A proton and a deutron (q=+e, m=2.0 u) having same kinetic energies enter a region of uniform magnetic field B⃗, moving perpendicular to B⃗. The ratio of the radius r_d of deutron path to the radius rₚ of the proton path is:

Solution

Physics sir · 4 steps

Write the radius in terms of kinetic energy: with the same charge and same K, r just goes as root m.

1. Radius of the circle. The magnetic force qvB supplies the centripetal force mv²/r. That gives r = mv/qB = p/qB.
2. Bring in kinetic energy. Since K = p²/2m, the momentum is p = √2mK. Put this into the radius formula.

   r = √2mK/qB

3. Compare the two. Both particles have the same q = +e, the same K and the same B. So r ∝ √m, and r_d/rₚ = √((2 u)/(1 u)) = √2.
4. So. r_d : rₚ = √2 : 1, which is option (c).

The idea to keep. In a magnetic field, r = p/qB. Equal kinetic energy means p = √2mK, so r ∝ √m/q. Equal momentum would make r depend on q alone.

Answer from the NTA final answer key. Solution by PrepWiser.