Aproton (charge +e, mass mp), a deuteron (charge +e, mass 2mp), and an alpha particle (charge +2e, mass 4mp) are accelerated from rest through a common potential difference δv. each of the particles enters a uniform magnetic field b, with its velocity in a direction perpendicular to b. the proton moves in a circular path of radius rp. (a) in terms of rp, determine the radius rd of the circular orbit for the deuteron. rd = (b) in terms of rp, determine the radius rα for the alpha particle. rα =

a) (2)^1/2*rp; b) ra= (2)^1/2*rp

Explanation: In order to explain this problem we have to consirer a circular motion for charged particles with velocity in a magnetic field.

The we know that:

the Ek=q*V=1/2*m*v^2 then

v=(2*q*V/m)^1/2

The magnetic force when v is perpendicular to B, is equal

Fm=q*v*B

when the charged particle is moving a circular motion, we have:

Fm=m*ac where ac is the centripetal acceleration which is equal,

ac=v^2/R  where R is the radius of the circular trajectory.

Finally we obtain;

q*v*B=m*v^2/R

R=m*v/(q*B) using that v= (2*q*V/m)^1/2

we have; R=(1/B)*(2*V*m/q)^1/2

For a proton we have

rp=(1/B)* (2*V*mp/e)^1/2

For a deuteron ( charge +e and mass 2mp)

rd=(1/B)* (2*V*2*mp/e)^1/2=(1/B)*(4*V*mp/e)^1/2=(2)^1/2*rp

For a alpha particle ( charge +2e, mass 4mp) =

ra=(1/B)*(2*V*4*mp/2*e)^1/2=(1/B)*(4*V*mp/e)^1/2=(2)^1/2*rp

jupiter is the largest planet in our solar system, having a mass and radius that are respectively, 318 and 11.2 times that of earth. suppose that an object falls from rest near the surface of each planet and that the acceleration due to gravity remains constant during the fall. each object falls the same distance before striking the ground. determine the ratio of the time of fall on jupiter to that on earth.

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