A proton with an initial velocity of (1.0x + 2.0y + 3.0z)* 10⁵ m/s enters a magnetic field of 0.50z T.Find the acceleration of the proton.The mass of the proton is 1.67 * 10^-27 kg.Where x,y and z are unit vectors in the x,y and z directions,respectively.

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The force on a charged particle in a magnetic field is given by F = q(v × B). The velocity v = (1.0x + 2.0y + 3.0z) * 10^5 m/s and the magnetic field B = 0.50z T. The cross product v × B = (1.0x + 2.0y) * 10^5 m/s × 0.50z T = (0.5 * 2.0x - 0.5 * 1.0y) * 10^5 T*m/s = (1.0x - 0.5y) * 10^5 T*m/s. The force F = q(v × B) = 1.6 * 10^-19 C * (1.0x - 0.5y) * 10^5 T*m/s = (1.6 * 10^-14x - 0.8 * 10^-14y) N. The acceleration a = F/m = (1.6 * 10^-14x - 0.8 * 10^-14y) N / 1.67 * 10^-27 kg = (9.58x - 4.79y) * 10^12 m/s^2.

A Proton with an Initial Velocity of (1