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?

1 Answer
Aug 8, 2018

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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