Determine the acceleration of a proton that is projected...

The magnetic force acting on a charged particle moving in a magnetic field is given by $F = qvB\sin\theta$.

Since the proton is projected perpendicularly, $\theta = 90^{{}^{{}^{\circ}}}$, which gives $F = qvB$.

The acceleration of the proton is $a = \dfrac{F}{m} = \dfrac{qvB}{m}$.

Substituting the standard values for a proton ($q = 1.6 \times 10^{-19} \text{ C}$, $m = 1.67 \times 10^{-27} \text{ kg}$) and the given values:

$a = \dfrac{1.6 \times 10^{-19} \times 3 \times 10^6 \times 0.6}{1.67 \times 10^{-27}}$

$a = \dfrac{2.88 \times 10^{-13}}{1.67 \times 10^{-27}}$

$a \approx 1.72 \times 10^{14} \text{ m s}^{-2}$