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A charged particle is moving with speed v perpendicular to a uniform magnetic field. A second identical charged particle is moving with speed 2v perpendicular to the same magnetic field. If the frequency of revolution of the first particle is f, the frequency of revolution of the second particle is

(a) f. (b) 2f. (c) 4f. (d) f/2. (e) f/4.

Respuesta :

oladayoademosu

Answer:

(a) f

Explanation:

Since the magnetic force = centripetal force,

Bqv = mv²/r

v = Bqr/m. Since B, q and m are constant,

v ∝ r.

Let v₁ = v, v₂ = 2v and r₁, r₂ be the speeds and radii of path of the first and second particles respectively. v₁/v₂ = r₁/r₂ = v/2v = 1/2

We know that the v = rω = 2πrf where f is frequency of revolution. Let f₁ and f₂ represent the frequencies of revolution of the first and second particles respectively.

so v₁ = 2πr₁f₁ and v₂ = 2πr₂f₂

v₁/v₂ = 2πr₁f₁/2πr₂f₂ = r₁f₁/r₂f₂

v₁/v₂ × r₂/r₁ = f₁/f₂= 1/2 × 2 = f₁/f₂ = 1 ⇒ f₁ = f₂. So, f₁ = f = f₂

So the frequency of revolution of the second particle  is f

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