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An alpha particle travels at a velocity of magnitude 460 m/s through a uniform magnetic field...

An alpha particle travels at a velocity of magnitude 460 m/s through a uniform magnetic field of magnitude 0.056 T. (An alpha particle has a charge of charge of +3.2 × 10-19 C and a mass 6.6 × 10-27 kg) The angle between the particle's direction of motion and the magnetic field is 61°. What is the magnitude of (a) the force acting on the particle due to the field, and (b) the acceleration of the particle due to this force? (c) Does the speed of the particle increase, decrease, or remain the same?

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Answer #1

PART A)

Force on moving charge is given by

F = qvBsin\theta

F =3.2*10^{-19}*460*0.056*sin61

F =7.21*10^{-18}N

PART B)

from Newton's II law we can write

F =ma

7.21*10^{-18}N =6.6*10^{-27}*a

a = 1.09*10^9 m/s^2

PART C)

Speed of the particle will remain the sameinsde magnetic field

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