In a nuclear experiment a proton with kinetic energy 3.0 mev moves in a circular path in a uniformmagnetic field. what energy must the following two particles haveif they are to circulate in the same orbit?
(a) an alpha particle (q =+2e, m = 4.0 u)
1 mev

(b) a deuteron (q = +e, m = 2.0 u)
2 mev

a)     K = 3 MeV   b)   K=  1.5 MeV

Explanation:

We can solve this experiment using the equation of the magnetic force with Newton's second law, where the acceleration is centripetal.

        F = q v x B

We can also write this equation based on the modules of the vectors

        F = qv B sin θ

With Newton's second law

       F = ma

       F = m v² / r

       q v B = m v² / r

       v = q B r / m

The kinetic energy is

       K = ½ m v²

Substituting

       K = ½ m (q B r/ m)²

       K = ½ B² r²  q² / m

       K = (½ B² R²)  q²/m

The amount in brackets does not change during the experiment

      K = A  q² / m

For the proton

     K = 3.0 10⁶eV (1.6 10⁻¹⁹ J / 1eV) = 4.8 10⁻¹³ J

With this data we can find the amount we call A

    A = K  m/q²

    A = 4.8 10⁻¹³ 1.67 10⁻²⁷ /(1.6 10⁻¹⁹)²

    A = 3.13 10⁻²

With this value we can write the equation

    K = 3.13 10⁻²  q² / m

Alpha particle

    m = 4 uma = 4 1.66 10⁻²⁷ kg

   K = 3.13 10⁻² (2 1.6 10⁻¹⁹)² / 4.0 1.66 10⁻²⁷

   K = 4.82 10⁻¹³ J ((1 eV / 1.6 10⁻¹⁹ J) = 3 10⁶ eV

   K = 3 MeV

Deuteron

   K = 3.13 10⁻² (1.6 10⁻¹⁹)²/2 1.66 10⁻²⁷

   K = 2.4 10⁻¹³ J (1eV / 1.6 10⁻¹⁹J)

   K = 1.5 10⁶ eV

   K=  1.5 MeV

Displacement is the distance from the final placement of the object from the initial point. So, since the swimmer started and ended in the same spot, it is zero.

Therefore your correct answer is:
A: 100m swimming 0 displacement