To determine the magnitude and direction of the magnetic field, we can use the formula for the magnetic force experienced by a charged particle moving through a magnetic field:
F = q * v * B * sin(θ)
Where:
F = Magnetic force
q = Charge of the particle (charge of a proton = 1.602 x 10^-19 C)
v = Velocity of the particle (2.40 x 10^7 m/s)
B = Magnetic field strength (what we want to find)
θ = Angle between the velocity vector and the magnetic field vector
In this case, the proton experiences an acceleration of 2.40 x 10^13 m/s² in the positive x direction. Since the magnetic force is the only force acting on it in the x direction, we can equate the magnetic force to the mass times acceleration:
F = m * a
Where:
F = Magnetic force (which we calculated above)
m = Mass of the proton (approximately 1.67 x 10^-27 kg)
a = Acceleration (2.40 x 10^13 m/s²)
Now, we can equate the magnetic force to the mass times acceleration:
q * v * B * sin(θ) = m * a
Plug in the known values:
(1.602 x 10^-19 C) * (2.40 x 10^7 m/s) * B * sin(90°) = (1.67 x 10^-27 kg) * (2.40 x 10^13 m/s²)
sin(90°) is equal to 1, so we can simplify the equation:
(1.602 x 10^-19 C) * (2.40 x 10^7 m/s) * B = (1.67 x 10^-27 kg) * (2.40 x 10^13 m/s²)
Now, solve for B:
B = [(1.67 x 10^-27 kg) * (2.40 x 10^13 m/s²)] / [(1.602 x 10^-19 C) * (2.40 x 10^7 m/s)]
B ≈ 5.24 x 10^-3 T (Tesla)
So, the magnitude of the magnetic field is approximately 5.24 x 10^-3 Tesla.
Now, let's determine the direction of the magnetic field. The force experienced by the proton is in the positive x direction, which means the magnetic field must be in the positive y direction (perpendicular to both the velocity and acceleration vectors) to produce this force. Therefore, the direction of the magnetic field is in the positive y direction.