An alpha particle (q = +2 m = 4.00) travels in a circular path of radius 5.72 cm

Question

An alpha particle (q = +2 m = 4.00) travels in a circular path of radius 5.72 cm in a uniform magnetic field with 8 = 1.62 T. Calculate (a) Its speed, (b) Its period of revolution, (c) Its kinetic energy, and (d) the potential difference through which it would have to be accelerated to achieve this energy. (a) Number Units (b) Number Units (c)Number Units (d) Number Units By accessing this Question Assistance, you will learn while you can points based on the Point Potential Policy set by your instructor.

Explanation / Answer

a) F= qvB and F= mv2/r for circular motion.

So, qB= mv/r.

v= qBr/m= 2× 1.6× 10-19× 1.62× 0.0572/ (6.64× 10-27)

v= 4.46× 106 m/s.

b) T= 2m/(qB)

Placing values, T= 8.04 × 10-8 s.

c) Kinetic Energy= 1/2 mv2= 6.6× 10-14 J.

d) 1/2 mv2= qV

V= 2.06× 105 V.

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