Answer:
B) the particle's momentum.
Step-by-step explanation:
We know that
The centripetal force on the particle when its moving in the radius R and velocity V
![F_c=(m* V^2)/(R)](https://img.qammunity.org/2021/formulas/physics/college/wewmkvwd2oshq8ttc8u6m9zwj3z9stqf7s.png)
The magnetic force on the particle when the its moving with velocity V in the magnetic filed B and having charge q
![F_m=q* V* B](https://img.qammunity.org/2021/formulas/physics/college/xuabapobeixwtrbysskw01d3gqfxnw9ona.png)
At the equilibrium condition
![F_m=F_c](https://img.qammunity.org/2021/formulas/physics/college/ov4gnkv8rg1d1qu632dam379ip6g7d0q0p.png)
![q* V* B=(m* V^2)/(R)](https://img.qammunity.org/2021/formulas/physics/college/gk0r6bd9c9r6n0sgwbkd0f1t9z6dby8stq.png)
![R=(m* V)/(q* B)](https://img.qammunity.org/2021/formulas/physics/college/bt424vd6zz971thd3xsh6cln96spjznc47.png)
Momentum = m V
Therefore we can say that the radius of curvature is directly proportional to the particle momentum.
B) the particle's momentum.