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An alpha particle (q = +2e, m = 4.00 u) travels in a circular path of radius 6.40 cm in a uniform magnetic field with B = 1.17 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.
An alpha particle (q = +2e, m = 4.00 u) travels in a circular path of radius 6.40 cm in a uniform magnetic field with B = 1.17 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)The magnetic force acts as centripetal force, so we can writewhereq is the charge of the particlev is its speedB is the magnetic field strengthm is the massr is the radius of the circular pathFor the alpha particle in the problem,B = 1.17 TRe-arranging the equation and solving for v, we find its speed:(b)The period of revolution is given by the ratio between the distance travelled in one circle (so, the circumference of the path) and the speed of the particle, sowherer is the radius of the pathv is the speedHere we haveSo the period of revolution is(c)The kinetic energy of a particle is given bywherem is its massv is its speedFor the alpha particle in the problem, we haveSo its kinetic energy is(d)When accelerated through a potential difference, a particle gains a kinetic energy equal to the change in electric potential energy - so we can write:where the term on the left is the change in electric potential energy, withq is the charge of the particleis the potential differenceHere we haveis the charge of the alpha particleis the kinetic energyRe-arranging the formula, we find...