Question:medium

An alpha particle moving with certain speed towards east enters a uniform magnetic field directed vertically up. The alpha particle will then move in

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The magnetic force on a charged particle, $\vec{F} = q(\vec{v} \times \vec{B})$, is always perpendicular to both the velocity $\vec{v}$ and the magnetic field $\vec{B}$. Because the force is perpendicular to the velocity, it does no work and cannot change the particle's speed or kinetic energy, only its direction.

Updated On: Mar 30, 2026

vertical circular path with the same speed
horizontal circular path with the same speed
vertical circular path with increased speed
vertical circular path with decreased speed

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The Correct Option is B

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