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Question:

A particle with a mass of 6.64 × 10^–27 kg and a charge of +3.20 × 10^–19 C is accelerated from rest through a potential difference of 2.45 × 10⁶ V. The particle then enters a uniform 1.60-T magnetic field. If the particle's velocity is perpendicular to the magnetic field at all times, what is the magnitude of the magnetic force exerted on the particle?

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Answers:

Kayen

Oracle

Karma Points: 21,902

(Ohio University Main Campus)

Date Posted: 7/24/2008 8:35:48 PM Status: Live

Asker's Rating: Lifesaver

Response:

Energy gained by the paricle in the el field,

K = 0.5m.v² = q.V

v² = 2q.V/m = 2.361x10¹⁴(on substituting the given values)

v = 1.537x10⁷m/s.

Force exerted by the mag field, (on substituting the given values)

F =qvB = 7.87x10^-12 N

Kayen (Enjoys helping you uderstand..)

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