL)]. Correct option: A.