Question

A proton with velocity v completes a circle with radius r in uniform magnetic field.what should be the velocity of alpha particle so that it completes a circle with same radius under the influence of the same magnetic field?

Answer 1

The force experienced by a particle of charge `q` moving in a magnetic field perpendicular to the direction of the field acts perpendicularly to both velocity and magnetic field and provides the centrpetal force required for the motion of the particle in a circular path.For the dynamic equilibrium established here we have the following relation

`(mv^2)/r=qvB_m`

`=>v=(qrB_m)/m`

Where

`m->"mass of the particle"`

`v->"speed of the particle"`

`r->"radius of circular path"`

`B_m->"intensity of magnetic field"`

In our problem both proton and `alpha` particle follow circular path of same radius under the influence of the same magnetic field.

For proton the above relation becomes

`=>v=(q_prB_m)/m_p`

And for `alpha` particle the above relation becomes

`=>v_alpha=(q_alpharB_m)/m_alpha`

So `v_alpha/v=(m_pq_alpha)/(m_alphaq_p)=1/4*2/1=1/2`

Hence `v_alpha=v/2`