Final answer:
The electromagnetic force on the proton at time t=0.32μs, with a velocity of 5.0Mm/s in the x-direction and a magnetic field of 0.896 mT in the z-direction, is 7.168 × 10^-16 N in the y-direction.
Step-by-step explanation:
The electromagnetic force on a proton in a magnetic field can be determined using the Lorentz force equation, which is F = q(v × B), where F is the force, q is the charge of the proton, v is the velocity of the proton, and B is the magnetic field. The charge of a proton is approximately 1.6 × 10-19 C. In this case, the magnetic field inside the solenoid is given by B = bt k, where b is a constant (2.8 T/ms) and t is the time (0.32μs).
To find the magnetic field at t=0.32μs, we calculate B = 2.8 T/ms × 0.32μs = 0.896 mT. The proton is moving with a velocity v = 5.0Mm/s in the i (or x) direction. The cross product v × B will give us the direction of the electromagnetic force, which will be perpendicular to both the velocity and the magnetic field vectors. Since the magnetic field is in the k (or z) direction and the proton is moving in the i (or x) direction, the force will be in the j (or y) direction. The magnitude of the force is F = qvB = (1.6 × 10-19 C)(5.0 × 106 m/s)(0.896 × 10-3 T) = 7.168 × 10-16 N in the y direction.