A proton is traveling at 2.3E6 m/s in a circular path in 0.75 T magnetic field. What is the magnitude of the force of the proton?
force=q*v*B
F = charge * velocity * field strength
F = 2.76E-13 N
To find the magnitude of the force on the proton, we can use the formula for the magnetic force on a charged particle in a magnetic field:
F = q * v * B
Where:
F is the force on the particle,
q is the charge of the particle,
v is the velocity of the particle, and
B is the magnetic field strength.
In this case, we are given:
v = 2.3E6 m/s (velocity of the proton)
B = 0.75 T (magnetic field strength)
The charge of a proton is +1.6E-19 C (Coulombs).
Substituting the given values into the formula, we have:
F = (1.6E-19 C) * (2.3E6 m/s) * (0.75 T)
Calculating this expression:
F = 2.07E-13 N
The magnitude of the force on the proton is 2.07E-13 Newtons.
To find the magnitude of the force acting on a charged particle moving in a magnetic field, you need to use the formula:
F = q * v * B * sinθ,
where F is the force, q is the charge of the particle, v is the velocity of the particle, B is the magnetic field strength, and θ is the angle between the velocity vector and the magnetic field vector.
In this case, the proton is traveling in a circular path, so its velocity vector is always perpendicular to the magnetic field vector. Therefore, the angle θ between them is 90 degrees, and sinθ equals 1.
The charge of a proton is q = +1.602 x 10^-19 C (coulombs).
The velocity of the proton is given as v = 2.3 x 10^6 m/s.
The magnetic field strength is B = 0.75 T (teslas).
Plugging in these values, we can calculate the magnitude of the force acting on the proton:
F = (1.602 x 10^-19 C) * (2.3 x 10^6 m/s) * (0.75 T) * (1)
= 2.4735 x 10^-13 N.
Therefore, the magnitude of the force acting on the proton is approximately 2.4735 x 10^-13 Newtons.