The force that curves the trajectory of a charged particle `q` in a magnetic field of intensity `B` is the Lorentz force:

`|F|=|q*(v xx B)| =qvB`

if vector speed `v` is perpendicular to vector induction `B` .

The radius of the curvature of the trajectory comes from the condition that the Lorentz force is equal to the centrifugal force.

`q*v*B =m*v^2/R rArr R = (m/q)*(v/B) = (m*v)/q *B`

Since in the problem, the linear momentum is the same for both electron and proton, apart from the fact that their path will be curved in a opposite directions (because the sign of the charge `q` in the denominator is different in the two cases) , the radius of curvature will be the same.

**Thus the correct answer is C) Both are equally curved**.

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