A 42 mT magnetic field points due west. If a proton of kinetic energy 9x10E-12 J enters this field in an upward direction, find the magnetic force acting on the proton in magnitude and direction.

To find the magnetic force acting on the proton, we will need to use the equation for the magnetic force on a charged particle in a magnetic field.

The equation for the magnetic force is given by:

F = q * v * B * sin(theta)

Where:
F is the force on the charged particle (proton)
q is the charge of the particle (proton charge = 1.602 x 10^-19 C)
v is the velocity of the particle
B is the magnitude of the magnetic field (42 mT = 42 x 10^-3 T)
theta is the angle between the velocity and the magnetic field direction.

In this case, the magnetic field is pointing due west, and the proton is entering the field in an upward direction. Since the magnetic field is perpendicular to the velocity of the proton, the angle theta will be 90 degrees.

Let's calculate the force:

F = (1.602 x 10^-19 C) * v * (42 x 10^-3 T) * sin(90°)

Now, let's determine the velocity of the proton using its kinetic energy.

Kinetic energy (KE) = (1/2) * m * v^2

Given: KE = 9 x 10^-12 J
Mass of proton (m) = 1.6726219 x 10^-27 kg

Rearranging the kinetic energy equation to solve for velocity (v):

v = sqrt((2 * KE) / m)

Substituting the values:

v = sqrt((2 * 9 x 10^-12) / 1.6726219 x 10^-27)

Now, with the value of the velocity, we can calculate the magnetic force using the equation:

F = (1.602 x 10^-19 C) * v * (42 x 10^-3 T) * sin(90°)

Calculating these values will give us the magnitude and direction of the magnetic force acting on the proton.

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