An electron enters a solenoid with speed 800 m s⁻¹ making 30° with its axis. Current in solenoid is 4 A, with 8000 turns along its length. Given e/m = √3 × 10¹¹ C kg⁻¹, calculate y if total revolutions are (y/5) × 10⁶. [See Image.]
Physics · Mixed Chapter
JEE Main Practice

Question

An electron enters a solenoid with speed 800 m s⁻¹ making 30° with its axis. Current in solenoid is 4 A, with 8000 turns along its length. Given e/m = √3 × 10¹¹ C kg⁻¹, calculate y if total revolutions are (y/5) × 10⁶. [See Image.]

Physics · Mixed Chapter

G1 Physics · Accepted Answer

Step 1 — Field inside a long solenoid is B = μ₀NI/ℓ, where N = 8000 turns and ℓ is its length. Step 2 — Electron cyclotron frequency f = (e/m)B/(2π); time inside solenoid = ℓ/(v cos30°). Step 3 — Number of revolutions = f t = (e/m)μ₀NI/(2π v cos30°). Length cancels. Step 4 — Substitute e/m = √3 × 10¹¹, μ₀ = 4π × 10⁻⁷, I = 4 A, v = 800 m s⁻¹: Nᵣ = 1.6 × 10⁶. Step 5 — Source writes Nᵣ = (y/5) × 10⁶; therefore y = 8.