A circular coil of radius 10π cm carries a steady current of 10 A ....

The magnetic field at the center of a circular coil is given by:

$B = \dfrac{\mu_0 I}{2R}$

Substituting the given values ($I = 10\text{ A}$, $R = 10\pi\text{ cm} = 0.1\pi\text{ m}$):

$B = \dfrac{4\pi \times 10^{-7} \times 10}{2 \times 0.1\pi} = \dfrac{40\pi \times 10^{-7}}{0.2\pi} = 2 \times 10^{-5}\text{ T}$

The direction of this magnetic field is perpendicular to the plane of the coil. Since the proton is projected in the plane of the coil, the angle $\theta$ between its velocity vector $\vec{v}$ and the magnetic field $\vec{B}$ is $90^\circ$.

The magnitude of the magnetic force experienced by the proton is:

$F = qvB\sin\theta$

Substituting $q = 1.6 \times 10^{-19}\text{ C}$, $v = 5 \times 10^6\text{ m s}^{-1}$, and $\theta = 90^\circ$:

$F = (1.6 \times 10^{-19}) \times (5 \times 10^6) \times (2 \times 10^{-5}) \times \sin(90^\circ)$

$F = 1.6 \times 10^{-19} \times 100$

$F = 1.6 \times 10^{-17}\text{ N}$

Answer: $1.6 \times 10^{-17}\text{ N}$