A proton travels through a region of uniform magnetic field at an angle θ relative to the magnetic field. The magnitude...

$$\begin{gathered} Mass of proton \left(m\right) = 1.67\times {10}^{-27} kg \\ Charge of proton \left(q\right) = 1.6\times {10}^{-19} C\hspace{0.17em} \\ Magnetic field strength \left(B\right) = 1.15 T\hspace{0.17em} \\ Velocity of proton \left(v\right)\hspace{0.17em}= 500000 \frac{m}{s} \\ Acceleration of proton \left(a\right)\hspace{0.17em}= 1.57\times {10}^{13} \frac{m}{s^2} \end{gathered}$$

$$\begin{gathered} U\sin g the formula for magnetic force : \\ F\hspace{0.17em}= q\times v\times B\times Sin\theta \\ U\sin g newton\text{'}s second law of motion : \\ net force = mass\times acceleration \\ mass\times acceleration =q\times v\times B\times Sin\theta \\ a = \frac{q\times v\times B\times Sin\theta }{m} \\ 1.57 \times {10}^{13} = \frac{1.6\times {10}^{-19}\times 500000\times 1.15\times Sin\theta }{1.67\times {10}^{-27}} \\ 1.57\times {10}^{13} = 5.508\times {10}^{13}\times Sin\theta \\ 0.285 = Sin\theta \\ 16.{55}^o = \theta \\ (Angle between velocity and magnetic field) \end{gathered}$$