Question

Two protons move with uniform circular motion in the presence of uniform magnetic fields. Proton one moves twice as fast as proton two. The magnitude of the magnetic field in which proton one is immersed is twice the magnitude of the magnetic field in which proton two is immersed. The radius of the circle around which proton one moves is r. What is the radius of the circle around which proton two moves

Answer 1

Answer:

r₂ = 4 r

Explanation:

For this exercise let's use Newton's second law with the magnetic force

          F = q v x B

bold letters indicate vectors, the magnitude of this expression is

          F = q v B sin θ  

in this case we assume that the angle is 90º between the speed and the magnetic field.

If we use the rule of the right hand with the positive charge, the thumb in the direction of the speed, the fingers extended in the direction of the magnetic field, the palm points in the direction of the force, which is towards the center of the circle, therefore the force is radial and the acceleration is centripetal

           a = v² / r

let's use Newton's second law

           F = ma

           q v B = m v² / r

           r = $\frac{qB}{mv}$

Let's apply this expression to our case.

Proton 1

             r = \frac{qB_1}{mv_1}

Proton 2

             r₂ = $\frac{q \ B_2}{m \ v_2}$

in the exercise indicate some relationships between the two protons

*    v₁ = 2 v₂

    v₂ = v₁ / 2

*   B₂ = 2B₁

we substitute

           r₂ = $\frac{q \ 2B_1}{m \ \frac{v_1}{2} }$

           r₂ = 4 $\frac{qB_1}{mv_1}$

           r₂ = 4 r