A proton with a kinetic energy of 4.7×10−16J moves perpendicular to a magnetic field of 0.25T .
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To determine the motion of a proton in a magnetic field, we need to use the principles of magnetic force. The force on a charged particle moving in a magnetic field is given by the equation:
F = q * v * B * sin(theta)
where:
F is the force experienced by the particle,
q is the charge of the particle (in this case, the charge of a proton, which is 1.6 x 10^-19 C),
v is the velocity of the particle,
B is the magnetic field strength, and
theta is the angle between the velocity and the magnetic field.
In this case, the proton moves perpendicular to the magnetic field, so theta is 90 degrees, and sin(theta) is equal to 1.
Since we are given the kinetic energy of the proton and the magnetic field strength, we need to find the velocity to determine the force acting on the proton.
The kinetic energy (KE) of a particle is given by the equation:
KE = (1/2) * m * v^2
where:
m is the mass of the particle (in this case, the mass of a proton, which is approximately 1.67 x 10^-27 kg),
v is the velocity of the particle.
We can rearrange this equation to solve for velocity:
v = sqrt((2 * KE) / m)
Using the given kinetic energy, we can substitute the values into the equation and calculate the velocity of the proton.
v = sqrt((2 * 4.7 x 10^-16 J) / (1.67 x 10^-27 kg))
After calculating the value of v, we can plug it into the equation for magnetic force to calculate the force experienced by the proton:
F = (1.6 x 10^-19 C) * v * (0.25 T) * 1
The result will give us the force experienced by the proton moving perpendicular to the magnetic field.