A proton experiences a force of 4 Newton when it enters perpendicular to the direction of the magnetic field with a speed 100 m/s. What is the force experienced by the proton if it enters parallel to the direction of the magnetic field with a speed of 200 m/s?
F=0 since F=qvBsinα (α=0)
To find the force experienced by the proton when it enters parallel to the direction of the magnetic field, we can use the formula for the magnetic force on a charged particle:
F = qvB sin(θ)
where:
F is the magnetic force
q is the charge of the proton (1.6 x 10^-19 C)
v is the speed of the proton (200 m/s)
B is the magnetic field strength
θ is the angle between the velocity vector and the magnetic field vector
In this case, when the proton enters parallel to the direction of the magnetic field, the angle between the velocity vector and the magnetic field vector is 0°, so sin(θ) = 0.
Therefore, the force experienced by the proton when it enters parallel to the direction of the magnetic field is:
F = qvB sin(θ) = (1.6 x 10^-19 C) x (200 m/s) x (0) = 0
Hence, the force experienced by the proton is 0 Newtons.
To determine the force experienced by the proton when it enters parallel to the direction of the magnetic field, we can use the formula for the magnetic force on a moving charged particle:
F = qvBsinθ,
where:
F is the magnetic force,
q is the charge of the proton,
v is the velocity of the proton,
B is the magnetic field strength, and
θ is the angle between the velocity vector and the magnetic field direction.
In the given scenario, the proton enters parallel to the direction of the magnetic field. When the magnetic field and velocity vectors are parallel, the angle θ between them is 0 degrees. Therefore, sinθ = 0, which simplifies the equation to:
F = qvBsinθ
= qvBsin0
= 0.
Hence, when the proton enters parallel to the direction of the magnetic field, it will experience zero force.