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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:

bodega

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Karma Points: 7

Date Posted: 7/24/2008 7:21:19 PM Status: Live

Asker's Rating: Helpful

Response:

1.Since the particle starts from rest then accelerates through the magnetic field, a force is exerted on it. To figure out the acceleration, we first must calculate its final and initial kitnetic and potential energies using the work-energy theorem for charges particles (KEi+EPEi=KEf+EPEf) where EPE=qV (voltage x charge) for initial and final states. Since the particle is accelerated from rest, the initial velocity is 0. Rearrange the equation and solve for v. Then follow the next step:
2. There is a force of 1.60 T. Knowing that F = v sinθ Bq, where B is the magnetude of the magnetic field and q is the charge of the particle, using Newton's 2nd law (f=ma), set F equal to the magnetic force and solve for a. θ=90 degrees since it is perpendicular. P.S. : The velocity is what you calculated earlier for part 1.

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