An alpha particle is moving with a velocity v. It enters a uniform magnetic field (B) as shown in…
An alpha particle is moving with a velocity v. It enters a uniform magnetic field (B) as shown in Figure 6 below. The magnetic field is perpendicular and into the plane of the paper. A uniform electric field is applied in the same region as the magnetic field so that the alpha particle passes undeviated through the combined fields.
(i)
State the direction of the applied electric field with a proper reason.
(ii)
Without any change in the electric and magnetic field, the alpha particle is replaced by a proton moving with a velocity v. Will there be any change in the path of the particle? Give a reason for your answer.
Answer
Answer (i)
AIWritten by AI (gemini) - it can contain mistakes.
The electric field $\vec{E}$ should be in the plane of the paper, pointing vertically downwards. (The magnetic force $\vec{F}_m = q(\vec{v} \times \vec{B})$ on the positive alpha particle is directed vertically upwards; thus, the electric force $\vec{F}_e = q\vec{E}$ must act vertically downwards to cancel it, which requires $\vec{E}$ to point downwards).
Answer (ii)
AIWritten by AI (gemini) - it can contain mistakes.
No, there will be no change in the path of the proton. For an undeviated path, $qE = qvB \implies v = E/B$, which is independent of the mass and charge of the particle. Since the velocity v, electric field E, and magnetic field B are unchanged, the proton also passes undeviated.
From ISC 2027 Specimen Physics Paper 1, question 11.